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IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
l = input() n = len(l) k = [(0) for i in range(n)] for t in range(1, n): if l[t] != l[t - 1]: k[t - 1] = 1 if l[-1] == "b": k[-1] = 0 else: k[-1] = 1 print(*k)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER NUMBER ASSIGN VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = str(input()) n = len(s) ansarr = [0] * len(s) i = 0 first = 0 while i < n: if s[i] == "b": while i < n and s[i] != "a": i += 1 if i == n: break elif first != 0: ansarr[i - 1] = 1 first = 0 else: while i < n and s[i] == "a": ansarr[i] = 0 i += 1 first = 1 ansarr[i - 1] = 1 print(*ansarr)
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR VAR STRING WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR STRING ASSIGN VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def reverse(s, index): s1 = s[: index + 1] s1 = s1[::-1] s1 = s1 + s[index + 1 :] return s1 def main(): s = input() n = len(s) ans = [] for i in range(len(s) - 1): if s[i] != s[i + 1]: s = reverse(s, i) ans.append(1) else: ans.append(0) if s[-1] == "a": s = reverse(s, n - 1) ans.append(1) else: ans.append(0) for i in ans: print(i, end=" ") main()
FUNC_DEF ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def rev(s, n): x = s[0:n][::-1] return str(x + s[n:]) s = input() n = len(s) ans = "" for i in range(n - 1): if s[i] == "a" and s[i + 1] == "b": s = rev(s, i + 1) ans += "1 " elif s[i] == "b" and s[i + 1] == "a": s = rev(s, i + 1) ans += "1 " else: ans += "0 " if s[n - 1] == "a": ans += "1" else: ans += "0" print(ans)
FUNC_DEF ASSIGN VAR VAR NUMBER VAR NUMBER RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR STRING IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR STRING VAR STRING IF VAR BIN_OP VAR NUMBER STRING VAR STRING VAR STRING EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
import sys a = input() res = ["0"] * len(a) i = 1 while i < len(a): if i > 1 and a[i - 1] != a[i]: res[i - 1] = "1" i += 1 if i != 1 and a[i - 1] == "a": res[i - 1] = "1" sys.stdout.write(" ".join(res))
IMPORT ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST STRING FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR NUMBER STRING VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = list(input()) res = [] for i in range(len(s) - 1): if s[i] != s[i + 1]: res.append(1) s = s[: i + 1][::-1] + s[i + 1 :] else: res.append(0) print(*res, 0 if s < s[::-1] else 1)
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR VAR NUMBER NUMBER NUMBER
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
gcd = lambda a, b: gcd(b, a % b) if b else a def main(): arr = input() sa = 0 for i in arr: if i == "a": sa += 1 brr = [(0) for i in range(len(arr))] for i in range(len(arr) - 1): if arr[i] != arr[i + 1]: arr = arr[: i + 1][::-1] + arr[i + 1 :] brr[i] = 1 if arr[-1] == "a": brr[-1] = 1 print(*brr) main()
ASSIGN VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() ans = [(0) for i in range(len(s))] for i in range(1, len(s)): if s[i] == "a" and s[i - 1] == "b" or s[i] == "b" and s[i - 1] == "a": ans[i - 1] = 1 s = s[0:i][::-1] + s[i:] if s[-1] == "a": ans[-1] = 1 print(*ans)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER VAR VAR IF VAR NUMBER STRING ASSIGN VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
from itertools import combinations from sys import setrecursionlimit, stdin setrecursionlimit(200000) def reverse_check_big(string, index): temp = string[: index + 1] forward = "".join(temp) reverse = "".join(temp[::-1]) if reverse > forward: return True else: return False def change(string, index): temp = string[: index + 1] temp = temp[::-1] for i in range(index + 1): string[i] = temp[i] def reverse_check_small(string, index): temp = string[: index + 1] forward = "".join(temp) reverse = "".join(temp[::-1]) if reverse < forward: return True else: return False def find_smaller_index(string, index): malph = "z" ans = -1 for i in range(index + 1): if string[i] <= malph: malph = string[i] ans = i return ans def find_bigger_index(string, index): malph = "a" ans = -1 for i in range(index + 1): if string[i] >= malph: malph = string[i] ans = i return ans string = list(stdin.readline().strip()) binary = [0] * len(string) flag = 0 index = len(string) - 1 process = [] while index >= 0: if flag == 0: new_index = find_smaller_index(string, index) if new_index > 0: process.append([new_index - 1, 1]) flag = 1 index = new_index - 1 else: new_index = find_bigger_index(string, index) if new_index > 0: process.append([new_index - 1, 0]) flag = 0 index = new_index - 1 index = -1 for i in range(len(string)): if len(process) == 0: break elif i > index: index, t = process.pop() if t == 0: if string[i] > string[0]: pass elif string[i] == string[0]: if reverse_check_small(string, i): change(string, i) binary[i] = 1 else: pass else: change(string, i) binary[i] = 1 elif string[i] < string[0]: pass elif string[i] == string[0]: if reverse_check_big(string, i): change(string, i) binary[i] = 1 else: pass else: change(string, i) binary[i] = 1 index = find_smaller_index(string, len(string) - 1) change(string, index) binary[index] = 1 print(*binary)
EXPR FUNC_CALL VAR NUMBER FUNC_DEF ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL STRING VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_DEF ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR FUNC_CALL STRING VAR NUMBER IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR LIST WHILE VAR NUMBER IF VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR LIST BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR LIST BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR FUNC_CALL VAR IF VAR NUMBER IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR NUMBER IF VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def ans(s): flag = True for i in s: if i == "b": flag = False if flag: return [(0) for i in s] h = [] for i in range(len(s)): j = len(s) - 1 - i if s[j] == "a": if j == 0: return [(0) for i in s] f = ans(s[j - 1 :: -1]) return f[::-1] + [1] + h h.append(0) return [(0) for i in s] def main(): s = input() h = [(0) for c in s] start = True last = s[0] == "a" for i in range(len(s)): if s[i] == "a": if start: continue if not last: h[i - 1] = 1 last = True else: start = False if last: h[i - 1] = 1 last = False if s[-1] == "a": h[-1] = 1 print(" ".join([str(x) for x in h])) main()
FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR NUMBER IF VAR RETURN NUMBER VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER VAR IF VAR VAR STRING IF VAR NUMBER RETURN NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER RETURN BIN_OP BIN_OP VAR NUMBER LIST NUMBER VAR EXPR FUNC_CALL VAR NUMBER RETURN NUMBER VAR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR IF VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s, out = list(input()), [] last = s[0] for i in range(len(s)): if i < len(s) - 1: if last == "a" and s[i] == "b" and s[i + 1] == "a": out.append(1) last = "b" elif last == "b" and s[i] == "a" and s[i + 1] == "b": out.append(1) last = "a" else: out.append(0) elif s[i] == "a" and last == "b": out.append(1) else: out.append(0) print(*out)
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR LIST ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR STRING VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR STRING IF VAR STRING VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR VAR STRING VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def g_couple(itr): a = next(itr) for b in itr: yield a, b a = b def g_input(str): nb = 0 for c in str: if c != " ": nb = nb * 10 + int(c) else: yield nb nb = 0 yield nb def main(): l_char = list(input()) l_char.reverse() l_bits = [int(char == "b") for char in l_char] l_switch = [] switch_bit = 0 for bit in l_bits: if bit == switch_bit: switch_bit = (switch_bit + 1) % 2 l_switch.append(1) else: l_switch.append(0) l_switch.reverse() l_switch = [str(e) for e in l_switch] print(" ".join(l_switch)) main()
FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR FOR VAR VAR EXPR VAR VAR ASSIGN VAR VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR EXPR VAR ASSIGN VAR NUMBER EXPR VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR STRING VAR VAR ASSIGN VAR LIST ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() o = ["0"] * len(s) a = s[0] for i in range(1, len(s)): if a == s[i]: o[i - 1] = "1" a = s[i - 1] if a == "b": o[-1] = "1" print(" ".join(o))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST STRING FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR STRING ASSIGN VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
st = input() ans = "" k = 0 while len(st) > k and st[k] == "a": k += 1 ans += "0 " while len(st) > k and st[k] == "b": k += 1 ans += "0 " for i in range(k, len(st)): if st[i - 1] != st[i]: ans += "1 " else: ans += "0 " if st[-1] == "a": ans += "1" else: ans += "0" ans = ans[2:] print(ans)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER WHILE FUNC_CALL VAR VAR VAR VAR VAR STRING VAR NUMBER VAR STRING WHILE FUNC_CALL VAR VAR VAR VAR VAR STRING VAR NUMBER VAR STRING FOR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER VAR VAR VAR STRING VAR STRING IF VAR NUMBER STRING VAR STRING VAR STRING ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() ANS = [] s = s + "b" now = s[0] if now == "a": i = s.index("b") elif "a" in s: i = s.index("a") else: i = 0 while i < len(s) - 1: if s[i] == "a" and s[i + 1] == "b": ANS.append(i) elif s[i] == "b" and s[i + 1] == "a": ANS.append(i) i += 1 LIST = [(0) for i in range(len(s) - 1)] for a in ANS: LIST[a] = 1 for l in LIST: print(l, end=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR STRING ASSIGN VAR VAR NUMBER IF VAR STRING ASSIGN VAR FUNC_CALL VAR STRING IF STRING VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR NUMBER WHILE VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) dp = [] dp.append([]) dp.append([]) dp[0].append((s[0], [0])) dp[1].append((s[0], [0])) for i in range(1, len(s)): t0, sw0 = dp[0][i - 1] t1, sw1 = dp[1][i - 1] g = (t1 + s[i])[::-1] if t0 + s[i] > g: sw = sw1.copy() sw.append(1) dp[0].append((g, sw)) else: sw = sw0.copy() sw.append(0) dp[0].append((t0 + s[i], sw)) g = (t0 + s[i])[::-1] if t1 + s[i] < g: sw = sw0.copy() sw.append(1) dp[1].append((g, sw)) else: sw = sw1.copy() sw.append(0) dp[1].append((t1 + s[i], sw)) print(*dp[0][n - 1][1])
