description stringlengths 171 4k | code stringlengths 94 3.98k | normalized_code stringlengths 57 4.99k |
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For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | for _ in range(int(input())):
xd = {}
x = input()
for i in range(len(x)):
xd[x[i]] = xd.setdefault(x[i], 0) + 1
xd = sorted(xd.values())
if len(xd) <= 2:
print("Dynamic")
continue
ans = "Dynamic"
if len(xd) >= 4 and xd[3] != xd[2] + xd[1]:
xd[0], xd[1] = xd[1], xd[0]
for j in range(2, len(xd)):
if xd[j - 2] + xd[j - 1] != xd[j]:
ans = "Not"
break
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR DICT ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR STRING IF FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR STRING EXPR FUNC_CALL VAR VAR |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | for _ in range(int(input())):
x = input()
y = list(set(x))
l = []
for i in y:
l.append(x.count(i))
temp = 0
l.sort()
if len(l) < 3:
print("Dynamic")
else:
for i in range(2, len(l)):
if i == 3:
if not l[3] == l[2] + l[1] and not l[3] == l[2] + l[0]:
temp = 1
break
elif l[i - 1] + l[i - 2] != l[i]:
temp = 1
break
if temp:
print("Not")
else:
print("Dynamic") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR NUMBER IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | t = int(input())
for _ in range(t):
s = input()
arr_dic = {}
for i in range(len(s)):
if s[i] not in arr_dic:
arr_dic[s[i]] = 1
else:
arr_dic[s[i]] += 1
counter = 0
arr = sorted(list(arr_dic.values()), reverse=True)
if len(arr) < 3:
print("Dynamic")
else:
for i in range(0, len(arr) - 2):
if i != len(arr) - 4:
if arr[i] != arr[i + 1] + arr[i + 2]:
counter = 1
elif arr[i] != arr[i + 1] + arr[i + 2]:
c = arr[i + 3]
arr[i + 3] = arr[i + 2]
arr[i + 2] = c
if arr[i] != arr[i + 1] + arr[i + 2]:
counter = 1
if counter == 0:
print("Dynamic")
else:
print("Not") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | for _ in range(int(input())):
s = input()
s2 = list(s)
s1 = list(set(s))
a = []
for i in s1:
a.append(s2.count(i))
a = sorted(a)
if len(a) < 3:
print("Dynamic")
else:
for i in range(len(a) - 1, 1, -1):
if a[i] == a[i - 1] + a[i - 2]:
print("Dynamic")
break
else:
print("Not")
break | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | for _ in range(int(input())):
s = input()
d = {}
for i in s:
if i in d:
d[i] += 1
else:
d[i] = 1
l = []
for i in d.values():
l.append(i)
l.sort()
if len(l) < 3:
print("Dynamic")
elif l[-1] == l[-2] + l[-3]:
print("Dynamic")
else:
print("Not") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | t = int(input())
for _ in range(t):
st = input()
s = set(st)
a = []
f1 = f2 = 0
for i in s:
a.append(st.count(i))
a.sort()
if len(a) >= 3:
for i in range(2, len(a)):
if a[i] != a[i - 1] + a[i - 2]:
f1 = 1
break
x = a[0]
a[0] = a[1]
a[1] = x
for i in range(2, len(a)):
if a[i] != a[i - 1] + a[i - 2]:
f2 = 1
break
if f1 == 1 and f2 == 1:
print("Not")
else:
print("Dynamic")
else:
print("Dynamic") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | t = int(input().strip())
for k in range(t):
s = input().strip()
d = {}
maxx = -10000
cu = c = 0
for i in s:
if i not in d:
cu += 1
d[i] = 1
else:
d[i] += 1
if cu < 3:
print("Dynamic")
else:
lii = list(d.values())
lii.sort()
for i in range(2, len(lii)):
if lii[i] == lii[i - 1] + lii[i - 2]:
c = 1
else:
c = 0
lii[0], lii[1] = lii[1], lii[0]
for i in range(2, len(lii)):
if lii[i] == lii[i - 1] + lii[i - 2]:
c = 1
else:
c = 0
if c == 0:
print("Not")
else:
print("Dynamic") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR DICT ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | t = int(input())
while t:
t -= 1
s = input()
l = []
l1 = []
lik = 2
for i in s:
if i not in l:
l.append(i)
for i in l:
l1.append(s.count(i))
if len(l) < 3:
lik = 0
if lik == 2:
l1.sort()
if l1[-2] + l1[-3] != l1[-1]:
lik = 1
if lik == 0 or lik == 2:
print("Dynamic")
if lik == 1:
print("Not") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | t = int(input())
for i in range(t):
set_str = set()
set_dir = dict()
str_in = input()
for j in str_in:
set_str.add(j)
set_dir = {k: str_in.count(k) for k in set_str}
a = sorted(set_dir.values())
flag = 1
if len(a) >= 4 and a[3] != a[2] + a[1]:
a[1], a[0] = a[0], a[1]
for i in range(2, len(a)):
if a[i] != a[i - 1] + a[i - 2]:
flag = 2
break
if flag == 1 or len(set_str) < 3:
print("Dynamic")
else:
print("Not") | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | def check_fib(lis):
is_fib = True
for i in range(2, len(lis)):
if lis[i] != lis[i - 1] + lis[i - 2]:
is_fib = False
break
if is_fib:
return True
lis[0], lis[1] = lis[1], lis[0]
for i in range(2, len(lis)):
if lis[i] != lis[i - 1] + lis[i - 2]:
return False
return True
t = int(input())
for _ in range(t):
s = input()
freq_list = [s.count(x) for x in set(s)]
if len(freq_list) < 3:
print("Dynamic")
continue
freq_list.sort()
if check_fib(freq_list):
print("Dynamic")
else:
print("Not") | FUNC_DEF ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER IF VAR RETURN NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
For a string $S$ let the unique set of characters that occur in it one or more times be $C$. Consider a permutation of the elements of $C$ as $(c_1, c_2, c_3 ... )$. Let $f(c)$ be the number of times $c$ occurs in $S$.
If any such permutation of the elements of $C$ satisfies $f(c_i) = f(c_{i-1}) + f(c_{i-2})$ for all $i \ge 3$, the string is said to be a dynamic string.
Mr Bancroft is given the task to check if the string is dynamic, but he is busy playing with sandpaper. Would you help him in such a state?
Note that if the number of distinct characters in the string is less than 3, i.e. if $|C| < 3$, then the string is always dynamic.
-----Input:-----
- First line will contain $T$, number of testcases. Then the testcases follow.
- Each testcase contains of a single line of input, a string $S$.
-----Output:-----
For each testcase, output in a single line "Dynamic" if the given string is dynamic, otherwise print "Not". (Note that the judge is case sensitive)
-----Constraints-----
- $1 \leq T \leq 10$
- $1 \leq |S| \leq 10^5$
- $S$ contains only lower case alphabets: $a$, $b$, …, $z$
-----Sample Input:-----
3
aaaabccc
aabbcc
ppppmmnnoooopp
-----Sample Output:-----
Dynamic
Not
Dynamic
-----Explanation:-----
- Testase 1: For the given string, $C = \{a, b, c\}$ and $f(a)=4, f(b)=1, f(c)=3$. $f(a) = f(c) + f(b)$ so the permutation $(b, c, a)$ satisfies the requirement.
