message stringlengths 2 11.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 137 108k | cluster float64 18 18 | __index_level_0__ int64 274 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc | instruction | 0 | 64,816 | 18 | 129,632 |
"Correct Solution:
```
s=input()
n=int(input())
for i in range(n):
l=list(input().split())
if l[0]=="print":
print(s[int(l[1]):int(l[2])+1])
elif l[0]=="reverse":
s=s[:int(l[1])]+s[int(l[1]):int(l[2])+1][::-1]+s[int(l[2])+1:]
elif l[0]=="replace":
s=s[:int(l[1])]+l[3]+s[int(l[2])+1:]
``` | output | 1 | 64,816 | 18 | 129,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
temp = input()
n = int(input())
for _ in range(n):
order,a,b,*value = input().split()
a = int(a)
b = int(b)
if order == "print":
print(temp[a:b+1])
elif order == "reverse":
rev = temp[a:b+1]
temp = temp[:a] + rev[::-1] + temp[b+1:]
else:#replace
temp = temp[:a] + value[0] + temp[b+1:]
``` | instruction | 0 | 64,817 | 18 | 129,634 |
Yes | output | 1 | 64,817 | 18 | 129,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
s = input()
n = int(input())
for _ in range(n):
li = input().split()
op = li[0]
a, b = map(int, li[1:3])
if op == "replace":
p = li[3]
s = s[:a] + p + s[b+1:]
elif op == "reverse":
s = s[:a] + s[a:b+1][::-1] + s[b+1:]
elif op == "print":
print(s[a:b+1])
``` | instruction | 0 | 64,818 | 18 | 129,636 |
Yes | output | 1 | 64,818 | 18 | 129,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
str = input()
q = int(input())
for i in range(q):
s = input().split(" ")
cmd,a,b = s[0],int(s[1]),int(s[2])
if(cmd=="print"):
print(str[a:b+1])
elif(cmd=="reverse"):
str = str[:a]+str[a:b+1][::-1]+str[b+1:]
else:
str = str[:a]+s[3]+str[b+1:]
``` | instruction | 0 | 64,819 | 18 | 129,638 |
Yes | output | 1 | 64,819 | 18 | 129,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
s = input()
query = int(input())
for i in range(query):
q = input().split()
if q[0] == 'print':
print(s[int(q[1]):int(q[2])+1])
elif q[0] == 'reverse':
s = s[0:int(q[1])] + s[int(q[1]):int(q[2])+1][::-1] + s[int(q[2])+1:]
else:
s = s[0:int(q[1])] + q[3] + s[int(q[2])+1:]
``` | instruction | 0 | 64,820 | 18 | 129,640 |
Yes | output | 1 | 64,820 | 18 | 129,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
t = input()
for i in range(int(input())):
cmd = list(input().split())
cmd[1] = int(cmd[1])
cmd[2] = int(cmd[2])
if cmd[0] == "print":
print(t[cmd[1]-1 : cmd[2]])
elif cmd[0] == "reverse":
tmp = t[cmd[1]-1 : cmd[2]]
tmp = tmp[::-1]
t = t[:cmd[1]-1] + tmp + t[cmd[2]:]
elif cmd[0] == "replace":
t = t[:cmd[1]-1] + cmd[3] + t[cmd[2]:]
``` | instruction | 0 | 64,821 | 18 | 129,642 |
No | output | 1 | 64,821 | 18 | 129,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
data = input()
q = int(input())
for _ in range(q):
op = list(map(str,input().split()))
a = int(op[1])
b = int(op[2]) + 1
if(op[0] == "print"):
print(data[a:b])
elif(op[0] == "reverse"):
data = data[:a] + data[a:b][::-1] + data[:b]
elif(op[0] == "replace"):
data = data[:a] + op[3] + data[b:]
``` | instruction | 0 | 64,822 | 18 | 129,644 |
No | output | 1 | 64,822 | 18 | 129,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
def op_print(data,a,b):
print(data[a:b])
def op_reverse(data,a,b):
return data[:a] + data[::-1][-b:-a] + data[:b]
def op_replace(data,a,b,q):
return data.replace(data[a:b],q)
data = input()
q = int(input())
for _ in range(q):
op = list(map(str,input().split()))
op[2] = str(int(op[2])+1)
if(op[0] == "print"):
op_print(data,int(op[1]),int(op[2]))
elif(op[0] == "reverse"):
data = op_reverse(data,int(op[1]),int(op[2]))
elif(op[0] == "replace"):
data = op_replace(data,int(op[1]),int(op[2]),op[3])
``` | instruction | 0 | 64,823 | 18 | 129,646 |
No | output | 1 | 64,823 | 18 | 129,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Transformation
Write a program which performs a sequence of commands to a given string $str$. The command is one of:
* print a b: print from the a-th character to the b-th character of $str$
* reverse a b: reverse from the a-th character to the b-th character of $str$
* replace a b p: replace from the a-th character to the b-th character of $str$ with p
Note that the indices of $str$ start with 0.
Constraints
* $1 \leq $ length of $str \leq 1000$
* $1 \leq q \leq 100$
* $0 \leq a \leq b < $ length of $str$
* for replace command, $b - a + 1 = $ length of $p$
Input
In the first line, a string $str$ is given. $str$ consists of lowercase letters. In the second line, the number of commands q is given. In the next q lines, each command is given in the above mentioned format.
Output
For each print command, print a string in a line.