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR LIST EXPR FUNC_CALL VAR LIST EXPR FUNC_CALL VAR NUMBER VAR NUMBER LIST NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER IF BIN_OP VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR NUMBER BIN_OP VAR NUMBER NUMBER
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
import sys input = sys.stdin.readline mii = lambda: map(int, input().split()) s = input().strip() a1 = "" a2 = [] b1 = "" b2 = [] for ch in s: got = [a1 + ch, b1 + ch, ch + a1[::-1], ch + b1[::-1]] arr = [a2 + [0], b2 + [0], a2 + [1], b2 + [1]] a3 = min(got) idx = got.index(a3) a4 = arr[idx] b3 = a3[::-1] b4 = arr[idx ^ 2] a1 = a3 a2 = a4 b1 = b3 b2 = b4 print(" ".join(map(str, a2)))
IMPORT ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR LIST ASSIGN VAR STRING ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR LIST BIN_OP VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR LIST BIN_OP VAR LIST NUMBER BIN_OP VAR LIST NUMBER BIN_OP VAR LIST NUMBER BIN_OP VAR LIST NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() c = "ab" j = 0 i = len(s) r = [0] * i try: while 1: i = s.rindex(c[j], 0, i) r[i] = 1 j ^= 1 except: print(*r)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR WHILE NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER VAR ASSIGN VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) ans = [] bpt = n - 1 while bpt > 0: if s[bpt] == "a": ans.append(1) bpt -= 1 while bpt > 0 and s[bpt] == "a": ans.append(0) bpt -= 1 if bpt > 0 and s[bpt] != "a": ans.append(1) bpt -= 1 else: ans.append(0) bpt -= 1 ans.append(0) print(*reversed(ans))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER IF VAR VAR STRING EXPR FUNC_CALL VAR NUMBER VAR NUMBER WHILE VAR NUMBER VAR VAR STRING EXPR FUNC_CALL VAR NUMBER VAR NUMBER IF VAR NUMBER VAR VAR STRING EXPR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
from sys import stdin, stdout def ii(): return int(stdin.readline()) def mi(): return map(int, stdin.readline().split()) def li(): return list(mi()) def si(): return stdin.readline() s = input() ans = [0] * 1007 for i in range(1, len(s)): if s[i] == "a": ans[i - 1] ^= 1 ans[i] = 1 print(*ans[: len(s)])
FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() i = 0 result = [] n = len(s) i = 0 j = 0 while i < n and j < n: if s[i] == "a": j = i + 1 while j < n and s[j] == "a": j += 1 if i - 1 >= 0: result.append(i - 1) result.append(j - 1) i = j + 1 else: i += 1 for i in range(n): if i in result: print(1, end=" ") else: print(0, end=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = list(input()) a = [(0) for i in range(len(s))] for i in range(1, len(s)): if s[i] == "a": a[i] = 1 a[i - 1] ^= 1 print(" ".join(list(map(str, a))))
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR FUNC_CALL VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) l = [(0) for i in range(n)] a = "a" b = "b" c = a for i in range(n - 1, -1, -1): if c == s[i]: l[i] = 1 if c == a: c = b else: c = a for i in l: print(i, end=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR STRING ASSIGN VAR STRING ASSIGN VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def not_last(c): if c == "a": return "b" return "a" s = input() last = "b" res = [] for i in range(len(s) - 1): if s[i + 1] == last: res += [0] else: last = not_last(last) res += [1] if last == "a": res += [1] else: res += [0] print(" ".join(list(map(str, res))))
FUNC_DEF IF VAR STRING RETURN STRING RETURN STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR LIST NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR LIST NUMBER IF VAR STRING VAR LIST NUMBER VAR LIST NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR FUNC_CALL VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() l = [] i = 0 while i < len(s): l.append(0) while i < len(s) and s[i] == "b": l[-1] += 1 i += 1 l.append(0) while i < len(s) and s[i] == "a": l[-1] += 1 i += 1 i = 0 res = [] for j, e in enumerate(l): i += e if j % 2 == 0: if i != e and l[j + 1] > 0: if e: res.append(i - 1) elif e: res.append(i - 1) v = ["0"] * len(s) for e in res: v[e] = "1" print(" ".join(v))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER NUMBER IF VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP LIST STRING FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) min_str = "".join(sorted(list(s))) res = [] for i in range(n - 1, -1, -1): if not s[i] == min_str[i]: tail = min_str[i + 1 :] s_lst = reversed(list(min_str[: i + 1])) new_s = "".join(s_lst) min_str = new_s + tail res.append(1) else: res.append(0) print(" ".join([str(x) for x in list(reversed(res))]))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL STRING FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL STRING VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def flip(wd): changes = [] last = wd[0] for i, x in enumerate(wd): if x != last: changes.append(i - 1) last = x changes.append(len(wd) - 1) if wd[-1] == "b": changes.pop(-1) res = [0] * len(wd) for ch in changes: res[ch] = 1 return res def test(wd): f = flip(wd) x = rev(wd, f) assert wd.count("a") == x.count("a") assert wd.count("b") == x.count("b") assert len(x) == len(wd) assert x == "a" * wd.count("a") + "b" * wd.count("b") wd = input() print(*flip(wd))
FUNC_DEF ASSIGN VAR LIST ASSIGN VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR VAR NUMBER RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING FUNC_CALL VAR STRING FUNC_CALL VAR STRING FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR BIN_OP BIN_OP STRING FUNC_CALL VAR STRING BIN_OP STRING FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() def swap(i): return s[:i:-1] + s[i + 1 :] for e in range(len(s) - 1): if s[e] != s[e + 1]: print(1, end=" ") else: print(0, end=" ") if s[-1] == "a": print(1, end=" ") else: print(0)
ASSIGN VAR FUNC_CALL VAR FUNC_DEF RETURN BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING IF VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def main(): s = list(input().strip()) i = -1 n = len(s) b = [0] * n for j in range(n): if s[j] == "a": if i != -1: b[i] ^= 1 b[j - 1] ^= 1 i = j if i != -1: b[i] = 1 for x in b: print(x, end=" ") print() main()
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING IF VAR NUMBER VAR VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR IF VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input().strip() n = len(s) cond = s[0] ans = [] for i in range(n - 1): if s[i + 1] == cond: ans.append("0") else: ans.append("1") cond = s[i + 1] if cond == "b": ans.append("0") else: ans.append("1") print(" ".join(ans))
ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR BIN_OP VAR NUMBER IF VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() ans = [] for i in range(len(s) - 1): if s[i] != s[i + 1]: ans.append(i) ans.append(len(s) - 1) if s[-1] == "b": ans = ans[: len(ans) - 1] an = [0] * len(s) for i in ans: an[i] = 1 print(*an, sep=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR STRING
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) ans = [] s += "b" for i in range(n): if s[i] == "b" and s[i + 1] == "a" or s[i] == "a" and s[i + 1] == "b": ans.append(1) else: ans.append(0) print(*ans)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST VAR STRING FOR VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() output = [] best = s[0] rev = False s2 = s[1:] for i in s2: if i <= best: output.append(rev) rev = True best = i else: output.append(not rev) rev = False output.append(not rev) for i in output: if i: print(0, end=" ") else: print(1, end=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR IF VAR EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) ans = [0] * n for i in range(1, n): if s[i] == "a": if ans[i - 1] == 0: ans[i - 1] = 1 else: ans[i - 1] = 0 ans[i] = 1 print(*ans)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING IF VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) revs = [0] * n c = "a" toc = {"a": "b", "b": "a"} for i in range(n - 1, -1, -1): if s[i] == c: c = toc[c] revs[i] = 1 print(" ".join(map(str, revs)))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR STRING ASSIGN VAR DICT STRING STRING STRING STRING FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER NUMBER IF VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def main(): l = list(input()) wyn = [0] * len(l) pos = 0 for i in range(1, len(l)): if l[i] == "a": wyn[i - 1] ^= 1 wyn[i] ^= 1 print(*wyn, sep=" ") main()
FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
from sys import stdin, stdout a = list(input()) b = sorted(a) l = len(a) ans = ["0" for i in range(l)] for i in range(1, l): if a == b: break if a[0] == "a": if a[i] == "a": a[0 : i - 1] = a[i - 1 : 0 : -1] ans[i - 1] = "1" elif a[i] == "b": a[0 : i - 1] = a[i - 1 : 0 : -1] ans[i - 1] = "1" if a[0] == "b": ans[l - 1] = "1" stdout.write(" ".join(ans))
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR STRING VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR IF VAR NUMBER STRING IF VAR VAR STRING ASSIGN VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER STRING IF VAR VAR STRING ASSIGN VAR NUMBER BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER STRING IF VAR NUMBER STRING ASSIGN VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