- Testcase 2: Here too $C = \{a, b, c\}$ but no permutation satisfies the requirement of a dynamic string.
- Testcase 3: Here $C = \{m, n, o, p\}$ and $(m, n, o, p)$ is a permutation that makes it a dynamic string. | for tc in range(int(input())):
s = input()
c = set(s)
l = []
for i in c:
l.append(s.count(i))
if len(l) < 3:
print("Dynamic")
else:
l.sort()
if l[-1] == l[-2] + l[-3]:
print("Dynamic")
else:
print("Not") | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def findReplaceString(
self, s: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
l = []
for i, tgt, rpl in zip(indexes, sources, targets):
if s[i : i + len(tgt)] == tgt:
l.append((i, tgt, rpl))
l.sort()
j = 0
s = list(s)
for i, tgt, rpl in l:
s[i + j : i + j + len(tgt)] = rpl
j += len(rpl) - len(tgt)
return "".join(s) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST FOR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL STRING VAR VAR |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def findReplaceString(
self, S: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
idx_og = 0
idx_res = 0
res = S
for index, source, target in sorted(zip(indexes, sources, targets)):
if S[index : index + len(source)] == source:
index2 = index - idx_og + idx_res
res = res[:index2] + target + res[index2 + len(source) :]
idx_og = index + len(source)
idx_res = index2 + len(target)
return res | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR RETURN VAR VAR |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def check_source(self, start_idx, s, source):
i = 0
while i < len(source):
if start_idx == len(s):
return False
if source[i] != s[start_idx]:
return False
start_idx += 1
i += 1
return True
def findReplaceString(
self, s: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
res = []
i = 0
while i < len(s):
if i in indexes:
idx = indexes.index(i)
source = sources[idx]
target = targets[idx]
if self.check_source(i, s, source):
i += len(source)
res.append(target)
continue
res.append(s[i])
i += 1
return "".join(res) | CLASS_DEF FUNC_DEF ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR RETURN NUMBER IF VAR VAR VAR VAR RETURN NUMBER VAR NUMBER VAR NUMBER RETURN NUMBER FUNC_DEF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN FUNC_CALL STRING VAR VAR |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def findReplaceString(
self, S: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
index_position = dict()
r_list = []
for pos, index in enumerate(indexes):
source_len = len(sources[pos])
if S[index : index + source_len] == sources[pos]:
index_position[index] = pos
next_index = 0
for i in range(len(S)):
if i < next_index:
continue
if i in index_position:
pos = index_position[i]
r_list.append(targets[pos])
next_index += len(sources[pos])
else:
r_list.append(S[i])
next_index += 1
return "".join(r_list) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR IF VAR VAR ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN FUNC_CALL STRING VAR VAR |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def findReplaceString(
self, S: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
sources_dict = {}
targets_dict = {}
for i, pos_index in enumerate(indexes):
sources_dict[pos_index] = sources[i]
targets_dict[pos_index] = targets[i]
indexes.sort(reverse=True)
for pos_index in indexes:
source = sources_dict[pos_index]
target = targets_dict[pos_index]
tmp_str = S[pos_index : pos_index + len(source)]
if tmp_str == source:
S = S[:pos_index] + target + S[pos_index + len(source) :]
return S | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER FOR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR RETURN VAR VAR |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def findReplaceString(
self, S: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
result, offset = S, 0
for i, s, t in sorted(zip(indexes, sources, targets)):
if S[i : i + len(s)] == s:
result = result[: offset + i] + t + result[offset + i + len(s) :]
offset += len(t) - len(s)
return result | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR IF VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR VAR VAR VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN VAR VAR |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def findReplaceString(
self, S: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
S = list(S)
for i, s, t in sorted(zip(indexes, sources, targets), reverse=True):
appl = all(
[(i + ind < len(S) and S[i + ind] == s[ind]) for ind in range(len(s))]
)
if appl:
S[i : i + len(s)] = list(t)
return "".join(S) | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR RETURN FUNC_CALL STRING VAR VAR |
To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).
Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y. The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y. If not, we do nothing.
For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".
Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.
All these operations occur simultaneously. It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.
Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".
Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee".
"ec" doesn't starts at index 2 in the original S, so we do nothing.
Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters. | class Solution:
def findReplaceString(
self, S: str, indexes: List[int], sources: List[str], targets: List[str]
) -> str:
for i, s, t in sorted(zip(indexes, sources, targets), reverse=True):
if S[i : i + len(s)] == s:
S = S[:i] + t + S[i + len(s) :]
return S | CLASS_DEF FUNC_DEF VAR VAR VAR VAR VAR VAR VAR FOR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR RETURN VAR VAR |
Filled with optimism, Hyunuk will host a conference about how great this new year will be!
The conference will have $n$ lectures. Hyunuk has two candidate venues $a$ and $b$. For each of the $n$ lectures, the speaker specified two time intervals $[sa_i, ea_i]$ ($sa_i \le ea_i$) and $[sb_i, eb_i]$ ($sb_i \le eb_i$). If the conference is situated in venue $a$, the lecture will be held from $sa_i$ to $ea_i$, and if the conference is situated in venue $b$, the lecture will be held from $sb_i$ to $eb_i$. Hyunuk will choose one of these venues and all lectures will be held at that venue.
Two lectures are said to overlap if they share any point in time in common. Formally, a lecture held in interval $[x, y]$ overlaps with a lecture held in interval $[u, v]$ if and only if $\max(x, u) \le \min(y, v)$.
We say that a participant can attend a subset $s$ of the lectures if the lectures in $s$ do not pairwise overlap (i.e. no two lectures overlap). Note that the possibility of attending may depend on whether Hyunuk selected venue $a$ or venue $b$ to hold the conference.
A subset of lectures $s$ is said to be venue-sensitive if, for one of the venues, the participant can attend $s$, but for the other venue, the participant cannot attend $s$.
A venue-sensitive set is problematic for a participant who is interested in attending the lectures in $s$ because the participant cannot be sure whether the lecture times will overlap. Hyunuk will be happy if and only if there are no venue-sensitive sets. Determine whether Hyunuk will be happy.
-----Input-----
The first line contains an integer $n$ ($1 \le n \le 100000$), the number of lectures held in the conference.
Each of the next $n$ lines contains four integers $sa_i$, $ea_i$, $sb_i$, $eb_i$ ($1 \le sa_i, ea_i, sb_i, eb_i \le 10^9$, $sa_i \le ea_i, sb_i \le eb_i$).
-----Output-----
Print "YES" if Hyunuk will be happy. Print "NO" otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2
1 2 3 6
3 4 7 8
Output
YES
Input
3
1 3 2 4
4 5 6 7
3 4 5 5
Output
NO
Input
6
1 5 2 9
2 4 5 8
3 6 7 11
7 10 12 16
8 11 13 17
9 12 14 18
Output
YES
-----Note-----
In second example, lecture set $\{1, 3\}$ is venue-sensitive. Because participant can't attend this lectures in venue $a$, but can attend in venue $b$.