Examples
Input
abcde
3
replace 1 3 xyz
reverse 0 2
print 1 4
Output
xaze
Input
xyz
3
print 0 2
replace 0 2 abc
print 0 2
Output
xyz
abc
Submitted Solution:
```
def op_print(data,a,b):
print(data[a:b])
def op_reverse(data,a,b):
return data[:a] + data[a:b][::-1] + data[:b]
def op_replace(data,a,b,q):
return data[:a] + q + data[b:]
data = input()
q = int(input())
for _ in range(q):
op = list(map(str,input().split()))
op[2] = str(int(op[2])+1)
if(op[0] == "print"):
op_print(data,int(op[1]),int(op[2]))
elif(op[0] == "reverse"):
data = op_reverse(data,int(op[1]),int(op[2]))
elif(op[0] == "replace"):
data = op_replace(data,int(op[1]),int(op[2]),op[3])
``` | instruction | 0 | 64,824 | 18 | 129,648 |
No | output | 1 | 64,824 | 18 | 129,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
m={a:0 for a in 'qwertyuiopasdfghjklzxcvbnmQWERTYUIOPLKJHGFDSAZXCVBNM'}
m1={a:0 for a in 'qwertyuiopasdfghjklzxcvbnmQWERTYUIOPLKJHGFDSAZXCVBNM'}
t,s=input(),input()
for i in range(len(t)):
m[t[i]]+=1
for i in range(len(s)):
m1[s[i]]+=1
ura=0
opa=0
for a in 'qwertyuiopasdfghjklzxcvbnmQWERTYUIOPLKJHGFDSAZXCVBNM':
if m[a]==0 or m1[a]==0:
continue
temple=min(m[a],m1[a])
#print(temple,a,m[a],m1[a])
ura+=temple
m[a]-=temple
m1[a]-=temple
print(ura,end=' ',sep='')
for a in 'qwertyuiopasdfghjklzxcvbnm':
opa+=min(m[a]+m[a.upper()],m1[a]+m1[a.upper()])
print(opa)
``` | instruction | 0 | 66,000 | 18 | 132,000 |
Yes | output | 1 | 66,000 | 18 | 132,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
from collections import Counter
def main():
letter = Counter(input())
journal = Counter(input())
a = b = 0
for k in letter:
d = min(letter[k], journal[k])
letter[k] -= d
journal[k] -= d
a += d
for k in letter:
c = k.swapcase()
d = min(letter[k], journal[c])
letter[k] -= d
journal[c] -= d
b += d
print(a, b)
if __name__ == "__main__":
main()
``` | instruction | 0 | 66,001 | 18 | 132,002 |
Yes | output | 1 | 66,001 | 18 | 132,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
# print ("Enter the desired text")
text = input()
# print ("Enter the letters available")
letters = input()
d = {}
# Create the dictionary (the clever way!)
for ch in letters:
d[ch] = d.get(ch, 0) + 1
# Make list of ones not to touch again the second time through
a = []
for ch in text:
a.append(False)
yaycount = 0
whoopscount = 0
# First go through and maximize YAYs
for i in range(len(text)):
ch = text[i]
if (ch in d) and (d[ch] > 0): # An exact match--add one to YAY and adjust dict
yaycount = yaycount + 1
d[ch] = d[ch] - 1
a[i] = True
# Then go through again and count WHOOPS
for i in range(len(text)):
ch = text[i]
if a[i] == False:
if ch.lower() == ch: # It's lower case--check for upper case match
upper = ch.upper()
if (upper in d) and (d[upper] > 0):
whoopscount = whoopscount + 1
d[upper] = d[upper] - 1
elif ch.upper() == ch:
lower = ch.lower()
if (lower in d) and (d[lower] > 0):
whoopscount = whoopscount + 1
d[lower] = d[lower] - 1
print (str(yaycount) + " " + str(whoopscount))
``` | instruction | 0 | 66,002 | 18 | 132,004 |
Yes | output | 1 | 66,002 | 18 | 132,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
def process(str):
d = {}
for c in str:
d[c] = d.setdefault(c, 0) + 1
return d
def match(d1, d2, case):
ans = 0
for c in d1:
if case(c) not in d2:
continue
m = min(d1[c], d2[case(c)])
ans += m
d1[c] -= m
d2[case(c)] -= m
return ans
sd = process(input())
td = process(input())
yay = match(sd, td, lambda x : x)
whoops = match(sd, td, lambda x : x.lower() if x.isupper() else x.upper())
print(yay, whoops)
``` | instruction | 0 | 66,003 | 18 | 132,006 |
Yes | output | 1 | 66,003 | 18 | 132,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
s = str(input())
t = str(input())
y, w = 0, 0
def oposto(caractere):
if caractere.isupper():
return caractere.lower()
else:
return caractere.upper()
hashmap = {}
hashcomp = {}
for i in t:
if i in hashmap:
hashmap[i] += 1
else:
hashmap[i] = 1
if i.isupper():
if i.lower() in hashcomp:
hashcomp[i.lower()] += 1
else:
hashcomp[i.lower()] = 1
else:
if i.upper() in hashcomp:
hashcomp[i.upper()] += 1
else:
hashcomp[i.upper()] = 1
for i in s:
if i in hashmap and hashmap[i] > 0:
y += 1
hashmap[i] -= 1
hashcomp[oposto(i)] -= 1
elif i in hashcomp and hashcomp[i] > 0:
w += 1
hashcomp[i] -= 1
hashmap[oposto(i)] -= 1
print(str(y) + " " + str(w))
# for i in s:
# if i in hashmap:
# if hashmap[i] > 0:
# hashmap[i.upper()] -= 1
# hashmap[i.lower()] -= 1
# y += 1
# elif i not in hashmap
``` | instruction | 0 | 66,004 | 18 | 132,008 |
No | output | 1 | 66,004 | 18 | 132,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
import string
s = input()
t = input()
dictT = {}
for c in string.ascii_letters:
dictT[c] = 0
for c in t:
dictT[c]+=1
yay = 0
whoops = 0
for i in range(len(s)):
c = s[i]
if dictT[c] > 0:
yay+=1
dictT[c]-=1
elif (c.islower() and c.upper() == t[i]) or (c.isupper() and c.lower() == t[i]):
whoops+=1
print(yay, whoops)
``` | instruction | 0 | 66,005 | 18 | 132,010 |
No | output | 1 | 66,005 | 18 | 132,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
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
print('{0} {1}'.format(yay, whoops))
``` | instruction | 0 | 66,006 | 18 | 132,012 |
No | output | 1 | 66,006 | 18 | 132,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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| ≤ 2·105), consisting of uppercase and lowercase English letters — the text of Tanya's message.