seq = input() result = [] len_ = len(seq) result.append("0") if len_ > 1: for i in range(1, len_ - 1): if seq[i] == "b" and seq[i + 1] == "a": result.append("1") elif seq[i] == "a" and seq[i + 1] == "b": result.append("1") else: result.append("0") if seq[-1] == "a": result.append("1") else: result.append("0") print(" ".join(result))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING IF VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() r = [] for i in s: if i == "a": if len(r) != 0: r[-1] = 1 - r[-1] r.append(1) else: r.append(0) print(" ".join(map(str, r)))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR VAR IF VAR STRING IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER BIN_OP NUMBER VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = list(input()) n = len(s) lis = [-1] * n lis[0] = 0 for i in range(n - 1): if s[i] == s[i + 1]: lis[i] = 0 i = n - 1 while i >= 0 and s[i] != "a": lis[i] = 0 i -= 1 while i >= 0: if lis[i] == -1: lis[i] = 1 i -= 1 print(*lis)
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR NUMBER VAR VAR STRING ASSIGN VAR VAR NUMBER VAR NUMBER WHILE VAR NUMBER IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def Main(): arr = input() n = len(arr) res = {} res[0] = 0 for i in range(1, n): if arr[i] == "a": res[i - 1] ^= 1 res[i] = 1 else: res[i] = 0 for x in res: print(res[x], end=" ") Main()
FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF VAR VAR STRING VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING EXPR FUNC_CALL VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() n = len(s) ans = [(0) for i in range(n)] i = 0 while True: while i != n and s[i] == "a": i += 1 if i != 0: ans[i - 1] = 1 if i == n: break while i != n and s[i] == "b": i += 1 if i == n: break else: ans[i - 1] = 1 for i in ans: print(i, end=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER IF VAR VAR WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR STRING
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input().strip() n = len(s) ans = [0] * n i = 0 while i < n: j = i + 1 if s[i] == "b": while j < n and s[j] == "b": j += 1 if j != n: ans[j - 1] = 1 while j < n and s[j] == "a": j += 1 ans[j - 1] = 1 i = j print(*ans)
ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR STRING WHILE VAR VAR VAR VAR STRING VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER NUMBER WHILE VAR VAR VAR VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() + "b" a = list(s) x = [0] for i in range(1, len(a) - 1): if a[i] != a[i + 1] and a[i + 1] == a[0]: h = a[: i + 1] k = list(reversed(h)) a = k + a[i + 1 :] x.append(1) else: x.append(0) print(*x)
ASSIGN VAR BIN_OP FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR LIST NUMBER FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() t = s[0] for i in range(1, len(s)): if s[i] == t: t = s[i - 1] print("1 ", end="") else: print("0 ", end="") if t == "b": print(1) else: print(0)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR STRING STRING IF VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() out = [0] * len(s) k2 = 0 if s[0] == "a": for i in range(len(s) - 1): if s[i] == "b" and s[i + 1] == "a": k1, k2 = i, i for j in range(k2 + 1, len(s)): if s[j] == "b": break k2 += 1 out[k1] = 1 out[k2] = 1 i = k2 else: f = 1 for i in range(len(s) - 1): if f and s[i] == "a" and s[i + 1] == "b": out[i] = 1 f = 0 elif f == 0 and s[i] == "b" and s[i + 1] == "a": k1, k2 = i, i for j in range(k2 + 1, len(s)): if s[j] == "b": break k2 += 1 out[k1] = 1 out[k2] = 1 i = k2 if sum(out) == 0 and s[-1] == "a": out[-1] = 1 print(" ".join(map(str, out)))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER STRING FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR VAR STRING VAR BIN_OP VAR NUMBER STRING ASSIGN VAR VAR VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR NUMBER STRING ASSIGN VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() c = 0 for i in range(len(s)): if s[i] == "a" and c == 0: c = 1 if i != len(s) - 1: if s[i + 1] == "a": print(0, end=" ") else: print(1, end=" ") else: print(1, end=" ") elif s[i] == "b" and c == 0: print(0, end=" ") elif s[i] == "a" and i != len(s) - 1: if s[i + 1] == "a": print(0, end=" ") else: print(1, end=" ") elif s[i] == "b" and i != len(s) - 1: if s[i + 1] == "a": print(1, end=" ") else: print(0, end=" ") elif s[i] == "b" and i == len(s) - 1: print(0, end=" ") elif s[i] == "a" and i == len(s) - 1: print(1, end=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING IF VAR VAR STRING VAR NUMBER EXPR FUNC_CALL VAR NUMBER STRING IF VAR VAR STRING VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING IF VAR VAR STRING VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING EXPR FUNC_CALL VAR NUMBER STRING IF VAR VAR STRING VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER STRING IF VAR VAR STRING VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER STRING
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
t = input() i = 0 res = [(0) for _ in range(len(t))] while i < len(t): if t[i] == "a": j = i + 1 while j < len(t) and t[j] == "a": j += 1 if i != 0: res[i - 1] = 1 res[j - 1] = 1 i = j else: i += 1 print(" ".join(map(str, res)))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR WHILE VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR STRING VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() ss = sorted(s) r = [0] * len(s) b_char = "b" for i, c in enumerate(reversed(s)): if c != b_char: r[len(s) - i - 1] = 1 b_char = "a" if b_char == "b" else "b" print(" ".join([str(i) for i in r]))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR STRING FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR STRING STRING STRING EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
inputs = [s for s in input()] array_a = [] result = [0] * len(inputs) for i in range(len(inputs)): if inputs[i] == "a": array_a.append(i) if len(array_a) == 0: pass else: i = 0 while i < len(array_a): counter = 1 while i + 1 < len(array_a) and array_a[i] + 1 == array_a[i + 1]: counter += 1 i += 1 if array_a[i] - counter >= 0: result[array_a[i] - counter] = 1 result[array_a[i]] = 1 i += 1 print(" ".join([str(s) for s in result]))
ASSIGN VAR VAR VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING EXPR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def solve(word): swaps = [False] * len(word) modified = word for i in range(1, len(word)): if word[i] != word[i - 1]: swaps[i - 1] = True modified = modified[i - 1 :: -1] + modified[i:] if modified[0] != "a": swaps[-1] = True modified = modified[::-1] return swaps, modified word = input() swaps, _ = solve(word) print(" ".join([("1" if x else "0") for x in swaps]))
FUNC_DEF ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR VAR IF VAR NUMBER STRING ASSIGN VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER RETURN VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR STRING STRING VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() o = "" for i, c in enumerate(s[:-1]): if c == "b": if s[i + 1] == "a": s = s[: i + 1][::-1] + s[i + 1 :] o += "1" else: o += "0" elif s[i + 1] == "b": o += "1" s = s[: i + 1][::-1] + s[i + 1 :] else: o += "0" if s[0] == "b": s = s[::-1] o += "1" else: o += "0" print(" ".join(o))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING FOR VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR STRING IF VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR STRING VAR STRING IF VAR BIN_OP VAR NUMBER STRING VAR STRING ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER VAR STRING IF VAR NUMBER STRING ASSIGN VAR VAR NUMBER VAR STRING VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
s = input() ans = [0] * len(s) for i in range(1, len(s) - 1): if s[0] == "a" and s[i : i + 2] == "ba" or s[0] == "b" and s[i : i + 2] == "ab": s = s[: i + 1][::-1] + s[i + 1 :] ans[i] = 1 if s[-1] == "a": s = s[::-1] ans[-1] = 1 print(" ".join(str(i) for i in ans))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER IF VAR NUMBER STRING VAR VAR BIN_OP VAR NUMBER STRING VAR NUMBER STRING VAR VAR BIN_OP VAR NUMBER STRING ASSIGN VAR BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR NUMBER STRING ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR
IA has so many colorful magnets on her fridge! Exactly one letter is written on each magnet, 'a' or 'b'. She loves to play with them, placing all magnets in a row. However, the girl is quickly bored and usually thinks how to make her entertainment more interesting. Today, when IA looked at the fridge, she noticed that the word formed by magnets is really messy. "It would look much better when I'll swap some of them!" — thought the girl — "but how to do it?". After a while, she got an idea. IA will look at all prefixes with lengths from $1$ to the length of the word and for each prefix she will either reverse this prefix or leave it as it is. She will consider the prefixes in the fixed order: from the shortest to the largest. She wants to get the lexicographically smallest possible word after she considers all prefixes. Can you help her, telling which prefixes should be chosen for reversing? A string $a$ is lexicographically smaller than a string $b$ if and only if one of the following holds: $a$ is a prefix of $b$, but $a \ne b$; in the first position where $a$ and $b$ differ, the string $a$ has a letter that appears earlier in the alphabet than the corresponding letter in $b$. -----Input----- The first and the only line contains a string $s$ ($1 \le |s| \le 1000$), describing the initial string formed by magnets. The string $s$ consists only of characters 'a' and 'b'. -----Output----- Output exactly $|s|$ integers. If IA should reverse the $i$-th prefix (that is, the substring from $1$ to $i$), the $i$-th integer should be equal to $1$, and it should be equal to $0$ otherwise. If there are multiple possible sequences leading to the optimal answer, print any of them. -----Examples----- Input bbab Output 0 1 1 0 Input aaaaa Output 1 0 0 0 1 -----Note----- In the first example, IA can reverse the second and the third prefix and get a string "abbb". She cannot get better result, since it is also lexicographically smallest string obtainable by permuting characters of the initial string. In the second example, she can reverse any subset of prefixes — all letters are 'a'.