In first and third example, venue-sensitive set does not exist. | import sys
input = lambda: sys.stdin.readline().rstrip()
N = 100101
n = int(input())
rand = [((i + 12345) ** 3 % 998244353) for i in range(N * 2 + 20)]
al, ar, bl, br = [], [], [], []
for _ in range(n):
a, b, c, d = map(int, input().split())
al.append(a)
ar.append(b)
bl.append(c)
br.append(d)
def calk(l, r):
ma = {s: idx for idx, s in enumerate(sorted(list(set(l + r))))}
hash = [0] * N * 2
for idx, v in enumerate(l):
hash[ma[v]] += rand[idx]
for i in range(1, N * 2):
hash[i] += hash[i - 1]
return sum([(hash[ma[v]] * rand[idx]) for idx, v in enumerate(r)])
print("YES" if calk(al, ar) == calk(bl, br) else "NO") | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR VAR LIST LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP LIST NUMBER VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR STRING STRING |
Filled with optimism, Hyunuk will host a conference about how great this new year will be!
The conference will have $n$ lectures. Hyunuk has two candidate venues $a$ and $b$. For each of the $n$ lectures, the speaker specified two time intervals $[sa_i, ea_i]$ ($sa_i \le ea_i$) and $[sb_i, eb_i]$ ($sb_i \le eb_i$). If the conference is situated in venue $a$, the lecture will be held from $sa_i$ to $ea_i$, and if the conference is situated in venue $b$, the lecture will be held from $sb_i$ to $eb_i$. Hyunuk will choose one of these venues and all lectures will be held at that venue.
Two lectures are said to overlap if they share any point in time in common. Formally, a lecture held in interval $[x, y]$ overlaps with a lecture held in interval $[u, v]$ if and only if $\max(x, u) \le \min(y, v)$.
We say that a participant can attend a subset $s$ of the lectures if the lectures in $s$ do not pairwise overlap (i.e. no two lectures overlap). Note that the possibility of attending may depend on whether Hyunuk selected venue $a$ or venue $b$ to hold the conference.
A subset of lectures $s$ is said to be venue-sensitive if, for one of the venues, the participant can attend $s$, but for the other venue, the participant cannot attend $s$.
A venue-sensitive set is problematic for a participant who is interested in attending the lectures in $s$ because the participant cannot be sure whether the lecture times will overlap. Hyunuk will be happy if and only if there are no venue-sensitive sets. Determine whether Hyunuk will be happy.
-----Input-----
The first line contains an integer $n$ ($1 \le n \le 100000$), the number of lectures held in the conference.
Each of the next $n$ lines contains four integers $sa_i$, $ea_i$, $sb_i$, $eb_i$ ($1 \le sa_i, ea_i, sb_i, eb_i \le 10^9$, $sa_i \le ea_i, sb_i \le eb_i$).
-----Output-----
Print "YES" if Hyunuk will be happy. Print "NO" otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2
1 2 3 6
3 4 7 8
Output
YES
Input
3
1 3 2 4
4 5 6 7
3 4 5 5
Output
NO
Input
6
1 5 2 9
2 4 5 8
3 6 7 11
7 10 12 16
8 11 13 17
9 12 14 18
Output
YES
-----Note-----
In second example, lecture set $\{1, 3\}$ is venue-sensitive. Because participant can't attend this lectures in venue $a$, but can attend in venue $b$.
In first and third example, venue-sensitive set does not exist. | import sys
input = lambda: sys.stdin.readline().rstrip()
ke = 133333333
pp = 1000000000001003
def rand():
global ke
ke = ke**2 % pp
return (ke >> 10) % (1 << 20) + (1 << 20)
N = int(input())
W = [rand() for _ in range(N)]
AL, AR, BL, BR = [], [], [], []
for i in range(N):
a, b, c, d = map(int, input().split())
AL.append(a)
AR.append(b)
BL.append(c)
BR.append(d)
def chk(L, R):
S = sorted(list(set(L + R)))
D = {s: i for i, s in enumerate(S)}
L = [D[a] for a in L]
R = [D[a] for a in R]
X = [0] * 200200
for i, l in enumerate(L):
X[l] += W[i]
for i in range(1, 200200):
X[i] += X[i - 1]
s = 0
for i, r in enumerate(R):
s += X[r] * W[i]
return s
print("YES" if chk(AL, AR) == chk(BL, BR) else "NO") | IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR RETURN BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR LIST LIST LIST LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR VAR VAR VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR STRING STRING |
Filled with optimism, Hyunuk will host a conference about how great this new year will be!
The conference will have $n$ lectures. Hyunuk has two candidate venues $a$ and $b$. For each of the $n$ lectures, the speaker specified two time intervals $[sa_i, ea_i]$ ($sa_i \le ea_i$) and $[sb_i, eb_i]$ ($sb_i \le eb_i$). If the conference is situated in venue $a$, the lecture will be held from $sa_i$ to $ea_i$, and if the conference is situated in venue $b$, the lecture will be held from $sb_i$ to $eb_i$. Hyunuk will choose one of these venues and all lectures will be held at that venue.
Two lectures are said to overlap if they share any point in time in common. Formally, a lecture held in interval $[x, y]$ overlaps with a lecture held in interval $[u, v]$ if and only if $\max(x, u) \le \min(y, v)$.
We say that a participant can attend a subset $s$ of the lectures if the lectures in $s$ do not pairwise overlap (i.e. no two lectures overlap). Note that the possibility of attending may depend on whether Hyunuk selected venue $a$ or venue $b$ to hold the conference.
A subset of lectures $s$ is said to be venue-sensitive if, for one of the venues, the participant can attend $s$, but for the other venue, the participant cannot attend $s$.
A venue-sensitive set is problematic for a participant who is interested in attending the lectures in $s$ because the participant cannot be sure whether the lecture times will overlap. Hyunuk will be happy if and only if there are no venue-sensitive sets. Determine whether Hyunuk will be happy.
-----Input-----
The first line contains an integer $n$ ($1 \le n \le 100000$), the number of lectures held in the conference.
Each of the next $n$ lines contains four integers $sa_i$, $ea_i$, $sb_i$, $eb_i$ ($1 \le sa_i, ea_i, sb_i, eb_i \le 10^9$, $sa_i \le ea_i, sb_i \le eb_i$).
-----Output-----
Print "YES" if Hyunuk will be happy. Print "NO" otherwise.
You can print each letter in any case (upper or lower).
-----Examples-----
Input
2
1 2 3 6
3 4 7 8
Output
YES
Input
3
1 3 2 4
4 5 6 7
3 4 5 5
Output
NO
Input
6
1 5 2 9
2 4 5 8
3 6 7 11
7 10 12 16
8 11 13 17
9 12 14 18
Output
YES
-----Note-----
In second example, lecture set $\{1, 3\}$ is venue-sensitive. Because participant can't attend this lectures in venue $a$, but can attend in venue $b$.