The second line contains line t (|s| ≤ |t| ≤ 2·105), 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
Submitted Solution:
```
from collections import Counter
def main():
letter = Counter(input())
journal = Counter(input())
a = b = 0
for k in letter:
d = min(letter[k], journal[k])
letter[k] -= d
journal[k] -= d
a += d
for k in letter:
c = k.swapcase()
d = min(letter[k], journal[c])
letter[k] -= d
journal[k] -= d
b += d
print(a, b)
if __name__ == "__main__":
main()
``` | instruction | 0 | 66,007 | 18 | 132,014 |
No | output | 1 | 66,007 | 18 | 132,015 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures | instruction | 0 | 66,372 | 18 | 132,744 |
"Correct Solution:
```
w=input().split()
print(max(w,key=w.count),max(w,key=len))
``` | output | 1 | 66,372 | 18 | 132,745 |
Provide a correct Python 3 solution for this coding contest problem.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures | instruction | 0 | 66,376 | 18 | 132,752 |
"Correct Solution:
```
a = input()
a = a.split()
ans1 = max(a,key=a.count)
ans2 = max(a,key=len)
print(ans1,ans2)
``` | output | 1 | 66,376 | 18 | 132,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
date = list(map(str,input().split()))
Num = 0
Size = 0
for tag in sorted(set(date), key=date.index):
if date.count(tag) > Num:
Num = date.count(tag)
Ooi = tag
if len(tag) > Size:
Size = len(tag)
Msiz = tag
print(Ooi,Msiz)
``` | instruction | 0 | 66,378 | 18 | 132,756 |
Yes | output | 1 | 66,378 | 18 | 132,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
import statistics
m = list(map(str,input().split()));
mode = statistics.mode(m)
a = "";
for i in range(0,len(m)):
if len(a) < len(m[i]):
a = m[i];
print(mode,a);
``` | instruction | 0 | 66,379 | 18 | 132,758 |
Yes | output | 1 | 66,379 | 18 | 132,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
from collections import Counter
word = list(input().split())
count = Counter(word)
x,y = 0,0
for i in word:
if x<len(i):
x=len(i)
y=i
print(count.most_common()[0][0],y)
``` | instruction | 0 | 66,380 | 18 | 132,760 |
Yes | output | 1 | 66,380 | 18 | 132,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
import collections
d = input()
D = d.split()
c = collections.Counter(D)
print(c.most_common()[0][0], max(D,key=len))
``` | instruction | 0 | 66,381 | 18 | 132,762 |
Yes | output | 1 | 66,381 | 18 | 132,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
import sys
def main():
input_line = []
# with open('test_data') as f:
# for line in f.readline().split():
# input_line.append(line)
for line in sys.stdin.readline().split():
input_line.append(line)
#print(input_line)
if __name__ == '__main__':
main()
``` | instruction | 0 | 66,382 | 18 | 132,764 |
No | output | 1 | 66,382 | 18 | 132,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
import sys
f = sys.stdin
rhombus = oblong = 0
for line in f:
a, b, c = map(int, line.split(','))
if a == b:
rhombus += 1
if a * a + b * b == c * c:
oblong += 1
print(oblong)
print(rhombus)
``` | instruction | 0 | 66,383 | 18 | 132,766 |
No | output | 1 | 66,383 | 18 | 132,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
s=input().split()
dict={}
max=""
max_num=0
max_index=""
for i in s:
if i in dict:
dict[i]+=1
if max_num<dict[i]:
max_index=i
else: dict[i]=0
if len(i)>len(max):
max=i
print(max_index,max)
``` | instruction | 0 | 66,384 | 18 | 132,768 |
No | output | 1 | 66,384 | 18 | 132,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your task is to write a program which reads a text and prints two words. The first one is the word which is arise most frequently in the text. The second one is the word which has the maximum number of letters.
The text includes only alphabetical characters and spaces. A word is a sequence of letters which is separated by the spaces.
Input
A text is given in a line. You can assume the following conditions:
* The number of letters in the text is less than or equal to 1000.
* The number of letters in a word is less than or equal to 32.
* There is only one word which is arise most frequently in given text.
* There is only one word which has the maximum number of letters in given text.
Output
The two words separated by a space.