def check(current, source_final_ch): if current[-1] != source_final_ch: current.reverse() return 1, current[:-1] return 0, current[:-1] def calc(source): actions = [] best = sorted(source) current = best for i in range(len(source)): res = check(current, source[-i - 1]) act, current = res actions.insert(0, act) return actions def get_ints(): return [int(n) for n in input().split()] def get_floats(): return [float(n) for n in input().split()] def seq2str(seq): return " ".join(str(item) for item in seq) s = input() res = calc(list(s)) print(seq2str(res))
FUNC_DEF IF VAR NUMBER VAR EXPR FUNC_CALL VAR RETURN NUMBER VAR NUMBER RETURN NUMBER VAR NUMBER FUNC_DEF ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR EXPR FUNC_CALL VAR NUMBER VAR RETURN VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() def count(s): d = {} for c in s: d[c] = d.get(c, 0) + 1 return d dt = count(t) maxY = 0 maxW = 0 arr = [False] * len(s) for i in range(len(s)): c = s[i] if c in dt and dt[c] > 0: maxY += 1 dt[c] -= 1 arr[i] = True for i in range(len(s)): c = s[i] opC = c.lower() if opC == c: opC = c.upper() if opC in dt and dt[opC] > 0 and arr[i] == False: dt[opC] -= 1 maxW += 1 print(maxY, maxW)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
msg = input() nwp = input() aa = [(0) for i in range(26)] bb = [(0) for i in range(26)] cc = [(0) for i in range(26)] dd = [(0) for i in range(26)] for i in msg: if ord(i) < 95: bb[ord(i) - 65] += 1 else: aa[ord(i) - 97] += 1 for i in nwp: if ord(i) < 95: dd[ord(i) - 65] += 1 else: cc[ord(i) - 97] += 1 daba, store = 0, 0 vr, rb = 0, 0 for i in range(26): vr = min(aa[i], cc[i]) rb = min(bb[i], dd[i]) daba += vr daba += rb store += min(aa[i] + bb[i] - vr - rb, cc[i] + dd[i] - vr - rb) print(daba, store)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def tanya_and_postcard(tanya_message, newspaper_letters): difference = ord("a") - ord("A") yays_count = 0 whoopses_count = 0 letters_counts = get_letters_counts(newspaper_letters) tanya_message = list(tanya_message) for index in range(0, len(tanya_message)): letter = tanya_message[index] if letter in letters_counts and letters_counts[letter] > 0: letters_counts[letter] -= 1 yays_count += 1 tanya_message[index] = chr(0) for letter in tanya_message: letter = get_letter_in_opposite_casing(letter) if letter in letters_counts and letters_counts[letter] > 0: letters_counts[letter] -= 1 whoopses_count += 1 return {"yays_count": yays_count, "whoopses_count": whoopses_count} def get_letters_counts(letters): letters_counts = {} for letter in letters: if letter in letters_counts: letters_counts[letter] += 1 else: letters_counts[letter] = 1 return letters_counts def get_letter_in_opposite_casing(letter): if letter.islower(): return letter.upper() else: return letter.lower() def main(): tanya_message = input() newspaper_letters = input() result = tanya_and_postcard(tanya_message, newspaper_letters) print("{0} {1}".format(result["yays_count"], result["whoopses_count"])) def __starting_point(): main() __starting_point()
FUNC_DEF ASSIGN VAR BIN_OP FUNC_CALL VAR STRING FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER RETURN DICT STRING STRING VAR VAR FUNC_DEF ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER RETURN VAR FUNC_DEF IF FUNC_CALL VAR RETURN FUNC_CALL VAR RETURN FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR STRING VAR STRING FUNC_DEF EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() kek, lol = dict(), dict() for x in s: kek[x] = kek.get(x, 0) + 1 for x in t: lol[x] = lol.get(x, 0) + 1 res11, res2 = 0, 0 for x in kek: val = min(kek[x], lol.get(x, 0)) res11 += val kek[x] -= val lol[x] = lol.get(x, 0) - val for x in kek: if "a" <= x <= "z": val = min(kek[x], lol.get(x.upper(), 0)) res2 += val lol[x.upper()] = lol.get(x.upper(), 0) - val else: val = min(kek[x], lol.get(x.lower(), 0)) res2 += val lol[x.lower()] = lol.get(x.lower(), 0) - val print(res11, res2)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR FOR VAR VAR IF STRING VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR VAR ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def main(s1, s2): def flip(c): if ord(c) in range(ord("a"), ord("z") + 1): return chr(ord("A") - ord("a") + ord(c)) else: return chr(ord("a") - ord("A") + ord(c)) m = [(0) for i in range(256)] for c in s2: m[ord(c)] += 1 a = 0 b = 0 marked = set() for i, c in enumerate(s1): if m[ord(c)] > 0: m[ord(c)] -= 1 a += 1 marked.add(i) for i, c in enumerate(s1): if i in marked: continue if m[ord(flip(c))] > 0: m[ord(flip(c))] -= 1 b += 1 return "{} {}".format(a, b) print(main(input(), input()))
FUNC_DEF FUNC_DEF IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER RETURN FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR STRING FUNC_CALL VAR STRING FUNC_CALL VAR VAR RETURN FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR STRING FUNC_CALL VAR STRING FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER RETURN FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
r = input() h = input() r_dic = {} h_dic = {} for c in r: h_dic[c] = 0 if c in r_dic: r_dic[c] += 1 else: r_dic[c] = 1 for c in h: if c in h_dic: h_dic[c] += 1 else: h_dic[c] = 1 y = 0 o = 0 for k, v in r_dic.items(): if r_dic[k] > h_dic[k]: r_dic[k] -= h_dic[k] y += h_dic[k] h_dic[k] = 0 elif r_dic[k] < h_dic[k]: h_dic[k] -= r_dic[k] y += r_dic[k] r_dic[k] = 0 else: y += r_dic[k] r_dic[k] = 0 h_dic[k] = 0 for k, v in r_dic.items(): if r_dic[k] > 0: k_t = k if k_t.islower(): tmep_k = k_t.upper() else: tmep_k = k_t.lower() if tmep_k in h_dic: if r_dic[k] < h_dic[tmep_k]: o += r_dic[k] h_dic[tmep_k] -= r_dic[k] r_dic[k] = 0 elif r_dic[k] > h_dic[tmep_k]: o += h_dic[tmep_k] r_dic[k] -= h_dic[tmep_k] h_dic[tmep_k] = 0 else: o += r_dic[k] r_dic[k] = 0 h_dic[tmep_k] = 0 print(y, end=" ") print(o, end=" ")
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR IF VAR VAR NUMBER ASSIGN VAR VAR IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR STRING EXPR FUNC_CALL VAR VAR STRING
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() s = list(s) d = {} for i in range(len(t)): if t[i] in d: d[t[i]] += 1 else: d[t[i]] = 1 yay, whoop = 0, 0 for i in range(len(s)): if s[i] in d and d[s[i]] > 0: yay += 1 d[s[i]] -= 1 s[i] = " " for i in range(len(s)): if 65 <= ord(s[i]) <= 90: x = chr(ord(s[i]) + 32) else: x = chr(ord(s[i]) - 32) if x in d and d[x] > 0: whoop += 1 d[x] -= 1 print(yay, whoop)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF NUMBER FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() alp = [] countalp = [] salp = [] scountalp = [] yay = 0 woops = 0 for i, x in enumerate(s): if x not in alp: alp.append(x) n = s.count(x) countalp.append(n) for i, x in enumerate(t): if x not in salp: salp.append(x) n = t.count(x) scountalp.append(n) for b, y in enumerate(alp): tcount = t.count(y) if tcount > 0: if abs(countalp[b] - tcount) == 0: yay += tcount scountalp[salp.index(y)] -= tcount countalp[b] -= tcount else: yay += min(countalp[b], tcount) if countalp[b] - tcount < 0: scountalp[salp.index(y)] -= countalp[b] countalp[b] = 0 else: countalp[b] -= tcount scountalp[salp.index(y)] = 0 for k, z in enumerate(alp): if ( z.swapcase() in salp and countalp[k] > 0 and scountalp[salp.index(z.swapcase())] > 0 ): woops += min(scountalp[salp.index(z.swapcase())], countalp[k]) print(yay, woops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER IF FUNC_CALL VAR BIN_OP VAR VAR VAR NUMBER VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER FOR VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