In first and third example, venue-sensitive set does not exist. | import sys
import time
input = lambda: sys.stdin.readline().rstrip()
a = time.time()
ke = int(a * 2**20) % 2**20 + 2**20
pp = 100001000000000000021
def rand():
global ke
ke = ke**2 % pp
return (ke >> 30) % (1 << 15) + (1 << 15)
N = int(input())
W = [rand() for _ in range(N)]
A = []
B = []
for i in range(N):
a, b, c, d = map(int, input().split())
A.append((a + 1, b + 2, i))
B.append((c + 1, d + 2, i))
def chk(L):
NN = 18
BIT = [0] * (2**NN + 1)
BITC = [0] * (2**NN + 1)
SS = [0]
CC = [0]
def addrange(r0, x=1):
r = r0
SS[0] += x
CC[0] += 1
while r <= 2**NN:
BIT[r] -= x
BITC[r] -= 1
r += r & -r
def getvalue(r):
a = 0
c = 0
while r != 0:
a += BIT[r]
c += BITC[r]
r -= r & -r
return SS[0] + a, CC[0] + c
S = []
for a, b, i in L:
S.append(a)
S.append(b)
S = sorted(list(set(S)))
D = {s: i for i, s in enumerate(S)}
L = [(D[a], D[b], i) for a, b, i in L]
s = 0
L = sorted(L)
m = -1
for a, b, i in L:
v, c = getvalue(a)
s += v + c * W[i]
addrange(b, W[i])
return s
print("YES" if chk(A) == chk(B) else "NO") | IMPORT IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR BIN_OP VAR BIN_OP NUMBER NUMBER BIN_OP NUMBER NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR BIN_OP BIN_OP VAR NUMBER VAR RETURN BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP NUMBER NUMBER BIN_OP NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP BIN_OP NUMBER VAR NUMBER ASSIGN VAR LIST NUMBER ASSIGN VAR LIST NUMBER FUNC_DEF NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR NUMBER NUMBER WHILE VAR BIN_OP NUMBER VAR VAR VAR VAR VAR VAR NUMBER VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR RETURN BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR LIST FOR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR STRING STRING |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | from sys import stdin
input = stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().strip().split()))
a.sort(reverse=True)
mn = a.index(min(a))
print(n * min(a) + mn) | ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def orican(N, A):
A.sort()
i = A[0]
count = N * i
for i in A:
if i > A[0]:
count += 1
return count
t = int(input())
for _ in range(t):
N = int(input())
A = list(map(int, input().split()))
result = orican(N, A)
print(result) | FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR FOR VAR VAR IF VAR VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
arr.sort(reverse=True)
ind = 0
mini = arr[0]
for i in range(n):
if arr[i] < mini:
mini = arr[i]
ind = i
t = n * mini + ind
print(t) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def _func(n, a):
if n == 1:
return "0"
if 0 not in a:
a.sort()
_min = min(a)
_count = a.count(_min)
return n * _min + (n - _count)
else:
return n - a.count(0)
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
print(_func(n, a))
main() | FUNC_DEF IF VAR NUMBER RETURN STRING IF NUMBER VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR RETURN BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR RETURN BIN_OP VAR FUNC_CALL VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for i in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
k = min(a)
c = a.count(k)
s = (k + 1) * n - c
print(s) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for test in range(t):
n = int(input())
l = list(map(int, input().split()))
l.sort()
l = l[::-1]
a = l[-1]
b = l.index(a)
print(n * a + b) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def solve():
n = int(input())
li = list(map(int, input().split()))
li.sort(reverse=True)
dic = {}
dic1 = {}
key = 0
for i in li:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
if i in dic1:
dic1[i] += 1
else:
dic1[i] = 1
for i, j in dic.items():
key = dic1.pop(i)
dic1[i] = j
key = li[-1]
print(n * key + n - dic[key])
t = int(input())
for i in range(t):
solve() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for test in range(t):
n = int(input())
arr = [int(x) for x in input().split()]
arr.sort()
arr = arr[::-1]
time = arr[-1] * n
for i in arr:
if i > arr[-1]:
time += 1
print(time) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR FOR VAR VAR IF VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split()]
m = min(a)
c = a.count(m)
ans = m * n + (n - c)
print(ans) | ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
sorted(arr, reverse=True)
m = min(arr)
t = 0
for i in arr:
if i > m:
t += 1
print(m * n + t) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
x = int(input())
zoop = list(map(int, input().split()))
zoop.sort(reverse=True)
ans = 0
if zoop[-1] != 0:
for i in range(zoop[-1]):
ans += len(zoop)
i = 0
mn = min(zoop)
while zoop[i] != mn:
ans += 1
i += 1
else:
i = 0
while zoop[i] != 0:
ans += 1
i += 1
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR WHILE VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
a = []
for i in range(2 * t):
a += [list(map(int, input().split()))]
def f(n):
n.sort()
l = [x for x in n if x == n[0]]
return len(n) - len(l) + len(n) * n[0]
for i in range(t):
print(f(a[2 * i + 1])) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR VAR LIST FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR VAR VAR VAR VAR VAR NUMBER RETURN BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
b = input().split()
b.sort(reverse=True)
m = min(b)
i = b.index(m)
print(int(m) * n + i) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
ans = a[-1]
print(ans * n + a.index(ans)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for i in range(int(input())):
N = int(input())
A = list(map(int, input().split()))
A.sort()
time = 0
x = 0
c = 0
for j in range(len(A) - 1):
if A[j] != A[j + 1]:
x = j
c = 0
break
else:
c = 1
time = N * A[x]
if c == 0:
time += N - x - 1
print(time) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR IF VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def solve():
n = int(input())
li = list(map(int, input().split()))
li.sort(reverse=True)
dic = {}
for i in li:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
key = list(dic)[-1]
print(n * key + n - dic[key])
t = int(input())
for i in range(t):
solve() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for i in range(t):
n = int(input())
array = input().split()
mini = int(array[0])
count = 0
for k in range(n):
val = int(array[k])
if val == mini:
count += 1
elif val < mini:
mini = val
count = 1
print(n - count + n * mini) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
lst = list(map(int, input().split()))
cnt = 0
for i in lst:
if i == 0:
cnt += 1
ans = 0
if cnt == 0:
mn = min(lst)
cmin = 0
for i in lst:
if i == mn:
cmin += 1
ans = mn * n
ans = ans + (n - cmin)
else:
ans = n - cnt
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for i in range(t):
n = int(input())
l = list(map(int, list(input().split())))
l = sorted(l)[::-1]
mi = l[-1]
for i in range(n):
l[i] = l[i] - mi
count = 0
for i in l:
if i > 0:
count = count + 1
print(n * mi + count) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for _ in range(t):
n = int(input())
arr = list(map(int, input().split()))
mini = min(arr)
arr.sort()
f = 0
for i in range(n):
if arr[i] > mini:
print(n * arr[0] + n - i)
f = 1
break
if not f:
print(arr[0] * n) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def solve():
n = int(input())
a = list(map(int, input().split()))
minValue = 1000000
for j in a:
minValue = min(minValue, j)
ans = minValue * n
for j in a:
if j > minValue:
ans += 1
print(ans)
t = int(input())
for i in range(t):
solve() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for i in range(t):
n = int(input())
arr = [int(i) for i in input().split(" ")]
arr.sort(reverse=True)
min_value = min(arr)
for i in range(n):
arr[i] -= min_value