Example
Input
Thank you for your mail and your lectures
Output
your lectures
Submitted Solution:
```
# -*- coding: utf-8 -*-
from collections import defaultdict
d = defaultdict(int)
_input = input()
sentence = [e for e in _input.split()]
l = []
for word in sentence:
d[word] += 1
times = sorted(d.items(), key=lambda item: item[1], reverse=True)
lengths = sorted(d.items(), key=lambda item: len(item[0]), reverse=True)
print(times[0][1], end=' ')
print(lengths[0][0])
``` | instruction | 0 | 66,385 | 18 | 132,770 |
No | output | 1 | 66,385 | 18 | 132,771 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,424 | 18 | 132,848 |
"Correct Solution:
```
class Node():
p = None
l = None
r = None
char = ''
count = 0
def __init__(self, char='', count=0):
self.char = char
self.count = count
def solve(n, h):
if n.char:
# print('{}:{}'.format(n.char, h))
return len(h) * n.count
else:
ret = 0
if n.l: ret += solve(n.l, h + '0')
if n.r: ret += solve(n.r, h + '1')
return ret
# --------------------
S = input()
C = {}
T = []
for s in S:
C.setdefault(s, 0)
C[s] += 1
T = [Node(*x) for x in C.items()]
T.sort(key=lambda x: x.count)
if len(T) == 1:
n = Node()
n.l = T.pop(0)
T.append(n)
else:
while len(T) != 1:
n = Node()
n.l = T.pop(0)
n.r = T.pop(0)
n.count = n.l.count + n.r.count
T.append(n)
T.sort(key=lambda x: x.count)
print(solve(T[0], ''))
``` | output | 1 | 66,424 | 18 | 132,849 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,425 | 18 | 132,850 |
"Correct Solution:
```
S = input()
dict_string = {}
dict_depth = {}
for s in S:
if s in dict_string:
dict_string[s] += 1
else:
dict_string[s] = 1
dict_depth[s] = 0
dict_string_calc = dict_string.copy()
while len(dict_string_calc) != 1:
x = None
y = None
for k, v in sorted(dict_string_calc.items(), key=lambda x: x[1]):
if x == None:
x = k
x_cnt = v
continue
if y == None:
y = k
y_cnt = v
break
for s in (x + y):
if s in dict_depth:
dict_depth[s] += 1
else:
raise
del dict_string_calc[x]
del dict_string_calc[y]
dict_string_calc[x + y] = x_cnt + y_cnt
ans = 0
for k, v in dict_depth.items():
ans += dict_string[k] * v
if len(dict_string) == 1:
ans = len(S)
print(ans)
``` | output | 1 | 66,425 | 18 | 132,851 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,426 | 18 | 132,852 |
"Correct Solution:
```
from heapq import *
class Node():
def __init__(self, key, char=None):
self.key = key
self.char = char
self.pare = None
st = input()
_st = list(set(st))
a = []
for i in _st:
a += [(st.count(i),i)]
heapify(a)
node = {}
for i in range(len(a)):
cnt, char = a[i]
node[char] = Node(cnt, char)
while len(a) > 1:
p1, p2 = heappop(a)
q1, q2 = heappop(a)
x = Node(p1+q1, p2+q2)
node[p2+q2] = x
if len(a) == 0:
root = x
break
node[p2].pare = x
node[q2].pare = x
heappush(a, (p1+q1,p2+q2))
def dfs(node, cnt=1):
if node.pare == None:
return cnt
a = dfs(node.pare, cnt+1)
return a
Sum = 0
for i in _st:
no = node[i]
depth = dfs(no)
Sum += depth*no.key
print(Sum)
``` | output | 1 | 66,426 | 18 | 132,853 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,427 | 18 | 132,854 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Greedy algorithms - Huffman Coding
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_15_D&lang=jp
"""
import sys
from collections import Counter
class Node:
def __init__(self, v, f):
self.val = v
self.freq = f
self.left = None
self.right = None
def solve(s):
res = [Node(k, v) for k, v in Counter(s).items()]
while len(res) > 1:
res.sort(key=lambda x: x.freq)
z = Node('@', res[0].freq + res[1].freq)
z.left, z.right = res[0], res[1]
res = res[2:]
res.append(z)
d = dict()
stack = [(res[0], '')]
while stack:
n, st = stack.pop()
if not n.left and not n.right:
st = '0' if not st else st
d[n.val] = st
if n.left:
stack.append((n.left, st+'0'))
if n.right:
stack.append((n.right, st+'1'))
return len(''.join([d[ch] for ch in s]))
def main(args):
s = input()
ans = solve(s)
print(ans)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 66,427 | 18 | 132,855 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,428 | 18 | 132,856 |
"Correct Solution:
```
# ハフマン符号
# https://onlinejudge.u-aizu.ac.jp/courses/lesson/1/ALDS1/15/ALDS1_15_D
from collections import Counter
import heapq
class Node:
def __init__(self, count, val=None, left=None, right=None):
self.val = val
self.count = count
self.left = left
self.right = right
def __lt__(self, other):
return self.count < other.count
class HuffmanCoding:
def __init__(self, s: str):
self.s = s
self.counts = Counter(s).most_common()
self.root = self.build_tree()
self.encode_table = self.gen_encode_table()
def build_tree(self):
hq = list()
for val, cnt in self.counts:
heapq.heappush(hq, Node(cnt, val))