S = input() T = input() N = len(S) deg = [(0) for i in range(300)] for i in range(len(T)): deg[ord(T[i])] += 1 yay = 0 whp = 0 used = [(False) for i in range(3 * 10**5)] for i in range(N): if deg[ord(S[i])] >= 1: yay += 1 deg[ord(S[i])] -= 1 used[i] = True for i in range(N): if used[i] == False: if S[i].isupper(): tmp = S[i].lower() if deg[ord(tmp)] >= 1: whp += 1 deg[ord(tmp)] -= 1 used[i] = True if S[i].islower(): tmp = S[i].upper() if deg[ord(tmp)] >= 1: whp += 1 deg[ord(tmp)] -= 1 used[i] = True print(yay, whp)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR BIN_OP NUMBER BIN_OP NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def to_dict(text): out = {} for letter in text: out[letter] = out.get(letter, 0) + 1 return out s = input() t = input() parsed_s = to_dict(s) parsed_t = to_dict(t) yays = 0 for letter in parsed_s: matches = min(parsed_s[letter], parsed_t.get(letter, 0)) if matches > 0: yays += matches parsed_s[letter] -= matches parsed_t[letter] -= matches whoopses = 0 for letter in parsed_s: missing = parsed_s[letter] if missing > 0: if letter.islower(): matches = min(missing, parsed_t.get(letter.upper(), 0)) if matches > 0: whoopses += matches parsed_s[letter] -= matches parsed_t[letter.upper()] -= matches else: matches = min(missing, parsed_t.get(letter.lower(), 0)) if matches > 0: whoopses += matches parsed_s[letter] -= matches parsed_t[letter.lower()] -= matches print("%s %s" % (yays, whoopses))
FUNC_DEF ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP STRING VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() arrS = [0] * 70 arrT = [0] * 70 for i in range(len(s)): arrS[ord(s[i]) - ord("A")] += 1 for i in range(len(t)): arrT[ord(t[i]) - ord("A")] += 1 yay = 0 whoops = 0 for i in range(70): temp = min(arrS[i], arrT[i]) yay += temp arrT[i] -= temp arrS[i] -= temp for i in range(28): temp = min(arrS[i], arrT[i + 32]) arrT[i + 32] -= temp arrS[i] -= temp whoops += temp for i in range(32, 68): temp = min(arrS[i], arrT[i - 32]) arrT[i - 32] -= temp arrS[i] -= temp whoops += temp print("{} {}".format(yay, whoops))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input().strip() t = input().strip() yay = 0 whoops = 0 yay_map = [0] * 58 whoops_map = [] * 58 slength = len(s) tlength = len(t) s_map = [0] * 58 t_map = [0] * 58 up_low_distance = 32 for i in range(slength): s_map[ord(s[i]) - 65] += 1 for i in range(tlength): t_map[ord(t[i]) - 65] += 1 for i in range(26): if t_map[i] >= s_map[i]: yay += s_map[i] t_map[i] -= s_map[i] s_map[i] = 0 else: yay += t_map[i] s_map[i] -= t_map[i] t_map[i] = 0 if t_map[i + up_low_distance] >= s_map[i + up_low_distance]: yay += s_map[i + up_low_distance] t_map[i + up_low_distance] -= s_map[i + up_low_distance] s_map[i + up_low_distance] = 0 else: yay += t_map[i + up_low_distance] s_map[i + up_low_distance] -= t_map[i + up_low_distance] t_map[i + up_low_distance] = 0 if t_map[i] >= s_map[i + up_low_distance]: whoops += s_map[i + up_low_distance] t_map[i] -= s_map[i + up_low_distance] s_map[i + up_low_distance] = 0 else: whoops += t_map[i] s_map[i + up_low_distance] -= t_map[i] t_map[i] = 0 if t_map[i + up_low_distance] >= s_map[i]: whoops += s_map[i] t_map[i + up_low_distance] -= s_map[i] s_map[i] = 0 else: whoops += t_map[i + up_low_distance] s_map[i] -= t_map[i + up_low_distance] t_map[i + up_low_distance] = 0 print(yay, whoops)
ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def muda_case(letra): if letra.isupper(): return letra.lower() else: return letra.upper() a = input() b = input() contagem1 = {i: (0) for i in a} contagem2 = {i: (0) for i in b} for e in a: contagem1[e] += 1 for e in b: contagem2[e] += 1 YAY = WHOOPS = 0 for e in contagem1: if e in contagem2: while contagem1[e] > 0 and contagem2[e] > 0: YAY += 1 contagem1[e] -= 1 contagem2[e] -= 1 for e in contagem1: casedLetter = muda_case(e) if casedLetter in contagem2: while contagem1[e] > 0 and contagem2[casedLetter] > 0: WHOOPS += 1 contagem1[e] -= 1 contagem2[casedLetter] -= 1 print(YAY, WHOOPS)
FUNC_DEF IF FUNC_CALL VAR RETURN FUNC_CALL VAR RETURN FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER VAR VAR FOR VAR VAR VAR VAR NUMBER FOR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR WHILE VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR WHILE VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
entrada1 = input() entrada2 = input() masc1 = [0] * 30 masc2 = [0] * 30 minc1 = [0] * 30 minc2 = [0] * 30 for letra in entrada1: if letra.islower(): minc1[ord(letra) - ord("a")] += 1 else: masc1[ord(letra) - ord("A")] += 1 for letra in entrada2: if letra.islower(): minc2[ord(letra) - ord("a")] += 1 else: masc2[ord(letra) - ord("A")] += 1 yay = 0 whoops = 0 for i in range(len(masc1)): caso = min(masc1[i], masc2[i]) yay += caso masc1[i] -= caso masc2[i] -= caso caso = min(minc1[i], minc2[i]) yay += caso minc1[i] -= caso minc2[i] -= caso whoops += min(masc1[i], minc2[i]) whoops += min(minc1[i], masc2[i]) print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING NUMBER VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING NUMBER VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = str(input()) t = str(input()) y, w = 0, 0 def oposto(caractere): if caractere.isupper(): return caractere.lower() else: return caractere.upper() hasht = {} for i in t: if i in hasht: hasht[i] += 1 else: hasht[i] = 1 hashs = {} for i in s: if i in hashs: hashs[i] += 1 else: hashs[i] = 1 for i in s: if i in hasht and hasht[i] > 0 and hashs[i] > 0: y += 1 hashs[i] -= 1 hasht[i] -= 1 for i in s: if oposto(i) in hasht and hasht[oposto(i)] > 0 and hashs[i] > 0: w += 1 hasht[oposto(i)] -= 1 hashs[i] -= 1 print(str(y) + " " + str(w))
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FUNC_DEF IF FUNC_CALL VAR RETURN FUNC_CALL VAR RETURN FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR STRING FUNC_CALL VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def cov(ch): if ch >= "a" and ch <= "z": return ord(ch) - ord("a") else: return ord(ch) - ord("A") + 26 lt1 = list(map(cov, input())) lt2 = list(map(cov, input())) cnt = [0] * 52 for x in lt2: cnt[x] += 1 ans0 = 0 ans1 = 0 for i in range(0, len(lt1)): x = lt1[i] if cnt[x]: cnt[x] -= 1 ans0 += 1 lt1[i] = -1 for i in range(0, len(lt1)): x = lt1[i] if x == -1: continue elif x >= 26: x -= 26 else: x += 26 if cnt[x]: cnt[x] -= 1 ans1 += 1 print(ans0, ans1)
FUNC_DEF IF VAR STRING VAR STRING RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING RETURN BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
u = input() v = input() a = [(0) for i in range(200)] b = [(0) for i in range(200)] m = 0 n = 0 d = ord("a") - ord("A") for i in u: a[ord(i)] += 1 for i in v: b[ord(i)] += 1 for i in range(ord("A"), ord("Z") + 1): r = min(a[i], b[i]) + min(a[i + d], b[i + d]) n += r m += min(a[i] + a[i + d], b[i] + b[i + d]) - r print(n, m)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER VAR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR STRING FUNC_CALL VAR STRING FOR VAR VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
a = list(input()) b = list(input()) cnt = 0 scnt = 0 a1, b1 = dict(), dict() for s in a: if s in a1.keys(): a1[s] += 1 else: a1[s] = 1 for s in b: if s in b1.keys(): b1[s] += 1 else: b1[s] = 1 for s in a1.keys(): if s in b1.keys(): if a1[s] < b1[s]: cnt += a1[s] b1[s] = b1[s] - a1[s] a1[s] = 0 else: cnt += b1[s] a1[s] = a1[s] - b1[s] b1[s] = 0 for s in a1.keys(): if chr(ord(s) + 32) in b1.keys(): scnt += min(a1[s], b1[chr(ord(s) + 32)]) elif chr(ord(s) - 32) in b1.keys(): scnt += min(a1[s], b1[chr(ord(s) - 32)]) print("%d %d" % (cnt, scnt))
ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR IF VAR FUNC_CALL VAR IF VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR IF FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP STRING VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() cnt = [0] * 257 used = [0] * (len(s) + 100) for i in range(len(t)): cnt[ord(t[i])] += 1 res1 = 0 res2 = 0 for i in range(len(s)): if cnt[ord(s[i])] > 0: cnt[ord(s[i])] -= 1 res1 += 1 used[i] = 1 for i in range(len(s)): if used[i] == 0: if cnt[ord(s[i]) ^ 32] > 0: cnt[ord(s[i]) ^ 32] -= 1 used[i] = 1 res2 += 1 print(res1, res2)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER IF VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER NUMBER ASSIGN VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