j = 0
while arr[j] > 0:
j += 1
print(j + min_value * n) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
while t != 0:
n = int(input())
s = input().split(" ")
a = []
mini = 10000007
for i in range(n):
ele = int(s[i])
a.append(ele)
mini = min(mini, a[i])
res = n * mini
for i in range(n):
if a[i] > mini:
res = res + 1
print(res)
t = t - 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
arr = [int(x) for x in input().split()]
mn = min(arr)
ans = mn * n
for i in range(n):
arr[i] -= mn
print(ans + n - arr.count(0)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR FUNC_CALL VAR NUMBER |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for t in range(int(input())):
n = int(input())
lis = list(map(int, input().split()))
lis.sort(reverse=True)
if n == 1:
print(0)
elif min(lis) == max(lis):
print(lis[0] * n)
else:
mini = min(lis)
for i in range(n):
lis[i] -= mini
res = mini * n
index = lis.index(0)
res += index
print(res) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
a = sorted(list(map(int, input().split())))
a = a[::-1]
minn = a[-1]
minnind = a.index(minn)
total = n * minn
total += minnind
print(total) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
cnt = arr[0] * n
k = arr[0]
for i in range(n):
arr[i] = arr[i] - k
j = n - 1
while arr[j] != 0:
cnt += 1
j -= 1
print(cnt) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for _ in range(t):
n = int(input())
A = list(map(int, input().split()))
A.sort()
if A[0] == 0:
i = 0
while A[i] == 0:
i += 1
if i == n:
break
print(n - i)
else:
count = 1
for i in range(1, n):
if A[0] == A[i]:
count += 1
else:
break
ans = (A[0] + 1) * n - count
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | test = int(input())
for i in range(test):
n = int(input())
arr = list(map(int, input().split()))
min_elem = min(arr)
passes = min_elem * n
for j in range(len(arr)):
arr[j] -= min_elem
count = 0
for k in arr:
if k > 0:
count += 1
print(count + passes) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
while t != 0:
t = t - 1
n = int(input())
a = list(map(int, input().split()))
min1 = min(a)
c = 0
for i in a:
if i == min1:
c = c + 1
print(n - c + min1 * n) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for q in range(t):
n = int(input())
a = list(map(int, input().split(" ")))
a.sort()
a.reverse()
if a[-1] == 0:
index = a.index(0)
print(index)
else:
index = a.index(a[-1])
time = index + a[-1] * n
print(time) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def x(l):
l1 = []
c = 0
for j in l:
if j > 0:
c = c + 1
return c
t = int(input())
for i in range(t):
n = int(input())
l = list(map(int, input().split()))
time = 0
c = 0
for j in l:
if j == 0:
c = 1
break
if c == l:
time = x(l)
else:
y = min(l)
time = time + min(l) * len(l)
for k in range(n):
l[k] = l[k] - y
time = time + x(l)
print(time) | FUNC_DEF ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
print(min(arr) * n + n - arr.count(min(arr))) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
m = min(a)
cnm = a.count(m)
c0 = a.count(0)
if c0 == 0:
a = m * n
print(a + (n - cnm))
else:
print(n - c0) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
ans = 0
p = 0
m = min(l)
ans = m * n
for i in range(n):
if l[i] > m:
p = p + 1
ans = ans + p
print(ans) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for i in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
k = min(a)
l = []
c = 0
for j in range(n):
if a[j] == k:
l.append(j)
c += 1
s = (k + 1) * n - c
print(s) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for i in range(0, int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
m = a[-1]
c = a.count(m)
rem = n - c
ans = m * n + rem
print(ans) | FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | T = int(input())
for _ in range(T):
N = int(input())
arrA = input().split()
arrA = [int(i) for i in arrA]
a = min(arrA)
b = arrA.count(a)
print(N * a + (N - b)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for j in range(t):
N = int(input())
a = list(map(int, input("").strip().split()))[:N]
a.sort()
min1 = min(a)
T = min(a) * N
a = a[::-1]
count = a.count(min1)
T = T + (N - count)
print(T) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for _ in range(t):
n = int(input())
ans = 0
temp = input().split()
A = [int(ele) for ele in temp]
Min = min(A)
MinCount = 0
for ele in A:
if ele == Min:
MinCount += 1
ans += Min * n
ans += n - MinCount
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def main():
test_cases = int(input())
for t in range(test_cases):
n = int(input())
arr = [int(i) for i in input().split()]
arr = sorted(arr, reverse=True)
ind, mini = 0, arr[0]
for i in range(n):
if arr[i] < mini:
mini = arr[i]
ind = i
ind = ind + 1
out = mini * n + ind - 1
print(out)
main() | FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for tc in range(0, t):
n = int(input())
arr = list(map(int, input().split()))
min = arr[0]
ct = 1
for i in range(1, n):
if arr[i] < min:
min = arr[i]
ct = 1
elif arr[i] == min:
ct += 1
val = min * n + n - ct
print(val) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for i in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a = sorted(a)
a.reverse()
e = 0
e += a[-1] * n
z = a.count(a[-1])
e += n - z
print(e) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def Time(arr, n):
arr.sort(reverse=True)
time = 0
lim = n - 1
for i in range(n - 2, -1, -1):
if arr[i] == arr[i + 1]:
lim = i
continue
else:
break
time = n * arr[lim]
if time < 0:
return lim
time += lim
return time
t = int(input())
for i in range(t):
n = int(input())
arr = list(map(int, input().rstrip().rsplit()))
print(Time(arr, n)) | FUNC_DEF EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR IF VAR NUMBER RETURN VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def ans(arr, n):
x = min(arr)
z = arr.count(x)
print(x * n + n - z)
return
x = int(input())
for i in range(x):
n = int(input())
arr = [int(i) for i in input().split()]
ans(arr, n) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR RETURN ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input().strip())
for _ in range(t):
n = int(input().strip())
a = sorted(list(map(int, input().strip().split())), key=lambda x: -x)
r = 0
if a[-1] > 0:
r = a[-1]
l = n - 2
while l >= 0 and (a[l] <= 0 or a[l] == a[l + 1]):
l -= 1
l += 1
print(r * n + l) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
x = min(a)
a = [(i - x) for i in a]
c = 0
for i in range(n):
if a[i] != 0:
c += 1
print(x * n + c) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def check(n, a):
mn = min(a)
if mn > 0:
total = mn * n
for i in range(n):
if a[i] - mn > 0:
total += 1
else:
total = 0
for i in range(n):
if a[i] > 0:
total += 1
return total
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
print(check(n, a)) | FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for i in range(t):
n = int(input())
a = list(map(int, input().split()))
v = [-1] * n
s = 0
m = -1
for j in range(n):
if m < 0 or m > a[j]:
m = a[j]
s = j
j = s
ans = n
for z in a:
if z == m:
ans -= 1
print(ans + m * n) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR FOR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | try:
t = int(input())
while t > 0:
n = int(input())
nn = 0
nnn = 0
nnnn = 0
l = list(map(int, input().strip().split()))
l.sort()
lt = 0
lt2 = 1
count_for_zero = 0
count_for_one = 1
if l[0] == 0:
while True and lt < n:
if l[lt] == 0:
lt += 1
count_for_zero += 1
continue
break
print(n - count_for_zero)
continue
elif l[0] >= 1:
while True and lt2 < n:
if l[0] == l[lt2]:
lt2 += 1
count_for_one += 1
continue
break
ans = n * l[0] + n - count_for_one
print(ans)
continue
t = t - 1
except EOFError as e:
pass | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER NUMBER WHILE NUMBER VAR VAR IF VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR IF VAR NUMBER NUMBER WHILE NUMBER VAR VAR IF VAR NUMBER VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for t in range(int(input())):
n = int(input())
P = sorted([int(x) for x in input().split(" ")], reverse=True)
print(n * P[-1] + P.index(P[-1])) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for i in range(t):
n = int(input())
ar = list(map(int, input().split()))
mini = min(ar)
more = 0
for num in ar:
if num > mini:
more += 1
ans = n * mini + more
print(ans) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | def SieveOfEratosthenes(n):
prime = [(True) for i in range(n + 1)]
p = 2
while p * p <= n:
if prime[p] == True:
for i in range(p**2, n + 1, p):
prime[i] = False
p += 1
prime[0] = False
prime[1] = False
for p in range(n + 1):
if prime[p]:
print(p)
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
time = 0
a.sort()
count = 0
if a[0] == 0:
print(n - a.count(0))
else:
time = a[0] * n
for i in range(n):
if a[i] > a[0]:
time += 1
print(time) | FUNC_DEF ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER WHILE BIN_OP VAR VAR VAR IF VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER VAR ASSIGN VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | T = int(input())
for i in range(T):
N = int(input())
AE = [int(o) for o in input().split(" ")]
if 0 in AE:
print(N - AE.count(0))
else:
s = sorted(AE, reverse=True)
l = s[-1]
print(N * l + N - s.count(l)) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR FUNC_CALL VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | for i in range(int(input())):
n = int(input())
lis = list(map(int, input().split(" ")))
count0 = 0
mini = min(lis)
for i in range(len(lis)):
if lis[i] == 0:
count0 += 1
if count0 > 0:
print(n - count0)
else:
mincount = 0
for i in lis:
if i == mini:
mincount += 1
print(mini * n + (n - mincount)) | FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
while t > 0:
n = int(input())
arr = list(map(int, input().strip().split()))
min_ = min(arr)
for i in range(n):
arr[i] -= min_
count_zero = 0
for i in range(n):
if arr[i] == 0:
count_zero += 1
ans = (min_ + 1) * n - count_zero
print(ans)
t -= 1 | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR NUMBER |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | import sys
input = sys.stdin.readline
t = int(input())
for i in range(t):
n = int(input())
arr = [int(x) for x in input().split()]
arr.sort(reverse=True)
x = arr[n - 1]
if x == 0:
y = arr.count(0)
print(n - y)
else:
y = arr.count(x)
r = x * n
print(r + (n - y)) | IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR BIN_OP VAR VAR |
Monica decides that she would like to get to know the neighbours in the apartment better. She makes a batch of wonderful chocolates and hangs them on the door in a basket hoping that her neighbors will take some and they can meet. The neighbours (including Joey) eventually go crazy over the candy and demand more. Eventually, she keeps a bowl full of chocolates at the door for the last time.
There are N neighbours. The i^{th} neigbhour has initial energy equal to A_{i}. There is one bowl filled with chocolates. The neighbours are made to stand in a row and the bowl is passed around by obeying the following rules:
Any person can hold the bowl initially.
If the person holding the bowl has positive energy, he/she passes the bowl to the person on the immediate right of him/her. The rightmost person in the row passes the bowl to the leftmost person in the row.
The act of passing the bowl takes 1 second.
If the person holding the bowl has non-positive energy, he/she drops the bowl.
After each pass, the energy of the person reduces by 1.
Among all possible ways in which the N neighbours start the game, find the maximum time until the bowl is dropped.
------ Input Format ------
- First line will contain T, number of testcases. Then the testcases follow.
- First line of each testcase contains one integer N.
- Second line of each testcase contains of N integers, denoting the elements of array A.
------ Output Format ------
For each testcase, output in a single line the maximum time until the bowl is dropped.
------ Constraints ------
$1 ≤ T ≤ 100$
$1 ≤ N ≤ 10^{5}$
$0 ≤ A[i] ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $2 \cdot 10^{5}$
----- Sample Input 1 ------
3
3
2 1 1
3
0 5 0
4
3 0 2 1
----- Sample Output 1 ------
4
1
3
----- explanation 1 ------
Test case 1: One of the optimal orders in which all the neighbours can stand in the row is:
$1$ $\rightarrow$ $2$ $\rightarrow$ $3$ $\rightarrow$ $1$. The bowl is initially with person $1$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $1$.
- Person $2$, in one second, passes the bowl to person $3$ and his/her own energy becomes $0$.
- Person $3$, in one second, passes the bowl to person $1$ and his/her own energy becomes $0$.
- Person $1$, in one second, passes the bowl to person $2$ and his/her own energy becomes $0$.
- Person $2$ has $0$ energy, so he/she drops the bowl.
Thus, the bowl is dropped after $4$ seconds.
Test case 2: One of the optimal orders in which all the neighbours can stand in the row is:
$2$ $\rightarrow$ $1$ $\rightarrow$ $3$ $\rightarrow$ $2$. The bowl is initially with person $2$. Thus, it would travel as $2$ $\rightarrow$ $1$. The bowl can not be passed further due to $0$ energy of person $1$. | t = int(input())
for tc in range(t):
n = int(input())
arr = list(map(int, input().split()))
arr.sort(reverse=True)
cnt = arr.count(arr[-1])
print(arr[-1] * n + n - cnt) | ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
transMap = dict()
invalids = set()
for t in transactions:
name, time, amount, city = t.split(",")
if int(amount) > 1000:
invalids.add(t)
if name in transMap:
for stime, scity, st in transMap[name]:
if abs(int(time) - int(stime)) <= 60 and city != scity:
invalids.add(st)
invalids.add(t)
else:
transMap[name] = []
transMap[name].append((time, city, t))
return list(invalids) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR FOR VAR VAR VAR VAR VAR IF FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR LIST EXPR FUNC_CALL VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
ans = []
n = len(transactions)
rank = [True] * n
name = []
time = []
amount = []
city = []
for t in transactions:
num = t.split(",")
name.append(num[0])
time.append(num[1])
amount.append(num[2])
city.append(num[3])
for i in range(n):
if int(amount[i]) > 1000:
rank[i] = False
for j in range(i + 1, n):
if (
name[i] == name[j]
and abs(int(time[i]) - int(time[j])) <= 60
and city[i] != city[j]
):
rank[i] = False
rank[j] = False
for t in range(n):
if not rank[t]:
ans.append(transactions[t])
return ans | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR IF VAR VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
def parseTransaction(t):
vals = t.split(",")
return vals[0], int(vals[1]), int(vals[2]), vals[3]
invalid = set()
names = {}
for t in transactions:
name, time, amt, city = parseTransaction(t)
if amt > 1000:
invalid.add(t)
if name in names:
for other_t, other_time, other_city in names[name]:
if other_city != city and abs(other_time - time) <= 60:
invalid.add(other_t)
invalid.add(t)
else:
names[name] = []
names[name].append((t, time, city))
return list(invalid) | CLASS_DEF FUNC_DEF VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING RETURN VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR FOR VAR VAR VAR VAR VAR IF VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR LIST EXPR FUNC_CALL VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
invalids = set()
for i in range(len(transactions)):
name, time, amount, city = transactions[i].split(",")
if int(amount) > 1000:
invalids.add(transactions[i])
j = i + 1
while j < len(transactions):
nameJ, timeJ, amountJ, cityJ = transactions[j].split(",")
if (
nameJ == name
and abs(int(timeJ) - int(time)) <= 60
and city != cityJ
):
invalids.add(transactions[i])
invalids.add(transactions[j])
j += 1
return list(invalids) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR STRING IF VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
dic = defaultdict(list)
ans = set()
for trans in transactions:
name, time, amt, loc = trans.split(",")
if int(amt) > 1000:
ans.add(trans)
for trans2 in dic[name]:
other = trans2.split(",")
if other[3] != loc and abs(int(other[1]) - int(time)) <= 60:
ans.add(trans)
ans.add(trans2)
dic[name].append(trans)
return ans | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING IF VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
splits = [t.split(",") for t in transactions]
lists = {}
res = set()
for time, p, v, loc, t in sorted(
[
[int(sp[1]), sp[0], int(sp[2]), sp[3], t]
for sp, t in zip(splits, transactions)
]
):
if v > 1000:
res.add(t)
tk = time % 61
if p not in lists:
lists[p] = [None] * 61
lst = lists[p]
for i in range(61):
if lst[i] and abs(lst[i][0] - time) <= 60 and lst[i][1] != loc:
res.add(t)
res.add(lists[p][i][2])
lst[tk] = [time, loc, t]
return res | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR STRING VAR VAR ASSIGN VAR DICT ASSIGN VAR FUNC_CALL VAR FOR VAR VAR VAR VAR VAR FUNC_CALL VAR LIST FUNC_CALL VAR VAR NUMBER VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR BIN_OP LIST NONE NUMBER ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR VAR LIST VAR VAR VAR RETURN VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, ts: List[str]) -> List[str]:
nts = [t.split(",") for t in ts]
nts = sorted([[a, int(b), int(c), d] for a, b, c, d in nts])
res = set()
for a in nts:
if a[2] > 1000:
res.add(",".join(map(str, a)))
for i in range(len(nts)):
for j in range(i + 1, len(nts)):
a, b = nts[i], nts[j]
if a[0] != b[0] or abs(a[1] - b[1]) > 60:
break
if a[3] != b[3]:
res.add(",".join(map(str, a)))
res.add(",".join(map(str, b)))
return list(res) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR STRING VAR VAR ASSIGN VAR FUNC_CALL VAR LIST VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR IF VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
res = []
for i in range(len(transactions)):
added = False
for j in range(i + 1, len(transactions)):
a = transactions[i].split(",")
b = transactions[j].split(",")
if a[0] == b[0] and a[-1] != b[-1] and abs(int(a[1]) - int(b[1])) <= 60:
res.append(",".join(a))
res.append(",".join(b))
added = True
if not added:
if int(transactions[i].split(",")[2]) > 1000:
res.append(transactions[i])
return list(set(res)) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR STRING IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR ASSIGN VAR NUMBER IF VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR STRING NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions):
transactions.sort()
n = len(transactions)
result = set()
for i in range(n):
i_name, i_time, i_amount, i_city = transactions[i].split(",")
if int(i_amount) > 1000:
result.add(transactions[i])
for j in range(i + 1, n):
j_name, j_time, j_amount, j_city = transactions[j].split(",")
if j_name != i_name:
break
if i_city != j_city and abs(int(i_time) - int(j_time)) <= 60:
result.add(transactions[i])
result.add(transactions[j])
return list(result) | CLASS_DEF FUNC_DEF EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR STRING IF VAR VAR IF VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Transaction:
def __init__(self, name, time, amount, city):
self.name = name
self.time = int(time)
self.amount = int(amount)
self.city = city
class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
for i in range(len(transactions)):
transact = transactions[i].split(",")
transactions[i] = Transaction(
transact[0], transact[1], transact[2], transact[3]
)
invalid = set()
transactions.sort(key=lambda t: t.time)
for i, t in enumerate(transactions):
if t.amount > 1000:
invalid.add(self.toString(t))
for j in range(i + 1, len(transactions)):
s = transactions[j]
if t.name == s.name and t.city != s.city and s.time - t.time <= 60:
invalid.add(self.toString(s))
invalid.add(self.toString(t))
return invalid
def toString(self, s):
return "%s,%d,%d,%s" % (s.name, s.time, s.amount, s.city) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR CLASS_DEF FUNC_DEF VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN VAR VAR VAR FUNC_DEF RETURN BIN_OP STRING VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
if not transactions:
return []
res = set()
n = len(transactions)
for i in range(n):
n1, t1, a1, d1 = transactions[i].split(",")
if int(a1) > 1000:
res.add(transactions[i])
for j in range(i + 1, n):
n0, t0, a0, d0 = transactions[j].split(",")
if n0 == n1 and d1 != d0 and abs(int(t1) - int(t0)) <= 60:
res.add(transactions[i])
res.add(transactions[j])
ans = []
for t in res:
ans.append(t)
print(res, ans)
return ans | CLASS_DEF FUNC_DEF VAR VAR IF VAR RETURN LIST ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR STRING IF VAR VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
res = set()
for i in range(len(transactions) - 1):
t1 = transactions[i].split(",")
if int(t1[2]) > 1000:
res.add(transactions[i])
for j in range(i + 1, len(transactions)):
t2 = transactions[j].split(",")
if (
t1[0] == t2[0]
and t1[3] != t2[3]
and abs(int(t2[1]) - int(t1[1])) <= 60
):
res.add(transactions[i])
res.add(transactions[j])
if int(t2[2]) > 1000:
res.add(transactions[j])
return list(res) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR STRING IF FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR STRING IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
parsed_ts = []
invalid_ts = set()
for t in transactions:
t_list = t.split(",")
t_dict = {}
t_dict["str"] = t
t_dict["name"] = t_list[0]
t_dict["time"] = int(t_list[1])
t_dict["amount"] = int(t_list[2])
t_dict["city"] = t_list[3]
if t_dict["amount"] > 1000:
invalid_ts.add(t)
for t2 in parsed_ts:
if (
abs(t_dict["time"] - t2["time"]) <= 60
and t_dict["city"] != t2["city"]
and t_dict["name"] == t2["name"]
):
invalid_ts.add(t_dict["str"])
invalid_ts.add(t2["str"])
parsed_ts.append(t_dict)
return list(invalid_ts) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR DICT ASSIGN VAR STRING VAR ASSIGN VAR STRING VAR NUMBER ASSIGN VAR STRING FUNC_CALL VAR VAR NUMBER ASSIGN VAR STRING FUNC_CALL VAR VAR NUMBER ASSIGN VAR STRING VAR NUMBER IF VAR STRING NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR IF FUNC_CALL VAR BIN_OP VAR STRING VAR STRING NUMBER VAR STRING VAR STRING VAR STRING VAR STRING EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
transactionDict = {}
invalid = []
for i in transactions:
transactionData = i.split(",")
name, time, amount, city = (
transactionData[0],
transactionData[1],
transactionData[2],
transactionData[3],
)
if name in transactionDict.keys():
transactionVal = transactionDict[name]
else:
transactionVal = []
transactionVal.append([time, amount, city])
transactionDict[name] = transactionVal
for i in transactionDict.keys():
for v in transactionDict[i]:
if int(v[1]) > 1000:
if "{},{},{},{}".format(i, v[0], v[1], v[2]) not in invalid:
invalid.append("{},{},{},{}".format(i, v[0], v[1], v[2]))
for listd in transactionDict[i]:
if listd != v:
if abs(int(v[0]) - int(listd[0])) <= 60 and v[2] != listd[2]:
if "{},{},{},{}".format(i, v[0], v[1], v[2]) not in invalid:
invalid.append(
"{},{},{},{}".format(i, v[0], v[1], v[2])
)
return invalid | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR FUNC_CALL VAR ASSIGN VAR VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR LIST VAR VAR VAR ASSIGN VAR VAR VAR FOR VAR FUNC_CALL VAR FOR VAR VAR VAR IF FUNC_CALL VAR VAR NUMBER NUMBER IF FUNC_CALL STRING VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR NUMBER VAR NUMBER VAR NUMBER FOR VAR VAR VAR IF VAR VAR IF FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER VAR NUMBER IF FUNC_CALL STRING VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR NUMBER VAR NUMBER VAR NUMBER RETURN VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
M = collections.defaultdict(dict)
ans = set()
for t in transactions:
name, time, amount, city = t.split(",")
if int(amount) > 1000:
ans.add(t)
Flag = False
if name in M:
for k, v in list(M[name].items()):
if abs(int(time) - k) <= 60 and v.split(",")[-1] != city:
ans.add(v)
Flag = True
M[name][int(time)] = t
if Flag:
ans.add(t)
return list(ans) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR VAR FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR STRING NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR IF VAR EXPR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
personalTransactions = {}
invalidTransactions = set()
for i in transactions:
name, time, amount, city = i.split(",")
if int(amount) > 1000:
invalidTransactions.add(i)
if name not in personalTransactions.keys():
personalTransactions[name] = [(int(time), city, amount)]
continue
else:
personalTransactions[name].append((int(time), city, amount))
for j in personalTransactions[name]:
if city == j[1]:
continue
elif abs(j[0] - int(time)) <= 60:
invalidTransactions.add(i)
invalidTransactions.add(
name + "," + str(j[0]) + "," + j[2] + "," + j[1]
)
return list(invalidTransactions) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR DICT ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR ASSIGN VAR VAR LIST FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR FOR VAR VAR VAR IF VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR STRING FUNC_CALL VAR VAR NUMBER STRING VAR NUMBER STRING VAR NUMBER RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class dsu:
def __init__(self, n):
self.parent = [x for x in range(n)]
self.rank = [(0) for x in range(n)]
self.valid = [(True) for x in range(n)]
self.n = n
def find(self, x):
if self.parent[x] != x:
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def union(self, s, t):
sp, tp = self.find(s), self.find(t)
if sp == tp:
return
if self.rank[sp] < self.rank[tp]:
self.parent[sp] = tp
elif self.rank[sp] > self.rank[tp]:
self.parent[tp] = sp
else:
self.parent[tp] = sp
self.rank[sp] += 1
def cluster(self):
for x in range(self.n):
self.find(x)
class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
arrlen = len(transactions)
net = dsu(arrlen)
tplist = [t.split(",") for t in transactions]
for i in range(arrlen):
tp = tplist[i]
name, time, amt, city = tp[0], tp[1], tp[2], tp[3]
if int(amt) > 1000:
net.valid[i] = False
for j in range(i + 1, arrlen):
jtp = tplist[j]
jname, jtime, jamt, jcity = jtp[0], jtp[1], jtp[2], jtp[3]
if (
jname == name
and jcity != city
and abs(int(jtime) - int(time)) <= 60
):
net.union(i, j)
net.cluster()
for i in range(arrlen):
if net.parent[i] != i or net.rank[i] != 0:
net.valid[i] = False
return [transactions[i] for i in range(arrlen) if not net.valid[i]] | CLASS_DEF FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_DEF IF VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR RETURN VAR VAR FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR VAR RETURN IF VAR VAR VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER FUNC_DEF FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER RETURN VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Transactions:
def __init__(self, name, time, amount, city):
self.name = name
self.time = time
self.amount = amount
self.city = city
class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
transact = []
for t in transactions:
t = t.split(",")
obj = Transactions(t[0], int(t[1]), int(t[2]), t[3])
transact.append(obj)
transact.sort(key=lambda t: t.time)
invalid = set()
for i in range(len(transact)):
first = transact[i]
if first.amount > 1000:
invalid.add(self.returnString(first))
for j in range(i + 1, len(transact)):
second = transact[j]
if (
first.name == second.name
and first.city != second.city
and second.time - first.time <= 60
):
invalid.add(self.returnString(first))
invalid.add(self.returnString(second))
return list(invalid)
def returnString(self, first):
return (
first.name
+ ","
+ str(first.time)
+ ","
+ str(first.amount)
+ ","
+ first.city
) | CLASS_DEF FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR FUNC_DEF RETURN BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP BIN_OP VAR STRING FUNC_CALL VAR VAR STRING FUNC_CALL VAR VAR STRING VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
if not transactions or len(transactions) == 0:
return []
d = collections.defaultdict(dict)
res = set()
for t in transactions:
name, time, amount, city = t.split(",")
if int(amount) > 1000:
res.add(t)
flag = False
if name in d:
for _time, _t in d[name].items():
if abs(_time - int(time)) <= 60 and city != _t.split(",")[-1]:
flag = True
res.add(_t)
if flag:
res.add(t)
d[name][int(time)] = t
return list(res) | CLASS_DEF FUNC_DEF VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER RETURN LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR STRING IF FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR BIN_OP VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR STRING NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
A transaction is possibly invalid if:
the amount exceeds $1000, or;
if it occurs within (and including) 60 minutes of another transaction with the same name in a different city.
Each transaction string transactions[i] consists of comma separated values representing the name, time (in minutes), amount, and city of the transaction.
Given a list of transactions, return a list of transactions that are possibly invalid. You may return the answer in any order.
Example 1:
Input: transactions = ["alice,20,800,mtv","alice,50,100,beijing"]
Output: ["alice,20,800,mtv","alice,50,100,beijing"]
Explanation: The first transaction is invalid because the second transaction occurs within a difference of 60 minutes, have the same name and is in a different city. Similarly the second one is invalid too.
Example 2:
Input: transactions = ["alice,20,800,mtv","alice,50,1200,mtv"]
Output: ["alice,50,1200,mtv"]
Example 3:
Input: transactions = ["alice,20,800,mtv","bob,50,1200,mtv"]
Output: ["bob,50,1200,mtv"]
Constraints:
transactions.length <= 1000
Each transactions[i] takes the form "{name},{time},{amount},{city}"
Each {name} and {city} consist of lowercase English letters, and have lengths between 1 and 10.
Each {time} consist of digits, and represent an integer between 0 and 1000.
Each {amount} consist of digits, and represent an integer between 0 and 2000. | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
invalid_trans = set()
d = {}
for transaction in transactions:
t_list = transaction.split(",")
person, time, amount, loc = (
t_list[0],
int(t_list[1]),
int(t_list[2]),
t_list[3],
)
if amount > 1000:
invalid_trans.add(transaction)
if person not in d:
d[person] = []
else:
for t in d[person]:
old_t = t.split(",")
old_time, old_loc = int(old_t[1]), old_t[3]
if abs(old_time - time) <= 60 and old_loc != loc:
invalid_trans.add(transaction)
invalid_trans.add(t)
d[person].append(transaction)
return list(invalid_trans) | CLASS_DEF FUNC_DEF VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR VAR VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR LIST FOR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR VAR |
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