while len(hq) > 1:
x = heapq.heappop(hq)
y = heapq.heappop(hq)
cnt = x.count + y.count
z = Node(cnt, None, y, x)
heapq.heappush(hq, z)
if hq:
return heapq.heappop(hq)
else:
return None
def gen_encode_table(self):
encode_table = {}
def traversal(node, key):
if node.val:
if key == "":
key = "0"
encode_table[node.val] = key
else:
if node.left:
traversal(node.left, key + "1")
if node.right:
traversal(node.right, key + "0")
key = ""
if self.root:
traversal(self.root, key)
return encode_table
def get_encode_size(self):
size = 0
for a in self.s:
size += len(self.encode_table[a])
return size
def gen_encode_bin(self):
enc = ""
for a in self.s:
enc += self.encode_table[a]
return enc
def main():
S = input().strip()
huffman = HuffmanCoding(S)
print(huffman.get_encode_size())
if __name__ == "__main__":
main()
``` | output | 1 | 66,428 | 18 | 132,857 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,429 | 18 | 132,858 |
"Correct Solution:
```
from collections import Counter
S=input()
freq=sorted(list(dict(Counter(S)).items()),key=lambda w:w[1])
length={chr(ord('a')+i):0 for i in range(26)}
while len(freq)>1:
minimum1=freq.pop(0)
minimum2=freq.pop(0)
for c in minimum1[0]+minimum2[0]:
length[c]=length[c]+1
freq.append((minimum1[0]+minimum2[0],minimum1[1]+minimum2[1]))
freq=sorted(freq,key=lambda w:w[1])
retval=0
for s in S:
retval+=max(length[s],1)
print(retval)
``` | output | 1 | 66,429 | 18 | 132,859 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,430 | 18 | 132,860 |
"Correct Solution:
```
import heapq
from collections import Counter
class HuffmanTreeNode:
def __init__(self, char, freq):
self.char = char
self.freq = freq
self.left = None
self.right = None
def __lt__(self, other):
return self.freq < other.freq
def huffman_code_helper(root, code='', codes={}):
if not root:
return codes
if not root.left and not root.right:
codes[root.char] = code if code else '0'
return codes
huffman_code_helper(root.left, code + '0', codes)
huffman_code_helper(root.right, code + '1', codes)
return codes
def huffman_code(text):
counter = Counter(text)
heap = []
for char, freq in counter.items():
heap.append(HuffmanTreeNode(char, freq))
heapq.heapify(heap)
while len(heap) > 1:
left, right = heapq.heappop(heap), heapq.heappop(heap)
node = HuffmanTreeNode('', left.freq + right.freq)
node.left, node.right = left, right
heapq.heappush(heap, node)
root = heap[0]
return huffman_code_helper(root)
def solution():
text = input()
codes = huffman_code(text)
encoded = []
for char in text:
encoded.append(codes[char])
print(len(''.join(encoded)))
solution()
``` | output | 1 | 66,430 | 18 | 132,861 |
Provide a correct Python 3 solution for this coding contest problem.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1 | instruction | 0 | 66,431 | 18 | 132,862 |
"Correct Solution:
```
from collections import Counter, deque
from heapq import heappush, heappop, heapify
C = Counter(input())
N = len(C)
if N == 1:
print(sum(C.values()))
exit(0)
G = [0] * (N * 2)
G[:N] = C
que = [(v, i) for i, v in enumerate(C.values())]
heapify(que)
cur = len(C)
while len(que) > 1:
v0, nd0 = heappop(que)
v1, nd1 = heappop(que)
G[cur] = (nd0, nd1)
heappush(que, (v0 + v1, cur))
cur += 1
root = que[0][1]
ans = 0
que = deque([(0, root)])
while que:
d, nd = que.popleft()
if nd < N:
ans += d * C[G[nd]]
continue
u, v = G[nd]
que.append((d + 1, u))
que.append((d + 1, v))
print(ans)
``` | output | 1 | 66,431 | 18 | 132,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1
Submitted Solution:
```
from typing import Dict
class Node():
left = None
right = None
parent = None
char = ''
value = 0
def __init__(self, char: str = '', value: int = 0) -> None:
self.char = char
self.value = value
def get_coded_str_len(node: Node, base_str: str) -> int:
if node.char:
# Return the length of the coded string.
return len(base_str) * node.value
else:
coded_str_len = 0
if node.left:
coded_str_len += get_coded_str_len(node.left, base_str + '0')
if node.right:
coded_str_len += get_coded_str_len(node.right, base_str + '1')
return coded_str_len
if __name__ == "__main__":
input_str = input()
char_dict: Dict[str, int] = {}
for s in input_str:
if s in char_dict:
char_dict[s] += 1
else:
char_dict[s] = 1
nodes = [Node(char=k, value=v) for k, v in char_dict.items()]
# Sort in the ascending order.
nodes.sort(key=lambda x: x.value)
if 1 == len(nodes):
node = Node()
node.left = nodes.pop(0) # type: ignore
nodes.append(node)
else:
while 1 != len(nodes):
node = Node()
node.left = nodes.pop(0) # type: ignore
node.right = nodes.pop(0) # type: ignore
node.value = node.left.value + node.right.value # type: ignore
nodes.append(node)
# Sort in the ascending order.
nodes.sort(key=lambda x: x.value)
print(f"{get_coded_str_len(nodes[0], '')}")
``` | instruction | 0 | 66,432 | 18 | 132,864 |
Yes | output | 1 | 66,432 | 18 | 132,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1
Submitted Solution:
```
import heapq
S = input()
# 数え上げ
chars = [[0, n] for n in range(256)]
for s in S:
chars[ord(s)][0] += 1
# ハフマン木生成
# ：ノードの構造：[count, char, left=None, right=None]
# ：[1]のcharは、heapqでユニークキーとして機能するように追加した
counts = []
for char in chars:
if not char[0]:
continue
heapq.heappush(counts, [char[0], char[1], None, None, char])
# ハフマン木編成
while 1 < len(counts):
a, b = heapq.heappop(counts), heapq.heappop(counts)
heapq.heappush(counts, [a[0] + b[0], -1, a, b])
root = heapq.heappop(counts)
# コード化(左：0、右：1)
codes = {};
def dfs(node, code):
# 節の処理
if node[1] < 0:
dfs(node[2], code + "0")
dfs(node[3], code + "1")
return
# 葉の処理
codes[chr(node[1])] = code
dfs(root, "")
# 集計
count = 0
for s in S:
count += len(codes[s])
# 1種類の場合のみ例外処理…いずれなんとか
if 1 == len(set(S)):
print(len(S))
else:
print(count)
``` | instruction | 0 | 66,433 | 18 | 132,866 |
Yes | output | 1 | 66,433 | 18 | 132,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1
Submitted Solution:
```
from heapq import heappop, heappush
from collections import Counter
class Node:
def __init__(self, weight):
self.weight = weight
def set_parent(self,parent):
self.parent = parent
def get_length(self):
try:
tmp = 1 + self.parent.get_length()
except AttributeError:
tmp = 0
return tmp
def get_weight(self):
return self.weight
def __lt__(self, other):
return self.weight < other.weight
S = input()
d = Counter(S)
tree_dic = {}
h = []
num = 0
for i in d:
tree_dic[i] = Node(d[i])
heappush(h,tree_dic[i])
while(len(h) > 1):
tmp0 = heappop(h)
tmp1 = heappop(h)
tree_dic[num] = Node(tmp0.get_weight() + tmp1.get_weight())
tmp0.set_parent(tree_dic[num])
tmp1.set_parent(tree_dic[num])
heappush(h,tree_dic[num])
num+=1
ans = 0
for i in S:
ans += tree_dic[i].get_length()
if len(S) == 1:
print(1)
elif len(d) == 1:
print(len(S))
else:
print(ans)
``` | instruction | 0 | 66,434 | 18 | 132,868 |
Yes | output | 1 | 66,434 | 18 | 132,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.
Huffman coding is known as one of ways to obtain a code table for such encoding.
For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:
| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100
We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.
The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.
<image>
To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:
1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.
Finally, we can find a "Huffman code" in a path from the root to leaves.
Task
Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.
Constraints
* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.
Input
$S$
A string $S$ is given in a line.
Output
Print the length of a binary string in a line.
Examples
Input
abca
Output
6
Input
aaabbcccdeeeffg
Output
41
Input
z
Output
1
Submitted Solution:
```
import sys, collections, heapq
input = lambda: sys.stdin.readline().rstrip()
sys.setrecursionlimit(10**7)
INF = 10**10
def I(): return int(input())
def F(): return float(input())
def SS(): return input()
def LI(): return [int(x) for x in input().split()]
def LI_(): return [int(x)-1 for x in input().split()]
def LF(): return [float(x) for x in input().split()]
def LSS(): return input().split()
def resolve():
S = SS()
# ハフマン木の構成
cnt = collections.Counter(S)
# (頻度, 文字, 左の子, 右の子)
freq = [(v, k, None, None) for k, v in cnt.items()]
heapq.heapify(freq)
while len(freq) >= 2:
l = heapq.heappop(freq)
r = heapq.heappop(freq)
heapq.heappush(freq, (l[0]+r[0], '', l, r))
root = heapq.heappop(freq)
# 符号長の導出
code_len = {}
for i in cnt.keys():
code_len[i] = 1
def dfs(node, depth):
if node[1]:
if depth == 0:
code_len[node[1]] = 1
else:
code_len[node[1]] = depth
else:
dfs(node[2], depth+1)
dfs(node[3], depth+1)
dfs(root, 0)
# 符号の長さの計算
# print(code_len)
ans = 0
for i in S:
ans += code_len[i]
print(ans)
if __name__ == '__main__':
resolve()
``` | instruction | 0 | 66,435 | 18 | 132,870 |
Yes | output | 1 | 66,435 | 18 | 132,871 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Here is a list of strings. Let's take a break and play with shiritori. Shiritori is performed according to the following rules.
1. First of all, select one of your favorite strings from the list and exclude that string from the list.
2. Next, select one character string from the list in which the last character of the previously selected character string is the first character.
3. Exclude the selected string from the list.
After this, steps 2 and 3 will be repeated alternately.
Now, let's assume that this shiritori is continued until there are no more strings to select from the list in 2. At this time, I want to list all the characters that can be the "last character" of the last selected character string.
Constraints
The input satisfies the following conditions.
* $ 1 \ le N \ lt 10 ^ 4 $
* $ 1 \ le | s_i | \ lt 100 $
* $ s_i \ neq s_j (i \ neq j) $
* The string contains only lowercase letters
Input
$ N $
$ s_1 $
$ s_2 $
$ s_3 $
::
$ s_N $
The number of strings $ N $ is given on the first line.
The following $ N $ line is given a list of strings.
Output
Output the characters that can be the "last character" of the last selected character string in ascending order from a to z in lexicographical order.
Examples
Input
5
you
ate
my
hum
toast
Output
e
t
u
Input
7
she
sells
sea
shells
by
the
seashore
Output
a
e
y | instruction | 0 | 67,217 | 18 | 134,434 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import Counter
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
class Dinic:
""" 最大流(Dinic) """
INF = 10 ** 18
def __init__(self, n):
self.n = n
self.links = [[] for _ in range(n)]
self.depth = None
self.progress = None
def add_link(self, _from, to, cap):
self.links[_from].append([cap, to, len(self.links[to])])
self.links[to].append([0, _from, len(self.links[_from]) - 1])
def bfs(self, s):
from collections import deque
depth = [-1] * self.n
depth[s] = 0
q = deque([s])
while q:
v = q.popleft()
for cap, to, _ in self.links[v]:
if cap > 0 and depth[to] < 0:
depth[to] = depth[v] + 1
q.append(to)
self.depth = depth
def dfs(self, v, t, flow):
if v == t:
return flow
links_v = self.links[v]
for i in range(self.progress[v], len(links_v)):
self.progress[v] = i
cap, to, rev = link = links_v[i]
if cap == 0 or self.depth[v] >= self.depth[to]:
continue
d = self.dfs(to, t, min(flow, cap))
if d == 0:
continue
link[0] -= d
self.links[to][rev][0] += d
return d
return 0
def max_flow(self, s, t):
INF = Dinic.INF
flow = 0
while True:
self.bfs(s)
if self.depth[t] < 0:
return flow
self.progress = [0] * self.n
current_flow = self.dfs(s, t, INF)
while current_flow > 0:
flow += current_flow
current_flow = self.dfs(s, t, INF)
N = INT()
A = [input() for i in range(N)]
M = 26
C = Counter()
outcnt = [0] * M
incnt = [0] * M
for s in A:
a, b = ord(s[0])-97, ord(s[-1])-97
# 辺(a, b)の容量、aから出る辺の数、bに入る辺の数
C[(a, b)] += 1
outcnt[a] += 1
incnt[b] += 1
dinic = Dinic(M*2)
for a in range(M):
# 各文字でin -> out間はINFで繋いでおく
dinic.add_link(a, M+a, INF)
for b in range(M):
# 文字aのout -> 文字bのin
dinic.add_link(M+a, b, C[(a, b)])
ans = []
for a in range(M):
# 文字aのoutからinに向かって流す
res = dinic.max_flow(M+a, a)
# aで終わる単語が1つ以上あって、aに戻ってこれる流量がaを出る辺の数以上にあればOK
if incnt[a] >= 1 and res >= outcnt[a]:
ans.append(chr(a+97))
[print(a) for a in ans]
``` | output | 1 | 67,217 | 18 | 134,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pushok the dog has been chasing Imp for a few hours already.
<image>
Fortunately, Imp knows that Pushok is afraid of a robot vacuum cleaner.
While moving, the robot generates a string t consisting of letters 's' and 'h', that produces a lot of noise. We define noise of string t as the number of occurrences of string "sh" as a subsequence in it, in other words, the number of such pairs (i, j), that i < j and <image> and <image>.
The robot is off at the moment. Imp knows that it has a sequence of strings ti in its memory, and he can arbitrary change their order. When the robot is started, it generates the string t as a concatenation of these strings in the given order. The noise of the resulting string equals the noise of this concatenation.
Help Imp to find the maximum noise he can achieve by changing the order of the strings.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of strings in robot's memory.
Next n lines contain the strings t1, t2, ..., tn, one per line. It is guaranteed that the strings are non-empty, contain only English letters 's' and 'h' and their total length does not exceed 105.
Output
Print a single integer — the maxumum possible noise Imp can achieve by changing the order of the strings.
Examples
Input
4
ssh
hs
s
hhhs
Output
18
Input
2
h
s
Output
1
Note
The optimal concatenation in the first sample is ssshhshhhs.
Submitted Solution:
```
import getpass
import sys
import math
import functools
def ria():
return [int(i) for i in input().split()]
files = True
if getpass.getuser() == 'frohenk' and files:
sys.stdin = open("test.in")
# sys.stdout = open('test.out', 'w')
n = ria()[0]
ar = []
for i in range(n):
st = input()
ar.append((st.count('s'), st.count('h'), st))
ar.sort(key=functools.cmp_to_key(lambda x, y: -x[0] * y[1] + x[1] * y[0]))
#print(ar)
fst = ''
for d, a, b in ar:
fst += b
suma = 0
kek = 0
for i in fst:
if i == 's':
kek += 1
else:
suma += kek
print(suma)
sys.stdout.close()
``` | instruction | 0 | 67,854 | 18 | 135,708 |
Yes | output | 1 | 67,854 | 18 | 135,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pushok the dog has been chasing Imp for a few hours already.
<image>
Fortunately, Imp knows that Pushok is afraid of a robot vacuum cleaner.
While moving, the robot generates a string t consisting of letters 's' and 'h', that produces a lot of noise. We define noise of string t as the number of occurrences of string "sh" as a subsequence in it, in other words, the number of such pairs (i, j), that i < j and <image> and <image>.
The robot is off at the moment. Imp knows that it has a sequence of strings ti in its memory, and he can arbitrary change their order. When the robot is started, it generates the string t as a concatenation of these strings in the given order. The noise of the resulting string equals the noise of this concatenation.
Help Imp to find the maximum noise he can achieve by changing the order of the strings.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of strings in robot's memory.
Next n lines contain the strings t1, t2, ..., tn, one per line. It is guaranteed that the strings are non-empty, contain only English letters 's' and 'h' and their total length does not exceed 105.
Output
Print a single integer — the maxumum possible noise Imp can achieve by changing the order of the strings.
Examples
Input
4
ssh
hs
s
hhhs
Output
18
Input
2
h
s
Output
1
Note
The optimal concatenation in the first sample is ssshhshhhs.
Submitted Solution:
```
import sys,os,io
from sys import stdin
from functools import cmp_to_key
from bisect import bisect_left , bisect_right
def ii():
return int(input())
def li():
return list(map(int,input().split()))
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
for _ in range(1):
n = ii()
l = []
ans=''
last=[]
for i in range(n):
s = list(input())
if s.count('s')==0:
last+=s
continue
l.append([s.count('h')/s.count('s'),s])
l.sort()
# print(l)
ans=[]
for i in l:
ans+=i[1]
ans+=list(last)
ans=''.join(ans)
s = ans[:]
pres = [0]*(len(s))
sufh = [0]*(len(s))
for i in range(len(s)):
if i>0:
pres[i]+=pres[i-1]
if s[i]=='s':
pres[i]+=1
for i in range(len(s)-1,-1,-1):
if i<len(s)-1:
sufh[i]+=sufh[i+1]
if s[i]=='h':
sufh[i]+=1
cnt = 0
for i in range(len(s)):
if s[i]=='s':
cnt+=sufh[i]
print(cnt)
``` | instruction | 0 | 67,856 | 18 | 135,712 |
Yes | output | 1 | 67,856 | 18 | 135,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pushok the dog has been chasing Imp for a few hours already.
<image>
Fortunately, Imp knows that Pushok is afraid of a robot vacuum cleaner.
While moving, the robot generates a string t consisting of letters 's' and 'h', that produces a lot of noise. We define noise of string t as the number of occurrences of string "sh" as a subsequence in it, in other words, the number of such pairs (i, j), that i < j and <image> and <image>.
The robot is off at the moment. Imp knows that it has a sequence of strings ti in its memory, and he can arbitrary change their order. When the robot is started, it generates the string t as a concatenation of these strings in the given order. The noise of the resulting string equals the noise of this concatenation.
Help Imp to find the maximum noise he can achieve by changing the order of the strings.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of strings in robot's memory.
Next n lines contain the strings t1, t2, ..., tn, one per line. It is guaranteed that the strings are non-empty, contain only English letters 's' and 'h' and their total length does not exceed 105.
Output
Print a single integer — the maxumum possible noise Imp can achieve by changing the order of the strings.
Examples
Input
4
ssh
hs
s
hhhs
Output
18
Input
2
h
s
Output
1
Note
The optimal concatenation in the first sample is ssshhshhhs.
Submitted Solution:
```
import itertools
n = int(input())
t = list()
for i in range(1,n+1):
t.append(input())
#print("Initial list = ",t)
def count(in_list: list):
word = ""
for i in range(len(in_list)):
word += in_list[i]
#print(word)
count = 0
for i in range(len(word)):
for j in range(len(word)):
if i < j and word[i] == "s" and word[j] == "h":
count +=1
return count
# def scoreword(inword:str):
# val = 0
# for i in range(len(inword)):
# if inword[i] == "s":
# if i == 0:
# val +=1
# elif inword[i-1] == "h":
# val += 1/2
# elif inword[i-1] == "s":
# val +=1
# val = val/(len(inword))
# #print("scoreword = ",val)
# return val
def scoreword(inword:str):
val = 0
for i in range(len(inword)):
if inword[i] == "h":
val +=1
return val
def bestorder(in_list: list):
scores = list()
for wrd in in_list:
score = scoreword(wrd)
scores.append(score)
print(scores)
orders = list(in_list)
counter = 0
while min(scores)<100000:
for i in range(len(scores)):
if scores[i] <= min(scores):
orders[i] = counter
counter +=1
scores[i] = 100000
if counter >= len(in_list):
break
#print(orders)
for i in range(len(in_list)):
scores[orders[i]] = in_list[i]
#print(scores)
return scores
print(count(bestorder(t)))
``` | instruction | 0 | 67,859 | 18 | 135,718 |
No | output | 1 | 67,859 | 18 | 135,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
print('YES' if input().count('1')+1>>1<<1 >= input().count('1') else 'NO')
``` | instruction | 0 | 68,437 | 18 | 136,874 |
Yes | output | 1 | 68,437 | 18 | 136,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
a = input()
b = input()
c = a.count("1")
if(c&1):
a+='1'
j = 0
err = 0
for i in range(len(b)):
if(b[i] == '1'):
while(True):
if(j>len(a)-1):
err = 1
break
if(a[j] == '1'):
j+=1
break
else:
j+=1
if(err):
break
if(err):
print("NO")
else:
print("YES")
``` | instruction | 0 | 68,438 | 18 | 136,876 |
Yes | output | 1 | 68,438 | 18 | 136,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
s1=input()
s2=input()
a1=s1.count('1')
a1+=(a1)%2
if a1>=s2.count('1'):print("YES")
else:print("NO")
``` | instruction | 0 | 68,439 | 18 | 136,878 |
Yes | output | 1 | 68,439 | 18 | 136,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
a, b = input(), input()
ca = a.count('1')
cb = b.count('1')
if ca + ca % 2 >= cb:
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,440 | 18 | 136,880 |
Yes | output | 1 | 68,440 | 18 | 136,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
a=input()
b=input()
count=0
for c in reversed(b):
if(c=='1'):
break
count+=1
if(b[:-count] in a):
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,441 | 18 | 136,882 |
No | output | 1 | 68,441 | 18 | 136,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
#python 3.6
a=str(input())
b=str(input())
tmp=""
kalimat=""
for i in range(0,len(a)):
if(a[i]=="1"):
if(kalimat!=""):
kalimat+=tmp;
tmp=""
kalimat+="1"
else:
tmp+="0"
if(a==""):
print("NO")
elif(a.find(kalimat)!=-1):
print("YES")
else:
print("NO")
``` | instruction | 0 | 68,442 | 18 | 136,884 |
No | output | 1 | 68,442 | 18 | 136,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def check(start):
if(b[:start].count('1')%2): ch=1
else: ch=0
# print(b[start],ch)
for i in b[start:]:
if(int(i)!=ch):
return False
ch^=1
return True
a=input()
b=input()
c=min(len(a),len(b))
for i in range(c,0,-1):
# print(b[:i],a[-i:],i)
if(b[:i]==a[-i:] and check(i)):
print("YES")
exit()
if(check(0)):
print("YES")
exit()
print("NO")
``` | instruction | 0 | 68,443 | 18 | 136,886 |
No | output | 1 | 68,443 | 18 | 136,887 |
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