x = input() t = input() d = dict() for i in t: if i in d: d[i] += 1 else: d[i] = 1 s = [] for i in x: s.append([i, 0]) s.sort(reverse=True) ura = 0 opa = 0 for i in range(len(s)): if s[i][0] in d: if d[s[i][0]] > 0: ura += 1 s[i][1] = 1 d[s[i][0]] -= 1 for i in range(len(s)): if chr(ord(s[i][0]) - ord("a") + ord("A")) in d and s[i][1] == 0: if d[chr(ord(s[i][0]) - ord("a") + ord("A"))] > 0: opa += 1 s[i][1] = 1 d[chr(ord(s[i][0]) - ord("a") + ord("A"))] -= 1 for i in range(len(s)): if chr(ord(s[i][0]) - ord("A") + ord("a")) in d and s[i][1] == 0: if d[chr(ord(s[i][0]) - ord("A") + ord("a"))] > 0: opa += 1 s[i][1] = 1 d[chr(ord(s[i][0]) - ord("A") + ord("a"))] -= 1 print(ura, opa)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR LIST FOR VAR VAR EXPR FUNC_CALL VAR LIST VAR NUMBER EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR IF VAR VAR VAR NUMBER NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER VAR VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR STRING VAR VAR VAR NUMBER NUMBER IF VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR STRING NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR STRING NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR STRING VAR VAR VAR NUMBER NUMBER IF VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR STRING NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR STRING FUNC_CALL VAR STRING NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() n = len(s) made = [False] * n a = [0] * 130 for c in t: a[ord(c)] += 1 yay = whoops = 0 for i in range(n): if a[ord(s[i])] > 0: a[ord(s[i])] -= 1 yay += 1 made[i] = True for i in range(n): if made[i] == False: c = s[i] if c.islower() and a[ord(c.upper())] > 0: a[ord(c.upper())] -= 1 whoops += 1 elif c.isupper() and a[ord(c.lower())] > 0: a[ord(c.lower())] -= 1 whoops += 1 print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s1 = input() s2 = input() TAM = 26 a_min = ord("a") a_mai = ord("A") ls1_min = [0] * TAM ls1_mai = [0] * TAM ls2_min = [0] * TAM ls2_mai = [0] * TAM for letra in s1: if letra.islower(): ls1_min[ord(letra) - a_min] += 1 else: ls1_mai[ord(letra) - a_mai] += 1 for letra in s2: if letra.islower(): ls2_min[ord(letra) - a_min] += 1 else: ls2_mai[ord(letra) - a_mai] += 1 yay = 0 whoops = 0 for letra in range(TAM): casos = min(ls1_min[letra], ls2_min[letra]) yay += casos ls1_min[letra] -= casos ls2_min[letra] -= casos casos = min(ls1_mai[letra], ls2_mai[letra]) yay += casos ls1_mai[letra] -= casos ls2_mai[letra] -= casos for letra in range(TAM): casos = min(ls1_min[letra], ls2_mai[letra]) casos += min(ls1_mai[letra], ls2_min[letra]) whoops += casos print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR ASSIGN VAR BIN_OP LIST NUMBER VAR FOR VAR VAR IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s1 = input() s2 = input() ms1 = [0] * 26 ms2 = [0] * 26 ms3 = [0] * 26 ms4 = [0] * 26 m = 0 n = 0 for i in range(len(s1)): if ord(s1[i]) in range(ord("A"), ord("Z") + 1): ms1[ord(s1[i]) - ord("A")] += 1 else: ms2[ord(s1[i]) - ord("a")] += 1 for i in range(len(s2)): if ord(s2[i]) in range(ord("A"), ord("Z") + 1): ms3[ord(s2[i]) - ord("A")] += 1 else: ms4[ord(s2[i]) - ord("a")] += 1 for i in range(len(ms1)): if ms3[i] and ms1[i]: mn = min(ms3[i], ms1[i]) m += mn ms3[i] -= mn ms1[i] -= mn if ms2[i] and ms4[i]: mn = min(ms2[i], ms4[i]) m += mn ms2[i] -= mn ms4[i] -= mn for i in range(len(ms1)): if ms3[i] and ms2[i]: mn = min(ms2[i], ms3[i]) n += mn ms3[i] -= mn ms2[i] -= mn if ms1[i] and ms4[i]: mn = min(ms1[i], ms4[i]) n += mn ms1[i] -= mn ms4[i] -= mn print(m, n)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def main(): s, t = input(), input() mappingS = list(s) mappingT = list(t) yay = 0 whoops = 0 offset = ord("A") - ord("a") for i in range(ord("a"), ord("z") + 1, 1): sLower = mappingS.count(chr(i)) tLower = mappingT.count(chr(i)) sUpper = mappingS.count(chr(i + offset)) tUpper = mappingT.count(chr(i + offset)) contribution = min(sLower, tLower) + min(sUpper, tUpper) yay += contribution whoops += min(tLower + tUpper, sLower + sUpper) - contribution print("{0} {1}".format(yay, whoops)) main()
FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR STRING FUNC_CALL VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR STRING BIN_OP FUNC_CALL VAR STRING NUMBER NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def tanya(s, t): def invert(ch): if ch.isupper(): return ch.lower() return ch.upper() yay = 0 whoops = 0 S = {} T = {} for x in s: if x in S: S[x] += 1 else: S[x] = 1 for x in t: if x in T: T[x] += 1 else: T[x] = 1 for x in S: if x in T: if T[x] < S[x]: yay += T[x] S[x] -= T[x] T[x] = 0 else: yay += S[x] temp = S[x] S[x] = 0 T[x] = T[x] - temp if S[x] < 0: S[x] = 0 for x in S: if invert(x) in T: if T[invert(x)] < S[x]: whoops += T[invert(x)] else: whoops += S[x] print(yay, whoops) a = input().strip() b = input().strip() tanya(a, b)
FUNC_DEF FUNC_DEF IF FUNC_CALL VAR RETURN FUNC_CALL VAR RETURN FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR IF VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR VAR IF VAR FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() occur = [(0) for i in range(len(s))] t_freq = {} for i in t: try: t_freq[i] += 1 except: t_freq[i] = 1 y_count = 0 w_count = 0 for i in range(len(s)): try: if t_freq[s[i]] > 0: y_count += 1 t_freq[s[i]] -= 1 occur[i] = 1 except: pass for i in range(len(s)): try: if s[i].islower() and t_freq[s[i].upper()] and not occur[i]: w_count += 1 t_freq[s[i].upper()] -= 1 except: pass try: if s[i].isupper() and t_freq[s[i].lower()] and not occur[i]: w_count += 1 t_freq[s[i].lower()] -= 1 except: pass print(y_count, w_count)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR DICT FOR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() dict_t = dict() for letter in t: dict_t[letter] = dict_t.get(letter, 0) + 1 yay = 0 whoops = 0 remaining = "" for letter in s: if letter in dict_t and dict_t[letter] > 0: yay += 1 dict_t[letter] -= 1 else: remaining += letter for letter in remaining: if letter.isupper() and letter.lower() in dict_t and dict_t[letter.lower()] > 0: whoops += 1 dict_t[letter.lower()] -= 1 if letter.islower() and letter.upper() in dict_t and dict_t[letter.upper()] > 0: whoops += 1 dict_t[letter.upper()] -= 1 print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR VAR FOR VAR VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() M = {} for c in t: if c not in M: M[c] = 0 M[c] += 1 marked = [False] * len(s) yay = 0 whoops = 0 for i in range(len(s)): c = s[i] if marked[i]: continue if c in M and M[c] > 0: yay += 1 M[c] -= 1 marked[i] = True for i in range(len(s)): c = s[i] if c.islower(): c = c.upper() else: c = c.lower() if marked[i]: continue if c in M and M[c] > 0: whoops += 1 M[c] -= 1 marked[i] = True print("{} {}".format(yay, whoops))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER ASSIGN VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def getIndex(ch): if ord("Z") >= ord(ch) >= ord("A"): return ord(ch) - ord("A") else: return 26 + ord(ch) - ord("a") s = input().strip() t = input().strip() l = [0] * 52 yay = 0 whoop = 0 ls = [0] * len(s) for i in range(len(t)): l[getIndex(t[i])] += 1 for i in range(len(s)): if l[getIndex(s[i])] > 0 and ls[i] > -1: yay += 1 l[getIndex(s[i])] -= 1 ls[i] = -1 for i in range(len(s)): if ls[i] > -1: tmp = "" if ord("Z") >= ord(s[i]) >= ord("A"): tmp = s[i].lower() else: tmp = s[i].upper() if l[getIndex(tmp)] > 0: whoop += 1 l[getIndex(tmp)] -= 1 print("{0} {1}".format(yay, whoop))
FUNC_DEF IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR FUNC_CALL VAR STRING RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING RETURN BIN_OP BIN_OP NUMBER FUNC_CALL VAR VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER ASSIGN VAR STRING IF FUNC_CALL VAR STRING FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() yay = 0 whoops = 0 dic = {} resto = "" def inverter(c): if c.isupper(): return c.lower() else: return c.upper() for c in t: try: dic[c] += 1 except: dic[c] = 1 for c in s: try: if dic[c] > 0: dic[c] -= 1 yay += 1 else: resto += c except: resto += c for c in resto: cInvertido = inverter(c) try: if dic[cInvertido] > 0: dic[cInvertido] -= 1 whoops += 1 except: continue print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR DICT ASSIGN VAR STRING FUNC_DEF IF FUNC_CALL VAR RETURN FUNC_CALL VAR RETURN FUNC_CALL VAR FOR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER VAR VAR VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
alpha = [ "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", ] dict_s = {} dict_news = {} s = list(input()) newspaper = list(input()) for i in alpha: dict_s.update({i: s.count(i)}) dict_news.update({i: newspaper.count(i)}) ny = 0 for k in alpha: if dict_s[k] <= dict_news[k]: ny += dict_s[k] dict_news[k] -= dict_s[k] dict_s[k] = 0 else: ny += dict_news[k] dict_s[k] -= dict_news[k] dict_news[k] = 0 nw = 0 for k in alpha: if k.isupper() and dict_s[k] != 0: kl = k.lower() nw += min([dict_s[k], dict_news[kl]]) if k.islower() and dict_s[k] != 0: ku = k.upper() nw += min([dict_s[k], dict_news[ku]]) print(ny, nw)
ASSIGN VAR LIST STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING STRING ASSIGN VAR DICT ASSIGN VAR DICT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR DICT VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR DICT VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL VAR LIST VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s1 = input() s2 = input() lista1 = [0] * 150 lista2 = [0] * 150 for i in range(len(s1)): lista1[ord(s1[i]) - ord("A")] += 1 for i in range(len(s2)): lista2[ord(s2[i]) - ord("A")] += 1 ans1 = 0 for i in range(100): matchs = min(lista1[i], lista2[i]) ans1 += matchs lista1[i] -= matchs lista2[i] -= matchs ans2 = 0 for i in range(100): ans2 += min(lista1[i], lista2[i + 32]) ans2 += min(lista1[i + 32], lista2[i]) print(ans1, ans2)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR STRING NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() cnt = [0] * 300 used = [0] * len(s) for c in t: cnt[ord(c)] += 1 c1 = 0 c2 = 0 for i in range(len(s)): c = s[i] if cnt[ord(c)] > 0: c1 += 1 cnt[ord(c)] -= 1 used[i] = 1 for i in range(len(s)): c = s[i] if used[i] == 0 and cnt[ord(c) ^ 32] > 0: c2 += 1 cnt[ord(c) ^ 32] -= 1 print(c1, c2)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR FOR VAR VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR IF VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s, t = input(), input() count = dict() for c in t + s: count[c.upper()] = 0 count[c.lower()] = 0 for c in t: count[c] += 1 yay, woops = 0, 0 temp = "" for c in s: if count[c] > 0: yay += 1 count[c] -= 1 else: temp += c for c in temp: if c == c.upper() and count[c.lower()] > 0: woops += 1 count[c.lower()] -= 1 elif c == c.lower() and count[c.upper()] > 0: woops += 1 count[c.upper()] -= 1 print(yay, woops)
ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FOR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR NUMBER ASSIGN VAR FUNC_CALL VAR NUMBER FOR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR STRING FOR VAR VAR IF VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER VAR VAR FOR VAR VAR IF VAR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR NUMBER IF VAR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
alpha = "abcdefghijklmnopqrstuvwxyz" alpha += alpha.upper() h1 = {i: (0) for i in alpha} h2 = {i: (0) for i in alpha} for i in input(""): h1[i] += 1 for i in input(""): h2[i] += 1 y = w = 0 for i in alpha: y += min(h1[i], h2[i]) h1[i], h2[i] = max(0, h1[i] - h2[i]), max(0, h2[i] - h1[i]) for i in alpha: j = i.upper() if i.islower() else i.lower() w += min(h1[i], h2[j]) h1[i], h2[j] = max(0, h1[i] - h2[j]), max(0, h2[j] - h1[i]) print("{} {}".format(y, w))
ASSIGN VAR STRING VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER VAR VAR ASSIGN VAR VAR NUMBER VAR VAR FOR VAR FUNC_CALL VAR STRING VAR VAR NUMBER FOR VAR FUNC_CALL VAR STRING VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR FUNC_CALL VAR NUMBER BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() n = len(s) yay = 0 whoops = 0 dict = {} for c in s: if c in dict: dict[c] += 1 else: dict[c] = 1 t2 = "" for c in t: if c in dict: yay += 1 dict[c] -= 1 if dict[c] == 0: del dict[c] else: t2 += c for c in t2: if c.islower(): c = c.upper() else: c = c.lower() if c in dict: whoops += 1 dict[c] -= 1 if dict[c] == 0: del dict[c] print("{0} {1}".format(yay, whoops))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR DICT FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR STRING FOR VAR VAR IF VAR VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR VAR VAR FOR VAR VAR IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR NUMBER VAR VAR NUMBER IF VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
message = input() newspaper = input() dict_letter = {} dict_letterBoolean = [] yay = 0 whoops = 0 for l in message: if l in dict_letter: dict_letter[l] = dict_letter[l] + 1 else: dict_letter[l] = 1 for l in newspaper: if l in dict_letter and dict_letter[l] > 0: dict_letter[l] = dict_letter[l] - 1 yay = yay + 1 dict_letterBoolean.append(True) else: dict_letterBoolean.append(False) for i in range(len(newspaper)): l = newspaper[i] lSwapedcase = l.swapcase() if ( lSwapedcase in dict_letter and dict_letter[lSwapedcase] > 0 and not dict_letterBoolean[i] ): dict_letter[lSwapedcase] = dict_letter[lSwapedcase] - 1 whoops = whoops + 1 print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT ASSIGN VAR LIST ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() def id(c): if "a" <= c <= "z": return ord(c) - ord("a") return ord(c) - ord("A") + 26 cnt = [0] * 52 Free = [0] * len(s) yay, whoops = 0, 0 for i in t: cnt[id(i)] += 1 for i in range(len(s)): if cnt[id(s[i])] > 0: yay += 1 cnt[id(s[i])] -= 1 Free[i] = 1 for i in range(len(s)): if Free[i] == 0: if cnt[id(s[i].upper())] > 0: whoops += 1 cnt[id(s[i].upper())] -= 1 if cnt[id(s[i].lower())] > 0: whoops += 1 cnt[id(s[i].lower())] -= 1 print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF IF STRING VAR STRING RETURN BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING RETURN BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING NUMBER ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR BIN_OP LIST NUMBER FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER IF VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR NUMBER VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() a, b = {}, [] for i in t: if i in a: a[i] += 1 else: a[i] = 1 for i in s: b.append(i) res1, res2 = 0, 0 for i in range(len(b)): if b[i] in a and a[b[i]] > 0: res1 += 1 a[b[i]] -= 1 b[i] = "0" for i in b: t = i if i.islower(): t = i.upper() elif i.isupper(): t = i.lower() if t in a and a[t] > 0: res2 += 1 a[t] -= 1 print(res1, res2)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR DICT LIST FOR VAR VAR IF VAR VAR VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR STRING FOR VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR VAR NUMBER VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() def count(s): d = {} for c in s: d[c] = d.get(c, 0) + 1 return d def compare(h1, h2): y = 0 for key, val in h2.items(): if key in h1: y += min(h1[key], val) return y dt = count(t) ds = count(s) maxY = compare(dt, ds) dt = count(t.lower()) ds = count(s.lower()) temp = compare(dt, ds) maxW = temp - maxY print(maxY, maxW)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN VAR FUNC_DEF ASSIGN VAR NUMBER FOR VAR VAR FUNC_CALL VAR IF VAR VAR VAR FUNC_CALL VAR VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
msg = input() nsp = input() yay = 0 whoops = 0 no_yay = [] nsp_l = {} for l in set(nsp): nsp_l[l] = nsp.count(l) for c in msg: if c in nsp_l.keys() and nsp_l[c] > 0: nsp_l[c] -= 1 yay += 1 else: no_yay.append(c) for c in no_yay: if c.islower(): if c.upper() in nsp_l.keys() and nsp_l[c.upper()] > 0: nsp_l[c.upper()] -= 1 whoops += 1 if c.isupper(): if c.lower() in nsp_l.keys() and nsp_l[c.lower()] > 0: nsp_l[c.lower()] -= 1 whoops += 1 print(yay, whoops)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST ASSIGN VAR DICT FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR FOR VAR VAR IF VAR FUNC_CALL VAR VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR IF FUNC_CALL VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR NUMBER IF FUNC_CALL VAR IF FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL VAR NUMBER VAR FUNC_CALL VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
desejadas = input() recortadas = input() YAY = 0 WHOOPS = 0 desejo = {} possuo = {} for rec in desejadas: if rec not in desejo: desejo[rec] = 1 else: desejo[rec] += 1 for rec in recortadas: if rec not in possuo: possuo[rec] = 1 else: possuo[rec] += 1 for chave in desejo: if chave in possuo: minimo = min(desejo[chave], possuo[chave]) desejo[chave] -= minimo possuo[chave] -= minimo YAY += minimo if len(desejo) < len(possuo): for chave in desejo: if chave.isupper(): chaveCase = chave.lower() else: chaveCase = chave.upper() if chaveCase in possuo: minimo = min(desejo[chave], possuo[chaveCase]) desejo[chave] -= minimo possuo[chaveCase] -= minimo WHOOPS += minimo else: for chave in possuo: if chave.isupper(): chaveCase = chave.lower() else: chaveCase = chave.upper() if chaveCase in desejo: minimo = min(desejo[chaveCase], possuo[chave]) desejo[chaveCase] -= minimo possuo[chave] -= minimo WHOOPS += minimo print(str(YAY) + " " + str(WHOOPS))
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR DICT ASSIGN VAR DICT FOR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR VAR NUMBER FOR VAR VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR FOR VAR VAR IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FOR VAR VAR IF FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR STRING FUNC_CALL VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def change(x): if x == x.upper(): return x.lower() if x == x.lower(): return x.upper() s, t = input(), input() L = [0] * 256 for i in t: L[ord(i)] += 1 ans1 = ans2 = 0 P = [] for i in s: x = ord(i) if L[x] > 0: L[x] -= 1 ans1 += 1 else: P.append(i) for i in P: x = ord(change(i)) if L[x] > 0: L[x] -= 1 ans2 += 1 print(ans1, ans2)
FUNC_DEF IF VAR FUNC_CALL VAR RETURN FUNC_CALL VAR IF VAR FUNC_CALL VAR RETURN FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR BIN_OP LIST NUMBER NUMBER FOR VAR VAR VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR LIST FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR NUMBER VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input() t = input() s_letters = dict() t_letters = dict() CAPS = "QWERTYUIOPASDFGHJKLZXCVBNM" SMALLS = "qwertyuiopasdfghjklzxcvbnm" for letter in CAPS: s_letters[letter] = 0 t_letters[letter] = 0 for letter in SMALLS: s_letters[letter] = 0 t_letters[letter] = 0 for letter in s: s_letters[letter] += 1 for letter in t: t_letters[letter] += 1 hurray = 0 hopa = 0 for i in range(26): Letter = CAPS[i] letter = SMALLS[i] tmp = min(s_letters[Letter], t_letters[Letter]) hurray += tmp s_letters[Letter] -= tmp t_letters[Letter] -= tmp tmp = min(s_letters[letter], t_letters[letter]) hurray += tmp s_letters[letter] -= tmp t_letters[letter] -= tmp hopa += min( s_letters[Letter] + s_letters[letter], t_letters[Letter] + t_letters[letter] ) print(hurray, hopa)
ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR STRING ASSIGN VAR STRING FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR VAR VAR NUMBER FOR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR ASSIGN VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR VAR FUNC_CALL VAR BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
def foo(): s = input() d = {} for key in s: d[key] = d.get(key, 0) + 1 return d def f(s): d = {} for key in s.keys(): if ord(key) > ord("Z"): c = chr(ord(key) - ord("a") + ord("A")) d[c] = d.get(c, 0) + s[key] else: d[key] = d.get(key, 0) + s[key] return d s = foo() t = foo() yr = 0 op = 0 for key in s.keys(): element = min(s[key], t.get(key, 0)) yr += element s[key] -= element if t.get(key): t[key] -= element s = f(s) t = f(t) for key in s.keys(): element = min(s[key], t.get(key, 0)) op += element s[key] -= element if t.get(key): t[key] -= element print(yr, op)
FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR DICT FOR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER RETURN VAR FUNC_DEF ASSIGN VAR DICT FOR VAR FUNC_CALL VAR IF FUNC_CALL VAR VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR STRING FUNC_CALL VAR STRING ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR NUMBER VAR VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Tanya decided to present her dad a postcard on his Birthday. She has already created a message — string s of length n, consisting of uppercase and lowercase English letters. Tanya can't write yet, so she found a newspaper and decided to cut out the letters and glue them into the postcard to achieve string s. The newspaper contains string t, consisting of uppercase and lowercase English letters. We know that the length of string t greater or equal to the length of the string s. The newspaper may possibly have too few of some letters needed to make the text and too many of some other letters. That's why Tanya wants to cut some n letters out of the newspaper and make a message of length exactly n, so that it looked as much as possible like s. If the letter in some position has correct value and correct letter case (in the string s and in the string that Tanya will make), then she shouts joyfully "YAY!", and if the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS". Tanya wants to make such message that lets her shout "YAY!" as much as possible. If there are multiple ways to do this, then her second priority is to maximize the number of times she says "WHOOPS". Your task is to help Tanya make the message. -----Input----- The first line contains line s (1 ≤ |s| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text of Tanya's message. The second line contains line t (|s| ≤ |t| ≤ 2·10^5), consisting of uppercase and lowercase English letters — the text written in the newspaper. Here |a| means the length of the string a. -----Output----- Print two integers separated by a space: the first number is the number of times Tanya shouts "YAY!" while making the message, the second number is the number of times Tanya says "WHOOPS" while making the message. -----Examples----- Input AbC DCbA Output 3 0 Input ABC abc Output 0 3 Input abacaba AbaCaBA Output 3 4
s = input().strip() t = input().strip() ls = len(s) lt = len(t) d = {} yays = 0 whoops = 0 missed = [] for c in t: if c not in d: d[c] = 0 d[c] += 1 for c in s: if c in d and d[c]: yays += 1 d[c] -= 1 else: missed.append(c) for c in missed: k = c.swapcase() if k in d and d[k]: whoops += 1 d[k] -= 1 print("{0} {1}".format(yays, whoops))
ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR DICT ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST FOR VAR VAR IF VAR VAR ASSIGN VAR VAR NUMBER VAR VAR NUMBER FOR VAR VAR IF VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR VAR FOR VAR VAR ASSIGN VAR FUNC_CALL VAR IF VAR VAR VAR VAR VAR NUMBER VAR VAR NUMBER EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR