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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, do you know "Yes, I have a number"?
Peter: I used to do it on TV the other day. You remember something by the number of characters in each word contained in a sentence. "Yes, I have a number", so it means "the number 3.14" and is a keyword for remembering pi.
Dr .: Peter, that's not the case. This should be interpreted as 3.1416. The circumference ratio is 3.14159 ... That's why.
Peter: Then, did you forcibly omit that program just because Japan teaches that the circumference ratio is 3.14? ??
Dr .: ... Let's just say that the pi taught in elementary school has finally returned from 3 to 3.14.
Peter: Do you really remember this in the first place?
Dr .: It may be difficult for Japanese people. It seems that English-speaking people use it, because it is difficult to make a ground ball in English.
Peter: Even so, it seems to be a hassle to check.
Dr .: Then, I want you to make a program that converts sentences into a sequence of the number of characters in a word.
Peter: I understand. Please tell me the detailed specifications.
Dr .: Let's enter one line of text. For simplicity, you can use sentences that contain only alphabets and blanks. For this sentence, output the length of the character string between the blank, the beginning of the sentence, and the end of the sentence in order. You can proceed if the number of characters in the string does not exceed 9.
For example, Yes in "Yes I have" has 3 characters because it is separated by the beginning and the first space, and I is separated by the first and second spaces and has 1 character.
Dr .: Then, don't forget the character string where the number of characters becomes 0 when there are consecutive blanks.
Input
Multiple datasets are given as input. For each dataset, a string containing alphabets and spaces is given on one line.
When the character string is "END OF INPUT", it is the end of input. Do not output to this input.
Output
For each dataset, output a sequence of the number of characters for the string on one line.
Example
Input
Yes I have a number
How I wish I could calculate an unused color for space
Thank you
END OF INPUT
Output
31416
31415926535
53
Submitted Solution:
```
while True:
dataset = input().split()
if dataset == ["END", "OF", "INPUT"]: break
print(''.join(map(str, [len(d) for d in dataset])))
``` | instruction | 0 | 15,894 | 18 | 31,788 |
No | output | 1 | 15,894 | 18 | 31,789 |
Provide a correct Python 3 solution for this coding contest problem.
H - RLE Replacement
Problem Statement
In JAG Kingdom, ICPC (Intentionally Compressible Programming Code) is one of the common programming languages. Programs in this language only contain uppercase English letters and the same letters often appear repeatedly in ICPC programs. Thus, programmers in JAG Kingdom prefer to compress ICPC programs by Run Length Encoding in order to manage very large-scale ICPC programs.
Run Length Encoding (RLE) is a string compression method such that each maximal sequence of the same letters is encoded by a pair of the letter and the length. For example, the string "RRRRLEEE" is represented as "R4L1E3" in RLE.
Now, you manage many ICPC programs encoded by RLE. You are developing an editor for ICPC programs encoded by RLE, and now you would like to implement a replacement function. Given three strings $A$, $B$, and $C$ that are encoded by RLE, your task is to implement a function replacing the first occurrence of the substring $B$ in $A$ with $C$, and outputting the edited string encoded by RLE. If $B$ does not occur in $A$, you must output $A$ encoded by RLE without changes.
Input
The input consists of three lines.
> $A$
> $B$
> $C$
The lines represent strings $A$, $B$, and $C$ that are encoded by RLE, respectively. Each of the lines has the following format:
> $c_1$ $l_1$ $c_2$ $l_2$ $\ldots$ $c_n$ $l_n$ \$
Each $c_i$ ($1 \leq i \leq n$) is an uppercase English letter (`A`-`Z`) and $l_i$ ($1 \leq i \leq n$, $1 \leq l_i \leq 10^8$) is an integer which represents the length of the repetition of $c_i$. The number $n$ of the pairs of a letter and an integer satisfies $1 \leq n \leq 10^3$. A terminal symbol `$` indicates the end of a string encoded by RLE. The letters and the integers are separated by a single space. It is guaranteed that $c_i \neq c_{i+1}$ holds for any $1 \leq i \leq n-1$.
Output
Replace the first occurrence of the substring $B$ in $A$ with $C$ if $B$ occurs in $A$, and output the string encoded by RLE. The output must have the following format:
> $c_1$ $l_1$ $c_2$ $l_2$ $\ldots$ $c_m$ $l_m$ \$
Here, $c_i \neq c_{i+1}$ for $1 \leq i \leq m-1$ and $l_i \gt 0$ for $1 \leq i \leq m$ must hold.
Sample Input 1
R 100 L 20 E 10 \$
R 5 L 10 \$
X 20 \$
Output for the Sample Input 1
R 95 X 20 L 10 E 10 \$
Sample Input 2
A 3 B 3 A 3 \$
A 1 B 3 A 1 \$
A 2 \$
Output for the Sample Input 2
A 6 \$
Example
Input
R 100 L 20 E 10 \$
R 5 L 10 \$
X 20 \$
Output
R 95 X 20 L 10 E 10 \$ | instruction | 0 | 15,900 | 18 | 31,800 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
def inp():
s = LS()
a = []
for i in range(len(s)-1):
if i % 2 == 0:
a.append(s[i])
else:
a.append(int(s[i]))
return a
def com(a):
r = a[:2]
for i in range(2,len(a), 2):
if a[i] == r[-2]:
r[-1] += a[i+1]
else:
r += a[i:i+2]
return r
a = com(inp())
b = com(inp())
c = com(inp())
r = []
ff = True
if len(b) == 2:
b0 = b[0]
b1 = b[1]
for i in range(0,len(a),2):
if a[i] == b0:
while a[i+1] >= b1 and ff:
r += c
a[i+1] -= b1
ff = False
if a[i+1] > 0:
r += a[i:i+2]
else:
r += a[i:i+2]
else:
i = 0
al = len(a)
bl = len(b)
be = bl - 2
while i < al:
f = True
for j in range(0,bl,2):
ii = i + j
if al <= ii or a[ii] != b[j] or (a[ii+1] < b[j+1] if j in [0, be] else a[ii+1] != b[j+1]) or not ff:
f = False
break
if f:
for j in range(0,bl,2):
ii = i + j
a[ii+1] -= b[j+1]
if a[i+1] > 0:
r += a[i:i+2]
if f:
r += c
ff = False
i += 2
r += '$'
return ' '.join(map(str,com(r)))
print(main())
``` | output | 1 | 15,900 | 18 | 31,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Remember to flush your output while communicating with the testing program. You may use fflush(stdout) in C++, system.out.flush() in Java, stdout.flush() in Python or flush(output) in Pascal to flush the output. If you use some other programming language, consult its documentation. You may also refer to the guide on interactive problems: <https://codeforces.com/blog/entry/45307>.
You are given a string t consisting of n lowercase Latin letters. This string was cyphered as follows: initially, the jury had a string s consisting of n lowercase Latin letters. Then they applied a sequence of no more than n (possibly zero) operations. i-th operation is denoted by two integers a_i and b_i (1 ≤ a_i, b_i ≤ n), and means swapping two elements of the string with indices a_i and b_i. All operations were done in the order they were placed in the sequence. For example, if s is xyz and 2 following operations are performed: a_1 = 1, b_1 = 2; a_2 = 2, b_2 = 3, then after the first operation the current string is yxz, and after the second operation the current string is yzx, so t is yzx.
You are asked to restore the original string s. Unfortunately, you have no information about the operations used in the algorithm (you don't even know if there were any operations in the sequence). But you may run the same sequence of operations on any string you want, provided that it contains only lowercase Latin letters and its length is n, and get the resulting string after those operations.
Can you guess the original string s asking the testing system to run the sequence of swaps no more than 3 times?
The string s and the sequence of swaps are fixed in each test; the interactor doesn't try to adapt the test to your solution.
Input
Initially the testing system sends one string t, consisting of lowercase Latin letters (1 ≤ |t| = n ≤ 10^4).
Output
To give the answer, your program should print one line ! s with a line break in the end. After that, it should flush the output and terminate gracefully.
Interaction
Before giving the answer, you may submit no more than 3 queries. To ask a query, print one line in the following format: ? s', where s' should be a string consisting of exaclty n lowercase Latin letters. The line should be ended with a line break character. After submitting a query, flush the output and read the answer to your query — a string t' consisting of n lowercase Latin letters, which is the result of applying the sequence of swaps to string s'. This string will be given on a separate line ended by a line break character.
If you submit an incorrect query (or ask more than 3 queries), the answer to it will be one string 0. After receiving such an answer, your program should terminate immediately — otherwise you may receive verdict "Runtime error", "Time limit exceeded" or some other verdict instead of "Wrong answer".
Example
Input
yzx
aab
baa
aba
Output
? baa
? aba
? aab
! xyz
Note
In the sample, the testcase described in the statement is used. The participant asks the first query with string baa, which is transformed to aab. The second query contains string aba, which is transformed to baa. The third query contains string aab, which is transformed to aba. The participant can deduce that the initial string s was xyz.
Note for hacking phase:
To submit a test in hacking phase, you should provide it in the following format:
The first line should contain the string s you guess, consisting of n ∈ [1, 10000] lowercase Latin letters.
The second line should contain k (0 ≤ k ≤ n) — the number of swap operations in the sequence.
Then k lines should follow, i-th of them should denote i-th operation with two integers a_i and b_i (1 ≤ a_i, b_i ≤ n).
For example, the sample test would look like that:
xyz
2
1 2
2 3
Submitted Solution:
```
from sys import stdout
s = input()
print('?', s)
stdout.flush()
t = input()
print('?', t)
stdout.flush()
q = input()
print('?', q)
stdout.flush()
j = input()
print('!', min(t, q, j))
``` | instruction | 0 | 16,000 | 18 | 32,000 |
No | output | 1 | 16,000 | 18 | 32,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Remember to flush your output while communicating with the testing program. You may use fflush(stdout) in C++, system.out.flush() in Java, stdout.flush() in Python or flush(output) in Pascal to flush the output. If you use some other programming language, consult its documentation. You may also refer to the guide on interactive problems: <https://codeforces.com/blog/entry/45307>.
You are given a string t consisting of n lowercase Latin letters. This string was cyphered as follows: initially, the jury had a string s consisting of n lowercase Latin letters. Then they applied a sequence of no more than n (possibly zero) operations. i-th operation is denoted by two integers a_i and b_i (1 ≤ a_i, b_i ≤ n), and means swapping two elements of the string with indices a_i and b_i. All operations were done in the order they were placed in the sequence. For example, if s is xyz and 2 following operations are performed: a_1 = 1, b_1 = 2; a_2 = 2, b_2 = 3, then after the first operation the current string is yxz, and after the second operation the current string is yzx, so t is yzx.
You are asked to restore the original string s. Unfortunately, you have no information about the operations used in the algorithm (you don't even know if there were any operations in the sequence). But you may run the same sequence of operations on any string you want, provided that it contains only lowercase Latin letters and its length is n, and get the resulting string after those operations.
Can you guess the original string s asking the testing system to run the sequence of swaps no more than 3 times?
The string s and the sequence of swaps are fixed in each test; the interactor doesn't try to adapt the test to your solution.
Input
Initially the testing system sends one string t, consisting of lowercase Latin letters (1 ≤ |t| = n ≤ 10^4).
Output
To give the answer, your program should print one line ! s with a line break in the end. After that, it should flush the output and terminate gracefully.
Interaction
Before giving the answer, you may submit no more than 3 queries. To ask a query, print one line in the following format: ? s', where s' should be a string consisting of exaclty n lowercase Latin letters. The line should be ended with a line break character. After submitting a query, flush the output and read the answer to your query — a string t' consisting of n lowercase Latin letters, which is the result of applying the sequence of swaps to string s'. This string will be given on a separate line ended by a line break character.
If you submit an incorrect query (or ask more than 3 queries), the answer to it will be one string 0. After receiving such an answer, your program should terminate immediately — otherwise you may receive verdict "Runtime error", "Time limit exceeded" or some other verdict instead of "Wrong answer".
Example
Input
yzx
aab
baa
aba
Output
? baa
? aba
? aab
! xyz
Note
In the sample, the testcase described in the statement is used. The participant asks the first query with string baa, which is transformed to aab. The second query contains string aba, which is transformed to baa. The third query contains string aab, which is transformed to aba. The participant can deduce that the initial string s was xyz.
Note for hacking phase:
To submit a test in hacking phase, you should provide it in the following format:
The first line should contain the string s you guess, consisting of n ∈ [1, 10000] lowercase Latin letters.
The second line should contain k (0 ≤ k ≤ n) — the number of swap operations in the sequence.
Then k lines should follow, i-th of them should denote i-th operation with two integers a_i and b_i (1 ≤ a_i, b_i ≤ n).
For example, the sample test would look like that:
xyz
2
1 2
2 3
Submitted Solution:
```
#import sys
#input=sys.stdin.readline
def bin(x):
l=[]
for k in range(3):
l.append(x%26)
x=x//26
l.reverse()
return l
t=input()
n=len(t)
a=[0 for i in range(n)]
b=[0 for i in range(n)]
c=[0 for i in range(n)]
for i in range(n):
l=bin(i)
a[i]=chr(l[0]+ord('a'))
b[i]=chr(l[1]+ord('a'))
c[i]=chr(l[2]+ord('a'))
if(ord(a[i])>ord('z')):
print(min([]))
if(ord(b[i])>ord('z')):
print(min([]))
if(ord(c[i])>ord('z')):
print(min([]))
print('?',end=" ")
for i in a:
print(i,end="")
if(ord(i)>ord('z')):
print(min([]))
print()
x=input()
print('?',end=" ")
for i in b:
print(i,end="")
if(ord(i)>ord('z')):
print(min([]))
print()
y=input()
print('?',end=" ")
for i in c:
print(i,end="")
if(ord(i)>ord('z')):
print(min([]))
print()
z=input()
s=[0 for i in range(n)]
for i in range(n):
p=ord(x[i])-ord('a')
q=ord(y[i])-ord('a')
r=ord(z[i])-ord('a')
s[576*p+26*q+r]=t[i]
print("!",end=" ")
for i in s:
print(i,end="")
``` | instruction | 0 | 16,001 | 18 | 32,002 |
No | output | 1 | 16,001 | 18 | 32,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Remember to flush your output while communicating with the testing program. You may use fflush(stdout) in C++, system.out.flush() in Java, stdout.flush() in Python or flush(output) in Pascal to flush the output. If you use some other programming language, consult its documentation. You may also refer to the guide on interactive problems: <https://codeforces.com/blog/entry/45307>.
You are given a string t consisting of n lowercase Latin letters. This string was cyphered as follows: initially, the jury had a string s consisting of n lowercase Latin letters. Then they applied a sequence of no more than n (possibly zero) operations. i-th operation is denoted by two integers a_i and b_i (1 ≤ a_i, b_i ≤ n), and means swapping two elements of the string with indices a_i and b_i. All operations were done in the order they were placed in the sequence. For example, if s is xyz and 2 following operations are performed: a_1 = 1, b_1 = 2; a_2 = 2, b_2 = 3, then after the first operation the current string is yxz, and after the second operation the current string is yzx, so t is yzx.
You are asked to restore the original string s. Unfortunately, you have no information about the operations used in the algorithm (you don't even know if there were any operations in the sequence). But you may run the same sequence of operations on any string you want, provided that it contains only lowercase Latin letters and its length is n, and get the resulting string after those operations.
Can you guess the original string s asking the testing system to run the sequence of swaps no more than 3 times?
The string s and the sequence of swaps are fixed in each test; the interactor doesn't try to adapt the test to your solution.
Input
Initially the testing system sends one string t, consisting of lowercase Latin letters (1 ≤ |t| = n ≤ 10^4).
Output
To give the answer, your program should print one line ! s with a line break in the end. After that, it should flush the output and terminate gracefully.
Interaction
Before giving the answer, you may submit no more than 3 queries. To ask a query, print one line in the following format: ? s', where s' should be a string consisting of exaclty n lowercase Latin letters. The line should be ended with a line break character. After submitting a query, flush the output and read the answer to your query — a string t' consisting of n lowercase Latin letters, which is the result of applying the sequence of swaps to string s'. This string will be given on a separate line ended by a line break character.
If you submit an incorrect query (or ask more than 3 queries), the answer to it will be one string 0. After receiving such an answer, your program should terminate immediately — otherwise you may receive verdict "Runtime error", "Time limit exceeded" or some other verdict instead of "Wrong answer".
Example
Input
yzx
aab
baa
aba
Output
? baa
? aba
? aab
! xyz
Note
In the sample, the testcase described in the statement is used. The participant asks the first query with string baa, which is transformed to aab. The second query contains string aba, which is transformed to baa. The third query contains string aab, which is transformed to aba. The participant can deduce that the initial string s was xyz.
Note for hacking phase:
To submit a test in hacking phase, you should provide it in the following format:
The first line should contain the string s you guess, consisting of n ∈ [1, 10000] lowercase Latin letters.
The second line should contain k (0 ≤ k ≤ n) — the number of swap operations in the sequence.
Then k lines should follow, i-th of them should denote i-th operation with two integers a_i and b_i (1 ≤ a_i, b_i ≤ n).
For example, the sample test would look like that:
xyz
2
1 2
2 3
Submitted Solution:
```
#import sys
#input=sys.stdin.readline
def bin(x):
l=[]
for k in range(3):
l.append(x%26)
x=x//26
l.reverse()
return l
t=input()
n=len(t)
a=[0 for i in range(n)]
b=[0 for i in range(n)]
c=[0 for i in range(n)]
for i in range(n):
l=bin(i)
a[i]=chr(l[0]+ord('a'))
b[i]=chr(l[1]+ord('a'))
c[i]=chr(l[2]+ord('a'))
print('?',end=" ")
for i in a:
print(i,end="")
print()
x=input()
print('?',end=" ")
for i in b:
print(i,end="")
print()
y=input()
print('?',end=" ")
for i in c:
print(i,end="")
print()
z=input()
s=[0 for i in range(n)]
for i in range(n):
p=ord(x[i])-ord('a')
q=ord(y[i])-ord('a')
r=ord(z[i])-ord('a')
s[576*p+26*q+r]=t[i]
print("!",end=" ")
for i in s:
print(i,end="")
``` | instruction | 0 | 16,002 | 18 | 32,004 |
No | output | 1 | 16,002 | 18 | 32,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Remember to flush your output while communicating with the testing program. You may use fflush(stdout) in C++, system.out.flush() in Java, stdout.flush() in Python or flush(output) in Pascal to flush the output. If you use some other programming language, consult its documentation. You may also refer to the guide on interactive problems: <https://codeforces.com/blog/entry/45307>.
You are given a string t consisting of n lowercase Latin letters. This string was cyphered as follows: initially, the jury had a string s consisting of n lowercase Latin letters. Then they applied a sequence of no more than n (possibly zero) operations. i-th operation is denoted by two integers a_i and b_i (1 ≤ a_i, b_i ≤ n), and means swapping two elements of the string with indices a_i and b_i. All operations were done in the order they were placed in the sequence. For example, if s is xyz and 2 following operations are performed: a_1 = 1, b_1 = 2; a_2 = 2, b_2 = 3, then after the first operation the current string is yxz, and after the second operation the current string is yzx, so t is yzx.
You are asked to restore the original string s. Unfortunately, you have no information about the operations used in the algorithm (you don't even know if there were any operations in the sequence). But you may run the same sequence of operations on any string you want, provided that it contains only lowercase Latin letters and its length is n, and get the resulting string after those operations.
Can you guess the original string s asking the testing system to run the sequence of swaps no more than 3 times?
The string s and the sequence of swaps are fixed in each test; the interactor doesn't try to adapt the test to your solution.
Input
Initially the testing system sends one string t, consisting of lowercase Latin letters (1 ≤ |t| = n ≤ 10^4).
Output
To give the answer, your program should print one line ! s with a line break in the end. After that, it should flush the output and terminate gracefully.
Interaction
Before giving the answer, you may submit no more than 3 queries. To ask a query, print one line in the following format: ? s', where s' should be a string consisting of exaclty n lowercase Latin letters. The line should be ended with a line break character. After submitting a query, flush the output and read the answer to your query — a string t' consisting of n lowercase Latin letters, which is the result of applying the sequence of swaps to string s'. This string will be given on a separate line ended by a line break character.
If you submit an incorrect query (or ask more than 3 queries), the answer to it will be one string 0. After receiving such an answer, your program should terminate immediately — otherwise you may receive verdict "Runtime error", "Time limit exceeded" or some other verdict instead of "Wrong answer".
Example
Input
yzx
aab
baa
aba
Output
? baa
? aba
? aab
! xyz
Note
In the sample, the testcase described in the statement is used. The participant asks the first query with string baa, which is transformed to aab. The second query contains string aba, which is transformed to baa. The third query contains string aab, which is transformed to aba. The participant can deduce that the initial string s was xyz.
Note for hacking phase:
To submit a test in hacking phase, you should provide it in the following format:
The first line should contain the string s you guess, consisting of n ∈ [1, 10000] lowercase Latin letters.
The second line should contain k (0 ≤ k ≤ n) — the number of swap operations in the sequence.
Then k lines should follow, i-th of them should denote i-th operation with two integers a_i and b_i (1 ≤ a_i, b_i ≤ n).
For example, the sample test would look like that:
xyz
2
1 2
2 3
Submitted Solution:
```
from sys import stdout
s = input()
print('?', s)
stdout.flush()
t = input()
print('?', t)
stdout.flush()
q = input()
print('?', q)
stdout.flush()
j = input()
print('!', t)
``` | instruction | 0 | 16,003 | 18 | 32,006 |
No | output | 1 | 16,003 | 18 | 32,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output
Print a single number — the least number of actions needed to make the message fancy.
Examples
Input
PRuvetSTAaYA
Output
5
Input
OYPROSTIYAOPECHATALSYAPRIVETSTASYA
Output
0
Input
helloworld
Output
0
Submitted Solution:
```
ALP="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
alp="abcdefghijklmnopqrstuvwxyz"
s=input()
n=len(s)
small=[0]*(n+1)
capital=[0]*(n+1)
for i in range(len(s)):
if s[i] in alp:
small[i+1]=small[i]+1
else:
small[i+1] = small[i]
for i in range(n-1,-1,-1):
if s[i] in ALP:
capital[i] =capital[i+1]+1
else:
capital[i] = capital[i+1]
#print(small)
#print(capital)
ans=float("inf")
for i in range(n):
ans=min(ans, small[i] +capital [i+1])
print(ans)
``` | instruction | 0 | 16,250 | 18 | 32,500 |
Yes | output | 1 | 16,250 | 18 | 32,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output
Print a single number — the least number of actions needed to make the message fancy.
Examples
Input
PRuvetSTAaYA
Output
5
Input
OYPROSTIYAOPECHATALSYAPRIVETSTASYA
Output
0
Input
helloworld
Output
0
Submitted Solution:
```
from math import inf
s=input().strip()
x=[]
for i in s:
x.append(1 if i.isupper() else 0)
p=0
q=sum(x)
m=q
for i in range(len(x)):
p+=x[i]
q-=x[i]
m=min(m,(i+1-p)+q )
print(m)
``` | instruction | 0 | 16,251 | 18 | 32,502 |
Yes | output | 1 | 16,251 | 18 | 32,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output
Print a single number — the least number of actions needed to make the message fancy.
Examples
Input
PRuvetSTAaYA
Output
5
Input
OYPROSTIYAOPECHATALSYAPRIVETSTASYA
Output
0
Input
helloworld
Output
0
Submitted Solution:
```
s=input()
a=0
for i in range(len(s)):
if s[i].isupper():
a+=1
b=a
for i in range(len(s)):
if s[i].isupper():
a-=1
else:
a+=1
b=min(a,b)
print(b)
``` | instruction | 0 | 16,252 | 18 | 32,504 |
Yes | output | 1 | 16,252 | 18 | 32,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output
Print a single number — the least number of actions needed to make the message fancy.
Examples
Input
PRuvetSTAaYA
Output
5
Input
OYPROSTIYAOPECHATALSYAPRIVETSTASYA
Output
0
Input
helloworld
Output
0
Submitted Solution:
```
##############--->>>>> Deepcoder Amit Kumar Bhuyan <<<<<---##############
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(1000000)
threading.stack_size(1024000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import log,sqrt,factorial,cos,tan,sin,radians
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *
#import threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
import sys
input = sys.stdin.readline
scanner = lambda: int(input())
string = lambda: input().rstrip()
get_list = lambda: list(read())
read = lambda: map(int, input().split())
get_float = lambda: map(float, input().split())
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## lcm function
def lcm(a, b):
return (a * b) // math.gcd(a, b)
def is_integer(n):
return math.ceil(n) == math.floor(n)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
if y == 0:
return 1
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
# Euler's Toitent Function phi
def phi(n) :
result = n
p = 2
while(p * p<= n) :
if (n % p == 0) :
while (n % p == 0) :
n = n // p
result = result * (1.0 - (1.0 / (float) (p)))
p = p + 1
if (n > 1) :
result = result * (1.0 - (1.0 / (float)(n)))
return (int)(result)
def is_prime(n):
if n == 0:
return False
if n == 1:
return True
for i in range(2, int(n ** (1 / 2)) + 1):
if not n % i:
return False
return True
def next_prime(n, primes):
while primes[n] != True:
n += 1
return n
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
def factors(n):
res = []
for i in range(1, int(n ** 0.5) + 1):
if n % i == 0:
res.append(i)
res.append(n // i)
return list(set(res))
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
def binary_search(low, high, w, h, n):
while low < high:
mid = low + (high - low) // 2
# print(low, mid, high)
if check(mid, w, h, n):
low = mid + 1
else:
high = mid
return low
## for checking any conditions
def check(moves, n):
val = (moves + 1) // 2
rem = moves - val
sol = (val + 1) * (rem + 1)
return sol < n
## for sorting according to second position
def sortSecond(val):
return val[1]
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
small = "abcdefghijklmnopqrstuvwxyz"
large = small.upper()
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
import bisect as bis
import random
import sys
import collections as collect
import functools as fnt
from decimal import Decimal
# from sys import stdout
# import numpy as np
"""
_______________
rough work here
_______________
a bookshelf that can fit n books
i position of bookshelf is ai = 1
if there is a book
else 0 otherwise
in one move you can choose some contiguous
segment[l: r] consisting of books
shift it to the right by 1
shift it to the left by 1
"""
def solve():
s = string()
n = len(s)
i = 0
j = len(s) - 1
while i < len(s) and s[i] in large:
i += 1
while j >= 0 and s[j] in small:
j -= 1
# print(i, j)
countbig = 0
countsmall = 0
for k in range(i, n):
if s[k] in large:
countbig += 1
for k in range(j, -1, -1):
if s[k] in small:
countsmall += 1
print(min(countbig, countsmall))
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
t = 1
for i in range(t):
solve()
#dmain()
# Comment Read()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | instruction | 0 | 16,254 | 18 | 32,508 |
No | output | 1 | 16,254 | 18 | 32,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output
Print a single number — the least number of actions needed to make the message fancy.
Examples
Input
PRuvetSTAaYA
Output
5
Input
OYPROSTIYAOPECHATALSYAPRIVETSTASYA
Output
0
Input
helloworld
Output
0
Submitted Solution:
```
string=input()
n=len(string)
upper=0
lower=0
i=0
k=n-1
while i<n and ord(string[i])<97:
i+=1
while k>=0 and ord(string[k])>=97:
k-=1
for j in range(i,k+1):
if ord(string[j])>=97:
lower+=1
else:
upper+=1
print(min(lower,upper))
``` | instruction | 0 | 16,255 | 18 | 32,510 |
No | output | 1 | 16,255 | 18 | 32,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output
Print a single number — the least number of actions needed to make the message fancy.
Examples
Input
PRuvetSTAaYA
Output
5
Input
OYPROSTIYAOPECHATALSYAPRIVETSTASYA
Output
0
Input
helloworld
Output
0
Submitted Solution:
```
s=input()
x=-1
t=0
dp=[[0 for i in range(2)] for i in range(len(s))]
for i in range(1,len(s)):
if s[i].islower():
x=1
dp[i][0]=dp[i-1][0]+1
dp[i][1]=dp[i-1][1]
else:
if x!=-1:
t=dp[i][0]=dp[i-1][0]
dp[i][1]=dp[i-1][1]+1
print(min(t,dp[-1][1]))
``` | instruction | 0 | 16,257 | 18 | 32,514 |
No | output | 1 | 16,257 | 18 | 32,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input
The first line contains a single integer N (1 ≤ N ≤ 5000) — the number of commands in the program. N lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109 + 7.
Examples
Input
4
s
f
f
s
Output
1
Input
4
f
s
f
s
Output
2
Note
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
simple statement
for statement
for statement
simple statement
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
for statement
simple statement
for statement
simple statement
or
for statement
simple statement
for statement
simple statement
Submitted Solution:
```
import sys
input = sys.stdin.readlines()
n = int(input[0].strip())
m = 10**9 + 7
curr = [0] * (n+20)
last = [0] * (n+20)
curr[0] = 1
for s in range(1,n):
last, curr = curr, last
if input[s]=='f\n':
curr[0] = 0
for i in range(len(last)-1):
curr[i+1] = last[i]
elif input[s]=='s\n':
curr[-1] = 0
for i in range(len(last)-2, -1, -1):
curr[i] = (curr[i+1] + last[i]) % m
s = 0
for x in curr:
s = (s + x) % m
print(s)
``` | instruction | 0 | 16,539 | 18 | 33,078 |
Yes | output | 1 | 16,539 | 18 | 33,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input
The first line contains a single integer N (1 ≤ N ≤ 5000) — the number of commands in the program. N lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109 + 7.
Examples
Input
4
s
f
f
s
Output
1
Input
4
f
s
f
s
Output
2
Note
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
simple statement
for statement
for statement
simple statement
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
for statement
simple statement
for statement
simple statement
or
for statement
simple statement
for statement
simple statement
Submitted Solution:
```
import sys
sys.setrecursionlimit(1000000)
read = sys.stdin.readline
MOD = 1000000007
N = int(read())
sts = [read().strip() for _ in range(N)]
ind = [0 for _ in range(N)]
ind[0] = 1
indent = 0
last = 'f'
for st in sts:
if last != 'f':
for i in range(indent, -1, -1):
ind[i] += ind[i+1]
ind[i] %= MOD
if st == 'f':
indent += 1
for i in range(indent, 0, -1):
ind[i] += ind[i-1]
ind[i-1] = 0
last = st
# print(ind)
ans = 0
for i in range(indent+1):
ans += ind[i]
ans %= MOD
print(ans)
``` | instruction | 0 | 16,541 | 18 | 33,082 |
Yes | output | 1 | 16,541 | 18 | 33,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input
The first line contains a single integer N (1 ≤ N ≤ 5000) — the number of commands in the program. N lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109 + 7.
Examples
Input
4
s
f
f
s
Output
1
Input
4
f
s
f
s
Output
2
Note
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
simple statement
for statement
for statement
simple statement
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
for statement
simple statement
for statement
simple statement
or
for statement
simple statement
for statement
simple statement
Submitted Solution:
```
mod = 10 ** 9 + 7
n = int(input())
dp = [[0 for i in range(n)] for i in range(n)]
dp[0][0] = 1
for i in range(1, n):
chr = input()
if chr == 'f':
for j in range(1, n):
dp[i][j] = dp[i - 1][j - 1] % mod
else:
s = 0
for j in range(n - 1, -1, -1):
s += dp[i - 1][j]
s %= mod
dp[i][j] = s
print(sum(dp[-1]) % mod)
``` | instruction | 0 | 16,542 | 18 | 33,084 |
Yes | output | 1 | 16,542 | 18 | 33,085 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input
The first line contains a single integer N (1 ≤ N ≤ 5000) — the number of commands in the program. N lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109 + 7.
Examples
Input
4
s
f
f
s
Output
1
Input
4
f
s
f
s
Output
2
Note
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
simple statement
for statement
for statement
simple statement
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
for statement
simple statement
for statement
simple statement
or
for statement
simple statement
for statement
simple statement
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=1000000007
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return stdin.read().split()
range = xrange # not for python 3.0+
inp=inp()
n=int(inp.pop(0))
dp=[0]*n
ln=1
dp[0]=1
for i in range(n):
v=inp[i]
if v=='f':
ln+=1
else:
for j in range(1,ln):
dp[j]=(dp[j-1]+dp[j])%mod
pr_num(dp[ln-1])
``` | instruction | 0 | 16,543 | 18 | 33,086 |
Yes | output | 1 | 16,543 | 18 | 33,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input
The first line contains a single integer N (1 ≤ N ≤ 5000) — the number of commands in the program. N lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109 + 7.
Examples
Input
4
s
f
f
s
Output
1
Input
4
f
s
f
s
Output
2
Note
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
simple statement
for statement
for statement
simple statement
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
for statement
simple statement
for statement
simple statement
or
for statement
simple statement
for statement
simple statement
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
c=[" "]+[input().rstrip() for i in range(n)]
mod=10**9+7
dp=[[0]*(n+1) for i in range(n+1)]
dp[1][0]=1
sdp=[0]*(n+1)
sdp[1]=1
for i in range(2,n+1):
if c[i-1]=="f":
for j in range(1,n):
dp[i][j]=dp[i-1][j-1]
dp[i][j]%=mod
dp[i][0]=0
else:
for j in range(n+1):
dp[i][j]=sdp[j]
dp[i][j]%=mod
sdp=[dp[i][j] for j in range(n+1)]
for j in range(1,n+1)[::-1]:
sdp[j-1]+=sdp[j]
sdp[j-1]%=mod
print(sdp[0]%mod)
``` | instruction | 0 | 16,544 | 18 | 33,088 |
No | output | 1 | 16,544 | 18 | 33,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
We will consider an extremely simplified subset of Python with only two types of statements.
Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with "for" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
Input
The first line contains a single integer N (1 ≤ N ≤ 5000) — the number of commands in the program. N lines of the program follow, each line describing a single command. Each command is either "f" (denoting "for statement") or "s" ("simple statement"). It is guaranteed that the last line is a simple statement.
Output
Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109 + 7.
Examples
Input
4
s
f
f
s
Output
1
Input
4
f
s
f
s
Output
2
Note
In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
simple statement
for statement
for statement
simple statement
In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
for statement
simple statement
for statement
simple statement
or
for statement
simple statement
for statement
simple statement
Submitted Solution:
```
s=int(input())
lis=[input() for x in range(s)]
count=1
boo=False
for i in range(s-1):
if lis[i]=='f':
boo=True
if str(lis[i]+lis[i+1])=='sf' and boo:
count*=2
elif str(lis[i]+lis[i+1])=='ss' and boo:
count*=2
else:
pass
print(count)
``` | instruction | 0 | 16,546 | 18 | 33,092 |
No | output | 1 | 16,546 | 18 | 33,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly:
* Insert a letter `x` to any position in s of his choice, including the beginning and end of s.
Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.
Constraints
* 1 \leq |s| \leq 10^5
* s consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
Output
If the objective is achievable, print the number of operations required. If it is not, print `-1` instead.
Examples
Input
xabxa
Output
2
Input
ab
Output
-1
Input
a
Output
0
Input
oxxx
Output
3
Submitted Solution:
```
s=input()
l=0
r=len(s)-1
x=0
while l<r:
if s[l]==s[r]:
l+=1
r-=1
elif s[l]=='x':
x+=1
l+=1
elif s[r]=='x':
x+=1
r-=1
else:
x=-1
break
print(x)
``` | instruction | 0 | 16,671 | 18 | 33,342 |
Yes | output | 1 | 16,671 | 18 | 33,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly:
* Insert a letter `x` to any position in s of his choice, including the beginning and end of s.
Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.
Constraints
* 1 \leq |s| \leq 10^5
* s consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
Output
If the objective is achievable, print the number of operations required. If it is not, print `-1` instead.
Examples
Input
xabxa
Output
2
Input
ab
Output
-1
Input
a
Output
0
Input
oxxx
Output
3
Submitted Solution:
```
s = str(input())
r = 0
l = len(s)-1
result = 0
h = True
while r < l:
if s[r] == s[l]:
r += 1
l -= 1
elif s[l] == 'x':
result += 1
l -= 1
elif s[r] == 'x':
result += 1
r += 1
else:
h = False
break
if h:
print(result)
else:
print(-1)
``` | instruction | 0 | 16,673 | 18 | 33,346 |
Yes | output | 1 | 16,673 | 18 | 33,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly:
* Insert a letter `x` to any position in s of his choice, including the beginning and end of s.
Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.
Constraints
* 1 \leq |s| \leq 10^5
* s consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
Output
If the objective is achievable, print the number of operations required. If it is not, print `-1` instead.
Examples
Input
xabxa
Output
2
Input
ab
Output
-1
Input
a
Output
0
Input
oxxx
Output
3
Submitted Solution:
```
S = input()
remove_x = S.replace('x','')
count = 0
index = 0
left = 0
right = 0
ans = 0
s_len = len(S)
if remove_x != remove_x[::-1]:
print(-1)
else:
if len(remove_x) % 2 == 0:
center = len(remove_x) // 2
for i, s in enumerate(S):
if s != 'x':
count += 1
if count == center:
index = i
break
left = index
x_count = 0
j = left+1
while S[j] == 'x' and j < s_len:
x_count += 1
j += 1
if x_count % 2 != 0:
ans += 1
right = j
#print(f'{right=}')
else:
center = len(remove_x) // 2 + 1
for i, s in enumerate(S):
if s != 'x':
count += 1
if count == center:
index = i
break
left = index - 1
right = index + 1
#print(f'{left}, {right}')
while left >= 0 and right < s_len:
if S[left] != S[right]:
if S[left] == 'x':
ans += 1
left -= 1
elif S[right] == 'x':
ans += 1
right += 1
else:
left -= 1
right += 1
if right != s_len:
ans += s_len - right
else:
ans += left + 1
print(ans)
``` | instruction | 0 | 16,675 | 18 | 33,350 |
No | output | 1 | 16,675 | 18 | 33,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly:
* Insert a letter `x` to any position in s of his choice, including the beginning and end of s.
Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.
Constraints
* 1 \leq |s| \leq 10^5
* s consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
Output
If the objective is achievable, print the number of operations required. If it is not, print `-1` instead.
Examples
Input
xabxa
Output
2
Input
ab
Output
-1
Input
a
Output
0
Input
oxxx
Output
3
Submitted Solution:
```
s=input()
i=0
j=len(s)-1
ans=0
while i<j:
while s[i]==s[j]=="x":
i+=1
j-=1
if i>=len(s) or j<0:break
while s[i]=="x":
i+=1
ans+=1
if i>=len(s):break
if i>j:break
while s[j]=="x":
j-=1
ans+=1
if j<=-1:break
if s[i]!=s[j]:
ans=-1
break
if i>j:break
i+=1
j-=1
if i>j:break
print(ans)
``` | instruction | 0 | 16,677 | 18 | 33,354 |
No | output | 1 | 16,677 | 18 | 33,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
n=int(input())
s=[input().strip() for i in range(n+n-2)]
a=sorted(list(range(n+n-2)), key=lambda x: len(s[x]))
for _ in range(2):
a[-1], a[-2]=a[-2], a[-1]
ans=[0 for i in range(n+n-2)]
if s[a[-1]][1:] != s[a[-2]][:-1]:
continue
o=s[a[-1]]+s[a[-2]][-1]
ans[a[-1]]='P'
ans[a[-2]]='S'
tkp=[False for i in range(n)]
tks=[False for i in range(n)]
gg=False
for i in range(n+n-4):
l=len(s[a[i]])
if s[a[i]] == o[:l] and not tkp[l]:
tkp[l]=True
ans[a[i]]='P'
continue
if s[a[i]] == o[-l:] and not tks[l]:
tks[l]=True
ans[a[i]]='S'
continue
gg=True
break
if gg: continue
print(''.join(x for x in ans))
break
``` | instruction | 0 | 16,870 | 18 | 33,740 |
Yes | output | 1 | 16,870 | 18 | 33,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
n=int(input())
sort_entry=[]
pos=[]
res=[]
for i in range(0,2*n-2):
sort_entry.append(" ")
res.append(" ")
pos.append(0)
for i in range(0,2*n-2):
temp=input()
l=len(temp)
if(sort_entry[2*l-2]==" "):
sort_entry[2*l-2]=temp
pos[2*l-2]=i
else:
sort_entry[2*l-1]=temp
pos[2*l-1]=i
word=sort_entry[2*n-4]+sort_entry[2*n-3][n-2]
i=0
while(i<2*n-3):
if(sort_entry[i]==word[0:len(sort_entry[i])] and sort_entry[i+1]==word[n-len(sort_entry[i+1]):n]):
res[pos[i]]='P'
res[pos[i+1]]='S'
elif(sort_entry[i+1]==word[0:len(sort_entry[i+1])] and sort_entry[i]==word[n-len(sort_entry[i]):n]):
res[pos[i]]='S'
res[pos[i+1]]='P'
else:
i=-2
word=sort_entry[2*n-3]+sort_entry[2*n-4][n-2]
i=i+2
result=""
for i in range(0,2*n-2):
result=result+res[i]
print(result)
``` | instruction | 0 | 16,871 | 18 | 33,742 |
Yes | output | 1 | 16,871 | 18 | 33,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
from collections import defaultdict
import sys
input = sys.stdin.readline
def solve(a, b):
ans = ["?"] * m
w = ["?"] * n
ok = 1
s1, k1 = x[n - 1][a]
s2, k2 = x[n - 1][b]
for i in range(n - 1):
w[i] = s1[i]
for i in range(1, n):
if not w[i] == "?" and not w[i] == s2[i - 1]:
ok = 0
break
w[i] = s2[i - 1]
if ok:
for i in range(1, n):
s1, k1 = x[i][0]
s2, k2 = x[i][1]
w0 = ["?"] * n
ok0 = 1
for j in range(i):
w0[j] = s1[j]
for j in range(n - i, n):
if not w0[j] == "?" and not w0[j] == s2[i + j - n]:
ok0 = 0
break
w0[j] = s2[i + j - n]
for j in range(n):
if not w0[j] == "?" and not w[j] == w0[j]:
ok0 = 0
break
if ok0:
ans[k1], ans[k2] = "P", "S"
continue
w0 = ["?"] * n
ok0 = 1
for j in range(i):
w0[j] = s2[j]
for j in range(n - i, n):
if not w0[j] == "?" and not w0[j] == s1[i + j - n]:
ok0 = 0
break
w0[j] = s1[i + j - n]
for j in range(n):
if not w0[j] == "?" and not w[j] == w0[j]:
ok0 = 0
break
if ok0:
ans[k1], ans[k2] = "S", "P"
else:
ok = 0
break
if ok:
print("".join(ans))
exit()
return
n = int(input())
m = 2 * n - 2
x = [[] for _ in range(n)]
for i in range(m):
s = list(input().rstrip())
l = len(s)
x[l].append((s, i))
solve(0, 1)
solve(1, 0)
``` | instruction | 0 | 16,872 | 18 | 33,744 |
Yes | output | 1 | 16,872 | 18 | 33,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
import sys
import collections as cc
#input=sys.stdin.buffer.readline
I=lambda:list(map(int,input().split()))
n,=I()
ar=[]
f=cc.defaultdict(list)
ans=cc.defaultdict(str)
for i in range((2*n)-2):
ar.append(input())
f[len(ar[-1])].append(ar[-1])
ans[ar[-1]]=''
a,b=f[n-1]
c,d=f[1]
tot=[a+c,a+d,c+a,d+a,b+c,b+d,c+b,d+b]
for i in tot:
now=i
#print("asf",now)
tf=1
for i in range(1,n):
temp={now[:i],now[-i:]}
#print(i,temp,set(f[i]))
if temp!=set(f[i]):
tf=0
break
if tf:
#print("Asfd")
#print(now)
for i in range(1,n):
x=now[:i]
y=now[-i:]
ans[x]='P'
ans[y]='S'
break
#print("-----")
#print(ans)
visi=cc.defaultdict(int)
for i in ar:
if not visi[i]:
print(ans[i],end='')
visi[i]=1
else:
print('P',end='')
``` | instruction | 0 | 16,873 | 18 | 33,746 |
Yes | output | 1 | 16,873 | 18 | 33,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
for _ in range(1):
n = int(input())
limit = (2*n)-2
ans = ["Z"]*(limit+1)
d = {}
ind = {}
cd = {}
cind = {}
for i in range(1,limit+1):
s = input()
x = len(s)
if x in d:
d[x].append(s)
else:
d[x] = [s]
if s in ind:
ind[s].append(i)
else:
ind[s] = [i]
if x in cd:
cd[x].append(s)
else:
cd[x] = [s]
if s in cind:
cind[s].append(i)
else:
cind[s] = [i]
cnt = limit-1
visited = set()
visited.add(1)
visited.add(limit)
w1 = d[1].pop()
w2 = d[1].pop()
i1 = ind[w1].pop()
i2 = ind[w2].pop()
ans[i1] = "P"
ans[i2] = "S"
prev = w1
for i in range(2,limit+1):
if i in visited or i not in d:
break
w1 = d[i].pop()
w2 = d[i].pop()
if prev == w1[0:len(w1)-1]:
i1 = ind[w1].pop()
i2 = ind[w2].pop()
ans[i1] = "P"
ans[i2] = "S"
prev = w1
visited.add(i)
visited.add(cnt)
cnt -= 1
continue
if prev == w2[0:len(w2)-1]:
w1,w2 = w2,w1
i1 = ind[w1].pop()
i2 = ind[w2].pop()
ans[i1] = "P"
ans[i2] = "S"
prev = w1
visited.add(i)
visited.add(cnt)
cnt -= 1
continue
break
# print(cind)
# print(cd)
if ans.count("Z") > 1:
cnt = limit - 1
d = cd
ind = cind
visited = set()
visited.add(1)
visited.add(limit)
w1 = d[1].pop()
w2 = d[1].pop()
i1 = ind[w1].pop()
i2 = ind[w2].pop()
ans[i1] = "S"
ans[i2] = "P"
prev = w2
for i in range(2, limit + 1):
if i in visited or i not in d:
break
w1 = d[i].pop()
w2 = d[i].pop()
if prev == w1[0:len(w1) - 1]:
i1 = ind[w1].pop()
i2 = ind[w2].pop()
ans[i1] = "P"
ans[i2] = "S"
prev = w1
visited.add(i)
visited.add(cnt)
cnt -= 1
continue
if prev == w2[0:len(w2) - 1]:
w1, w2 = w2, w1
i1 = ind[w1].pop()
i2 = ind[w2].pop()
ans[i1] = "P"
ans[i2] = "S"
prev = w1
visited.add(i)
visited.add(cnt)
cnt -= 1
continue
break
if ans.count("Z") == 1:
print("".join(ans[1:]))
continue
print("".join(ans[1:]))
``` | instruction | 0 | 16,874 | 18 | 33,748 |
No | output | 1 | 16,874 | 18 | 33,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
try:
n = int(input())
a = []
pref, suf = None, None
for i in range(2 * n - 2):
a.append(input())
if len(a[-1]) == n - 1:
if pref is None:
pref = a[-1]
else:
suf = a[-1]
s = pref + suf[-1]
ans = [None] * (2 * n - 2)
pref = ""
for i in s:
pref += i
if len(pref) == len(s):
break
ind = a.index(pref)
ans[ind] = 'P'
a[ind] = None
s = s[::-1]
pref = ""
for i in s:
pref = i + pref
if len(pref) == len(s):
break
ind = a.index(pref)
ans[ind] = 'S'
a[ind] = None
print(''.join(ans))
except Exception as e:
print(e)
``` | instruction | 0 | 16,875 | 18 | 33,750 |
No | output | 1 | 16,875 | 18 | 33,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
n=int(input())
ml=-1
s=[]
ms=[]
for i in range(2*n-2):
temp=input()
s.append(temp)
if(len(temp)==n-1):
ms.append(temp)
if n==2:
pref=ms[0]
suff=ms[1]
else:
if ms[0][1]==ms[1][0]:
pref=ms[0]
suff=ms[1]
else:
suff=ms[0]
pref=ms[1]
io=[-1]*101
last=False
for i in range(2*n-2):
tl=len(s[i])
f=True
if io[tl]==0:
print('P',end='')
continue
elif io[tl]==1:
print('S',end='')
continue
for j in range(tl):
if s[i][j]!=pref[j]:
print("S",end='')
io[tl]=0
f=False
break
if f:
print('P',end='')
io[tl]=1
``` | instruction | 0 | 16,876 | 18 | 33,752 |
No | output | 1 | 16,876 | 18 | 33,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan wants to play a game with you. He picked some string s of length n consisting only of lowercase Latin letters.
You don't know this string. Ivan has informed you about all its improper prefixes and suffixes (i.e. prefixes and suffixes of lengths from 1 to n-1), but he didn't tell you which strings are prefixes and which are suffixes.
Ivan wants you to guess which of the given 2n-2 strings are prefixes of the given string and which are suffixes. It may be impossible to guess the string Ivan picked (since multiple strings may give the same set of suffixes and prefixes), but Ivan will accept your answer if there is at least one string that is consistent with it. Let the game begin!
Input
The first line of the input contains one integer number n (2 ≤ n ≤ 100) — the length of the guessed string s.
The next 2n-2 lines are contain prefixes and suffixes, one per line. Each of them is the string of length from 1 to n-1 consisting only of lowercase Latin letters. They can be given in arbitrary order.
It is guaranteed that there are exactly 2 strings of each length from 1 to n-1. It is also guaranteed that these strings are prefixes and suffixes of some existing string of length n.
Output
Print one string of length 2n-2 — the string consisting only of characters 'P' and 'S'. The number of characters 'P' should be equal to the number of characters 'S'. The i-th character of this string should be 'P' if the i-th of the input strings is the prefix and 'S' otherwise.
If there are several possible answers, you can print any.
Examples
Input
5
ba
a
abab
a
aba
baba
ab
aba
Output
SPPSPSPS
Input
3
a
aa
aa
a
Output
PPSS
Input
2
a
c
Output
PS
Note
The only string which Ivan can guess in the first example is "ababa".
The only string which Ivan can guess in the second example is "aaa". Answers "SPSP", "SSPP" and "PSPS" are also acceptable.
In the third example Ivan can guess the string "ac" or the string "ca". The answer "SP" is also acceptable.
Submitted Solution:
```
n = int(input())
a = []
b = []
for i in range(2*n-2):
f= input()
b.append([len(f), [f, i]])
a.append([len(f), f])
a.sort()
b.sort()
for i in range(len(a)):
a[i] = a[i][1]
b[i] = b[i][1]
s=''
x=['']*(2*n-2)
k=1
if a[-2][:-1] != a[-1][1:]:
s=a[-2][0]+a[-1]
elif a[-1][:-1] != a[-2][1:]:
s=a[-1][0]+a[-2]
else:
s=a[-1][0]+a[-2]
k = 0
for i in range(1, 2*n-2, 2):
if a[i] == s[:len(a[i])] and a[i-1] == s[-len(a[i])-1:]:
x[b[i][1]]='s'
x[b[i-1][1]]='p'
elif a[i-1] == s[:len(a[i])] and a[i] == s[-len(a[i])-1:]:
x[b[i-1][1]]='s'
x[b[i][1]]='p'
else:
k=1
s=a[-2][0]+a[-1]
x=['']*(2*n-2)
if k == 1:
for i in range(1, 2*n-2, 2):
if a[i] == s[:len(a[i])] and a[i-1] == s[-len(a[i]):]:
x[b[i][1]]='P'
x[b[i-1][1]]='S'
elif a[i-1] == s[:len(a[i])] and a[i] == s[-len(a[i]):]:
x[b[i-1][1]]='P'
x[b[i][1]]='S'
print(*x, sep='')
print(s)
``` | instruction | 0 | 16,877 | 18 | 33,754 |
No | output | 1 | 16,877 | 18 | 33,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
def func():
s=input()
t=input()
n=len(s)
vow='aeiou'
if len(s)!=len(t):
return "NO"
for i in range(n):
if s[i] in vow and t[i] not in vow:
return "NO"
if s[i] not in vow and t[i] in vow:
return "NO"
return "YES"
print(func())
``` | instruction | 0 | 16,886 | 18 | 33,772 |
Yes | output | 1 | 16,886 | 18 | 33,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
s = input()
t = input()
vowel = 'ueoai'
if len(s) != len(t):
print('No')
else:
for i in range(len(s)):
if (s[i] in vowel and t[i] not in vowel) or (s[i] not in vowel and t[i] in vowel):
print("No")
exit()
print('Yes')
``` | instruction | 0 | 16,887 | 18 | 33,774 |
Yes | output | 1 | 16,887 | 18 | 33,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
from sys import exit
a = ['a', 'e', 'i', 'o','u']
b = ['b','c','d','f','g','h','j','k','l','m','n','p','q','r','s','t','v','w','x','y','z']
s1,s2 = input(),input()
if len(s1)!=len(s2):
print('No')
else:
for i in range(len(s1)):
if not ((s1[i] in a and s2[i] in a) or (s1[i] in b and s2[i] in b)):
print('No')
exit()
print('Yes')
``` | instruction | 0 | 16,888 | 18 | 33,776 |
Yes | output | 1 | 16,888 | 18 | 33,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
a = list(input())
b = list(input())
c = ['a','e','i','o','u']
k=0
if(len(a) == len(b)):
for i in range(len(a)):
if(a[i] in c and b[i] in c):
k+=1
elif(a[i] not in c and b[i] not in c):
k+=1
if(k==len(a)):
print('Yes')
else:
print('No')
else:
print('No')
``` | instruction | 0 | 16,889 | 18 | 33,778 |
Yes | output | 1 | 16,889 | 18 | 33,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
g = ['a', 'e', 'i', 'o', 'u']
gs = [0]*26
gt = [0]*26
sgs = [0]*26
sgt = [0]*26
s = input()
t = input()
for c in s:
if c in g:
gs[ord(c) - ord('a')] += 1
else:
sgs[ord(c) - ord('a')] += 1
for c in t:
if c in g:
gt[ord(c) - ord('a')] += 1
else:
sgt[ord(c) - ord('a')] += 1
gs.sort()
gt.sort()
sgs.sort()
sgt.sort()
if gs == gt and sgs == sgt:
print('Yes')
else:
print('No')
``` | instruction | 0 | 16,890 | 18 | 33,780 |
No | output | 1 | 16,890 | 18 | 33,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
g = ['a', 'e', 'i', 'o', 'u']
s = input()
t = input()
if len(s) != len(t):
print('No')
else:
for i in range(len(s)):
if s[i] in g and t[i] not in g or s[i] in g and t[i] not in g:
print('No')
exit()
print('Yes')
``` | instruction | 0 | 16,891 | 18 | 33,782 |
No | output | 1 | 16,891 | 18 | 33,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
print("yes" if all([(a in "aeiou") == (b in "aeiou") for a,b in zip(input(),input())]) else "no")
``` | instruction | 0 | 16,892 | 18 | 33,784 |
No | output | 1 | 16,892 | 18 | 33,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We all know that a superhero can transform to certain other superheroes. But not all Superheroes can transform to any other superhero. A superhero with name s can transform to another superhero with name t if s can be made equal to t by changing any vowel in s to any other vowel and any consonant in s to any other consonant. Multiple changes can be made.
In this problem, we consider the letters 'a', 'e', 'i', 'o' and 'u' to be vowels and all the other letters to be consonants.
Given the names of two superheroes, determine if the superhero with name s can be transformed to the Superhero with name t.
Input
The first line contains the string s having length between 1 and 1000, inclusive.
The second line contains the string t having length between 1 and 1000, inclusive.
Both strings s and t are guaranteed to be different and consist of lowercase English letters only.
Output
Output "Yes" (without quotes) if the superhero with name s can be transformed to the superhero with name t and "No" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
Examples
Input
a
u
Output
Yes
Input
abc
ukm
Output
Yes
Input
akm
ua
Output
No
Note
In the first sample, since both 'a' and 'u' are vowels, it is possible to convert string s to t.
In the third sample, 'k' is a consonant, whereas 'a' is a vowel, so it is not possible to convert string s to t.
Submitted Solution:
```
def main():
a = input()
b = input()
l = ['a', 'e', 'i', 'o', 'u']
for i in range(min(len(a), len(b))):
if (a[i] in l and a[i] not in l) or (a[i] not in l and b[i] in l):
print("No")
return
print("Yes")
return
main()
``` | instruction | 0 | 16,893 | 18 | 33,786 |
No | output | 1 | 16,893 | 18 | 33,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
* the password length is at least 5 characters;
* the password contains at least one large English letter;
* the password contains at least one small English letter;
* the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Examples
Input
abacaba
Output
Too weak
Input
X12345
Output
Too weak
Input
CONTEST_is_STARTED!!11
Output
Correct
Submitted Solution:
```
s = input()
ans = "Correct"
if len(s) < 5:
ans = "Too weak"
a = 0
b = 0
c = 0
for i in range(len(s)):
if s[i].isalpha():
if s[i].isupper():
a += 1
else:
b += 1
if s[i].isdigit():
c += 1
if a == 0 or b == 0 or c == 0:
ans = "Too weak"
print(ans)
``` | instruction | 0 | 17,165 | 18 | 34,330 |
Yes | output | 1 | 17,165 | 18 | 34,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
* the password length is at least 5 characters;
* the password contains at least one large English letter;
* the password contains at least one small English letter;
* the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Examples
Input
abacaba
Output
Too weak
Input
X12345
Output
Too weak
Input
CONTEST_is_STARTED!!11
Output
Correct
Submitted Solution:
```
def chkupper(password):
for i in password:
if(i.isupper()):
return True
return False
def chklower(password):
for i in password:
if(i.islower()):
return True
return False
def chkdigit(password):
for i in password:
if(i.isdigit()):
return True
return False
password =input()
#print(password)
#print(len(password))
if(len(password)>=5):
#print('a',len(password))
if(chkupper(password)):
#print('b')
if(chklower(password)):
#print('c')
if(chkdigit(password)):
print('Correct')
else:
print('Too weak')
else:
print('Too weak')
else:
print('Too weak')
else:
print('Too weak')
``` | instruction | 0 | 17,167 | 18 | 34,334 |
Yes | output | 1 | 17,167 | 18 | 34,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
* the password length is at least 5 characters;
* the password contains at least one large English letter;
* the password contains at least one small English letter;
* the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Examples
Input
abacaba
Output
Too weak
Input
X12345
Output
Too weak
Input
CONTEST_is_STARTED!!11
Output
Correct
Submitted Solution:
```
def solve(s):
if len(s) < 5:
return "Too weak"
capital = 0
lower = 0
digit = 0
capitals = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
lowercase = 'abcdefghijklmnopqrstuvwxyz'
digits = '1234567890'
for i in range(len(s)):
if s[i] in capitals:
capital += 1
elif s[i] in lowercase:
lower += 1
elif s[i] in digits:
digit += 1
if capital > 0 and lower > 0 and digit > 0:
return "Correct"
return "Too weak"
s = input()
print(solve(s))
``` | instruction | 0 | 17,168 | 18 | 34,336 |
Yes | output | 1 | 17,168 | 18 | 34,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
* the password length is at least 5 characters;
* the password contains at least one large English letter;
* the password contains at least one small English letter;
* the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Examples
Input
abacaba
Output
Too weak
Input
X12345
Output
Too weak
Input
CONTEST_is_STARTED!!11
Output
Correct
Submitted Solution:
```
s = input()
l, b, d = False, False, False
for i in s:
if 'a' <= i <= 'z':
l = True
elif 'A' <= i <= 'Z':
b = True
elif '0' <= i <= '9':
d = True
print("Correct" if l and b and d else "Too weak")
``` | instruction | 0 | 17,169 | 18 | 34,338 |
No | output | 1 | 17,169 | 18 | 34,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
* the password length is at least 5 characters;
* the password contains at least one large English letter;
* the password contains at least one small English letter;
* the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Examples
Input
abacaba
Output
Too weak
Input
X12345
Output
Too weak
Input
CONTEST_is_STARTED!!11
Output
Correct
Submitted Solution:
```
s=input()
c=0
d={'caps':0,'small':0,'digit':0}
if(len(s)<5):
print('Too weak')
else:
f=0
for i in s:
m=ord(i)
if(d['small']==0 and m>=97 and m<=123):
d['small']=1
c+=1
elif(d['caps']==0 and m>=65 and m<=91):
d['caps']=1
c+=1
elif(d['digit']==0 and m>=48 and m<=57):
d['digit']=0
c+=1
if(c==3):
f=1
break
if(f==1):
print('Correct')
else:
print('Too weak')
``` | instruction | 0 | 17,170 | 18 | 34,340 |
No | output | 1 | 17,170 | 18 | 34,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
* the password length is at least 5 characters;
* the password contains at least one large English letter;
* the password contains at least one small English letter;
* the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Examples
Input
abacaba
Output
Too weak
Input
X12345
Output
Too weak
Input
CONTEST_is_STARTED!!11
Output
Correct
Submitted Solution:
```
s = input()
le, se, dig = False, False, False
for i in s:
o = ord(i)
if 48 <= o <= 57:
dig = True
if 65 <= o <= 90:
le = True
if 97 <= o <= 120:
se = True
if le and se and dig:
break
print('Correct' if le and se and dig and len(s) >= 5 else 'Too weak')
``` | instruction | 0 | 17,171 | 18 | 34,342 |
No | output | 1 | 17,171 | 18 | 34,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
* the password length is at least 5 characters;
* the password contains at least one large English letter;
* the password contains at least one small English letter;
* the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Examples
Input
abacaba
Output
Too weak
Input
X12345
Output
Too weak
Input
CONTEST_is_STARTED!!11
Output
Correct
Submitted Solution:
```
def f(t): return len(t) > 5 and any('0'<= i <= '9' for i in t) and any('A' <= i <= 'Z' for i in t) and any('A' <= i <= 'Z' for i in t)
print('Correct' if f(input()) else 'Too weak')
``` | instruction | 0 | 17,172 | 18 | 34,344 |
No | output | 1 | 17,172 | 18 | 34,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia <https://en.wikipedia.org/wiki/Japanese_crossword>).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.
<image> The example of encrypting of a single row of japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the length of the row. The second line of the input contains a single string consisting of n characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output
The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Examples
Input
3
BBW
Output
1
2
Input
5
BWBWB
Output
3
1 1 1
Input
4
WWWW
Output
0
Input
4
BBBB
Output
1
4
Input
13
WBBBBWWBWBBBW
Output
3
4 1 3
Note
The last sample case correspond to the picture in the statement.
Submitted Solution:
```
input()
inp = list(input())
num = int()
matrix = []
if inp[-1] == "B":
inp.append("W")
for x in range(len(inp)):
if inp[x] == "B":
num = num + 1
elif inp[x] == "W":
matrix.append(str(num))
num = int()
if matrix[-1] == "0":
matrix.pop(-1)
print(len(matrix))
print(" ".join(matrix))
``` | instruction | 0 | 17,290 | 18 | 34,580 |
Yes | output | 1 | 17,290 | 18 | 34,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia <https://en.wikipedia.org/wiki/Japanese_crossword>).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.
<image> The example of encrypting of a single row of japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the length of the row. The second line of the input contains a single string consisting of n characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output
The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Examples
Input
3
BBW
Output
1
2
Input
5
BWBWB
Output
3
1 1 1
Input
4
WWWW
Output
0
Input
4
BBBB
Output
1
4
Input
13
WBBBBWWBWBBBW
Output
3
4 1 3
Note
The last sample case correspond to the picture in the statement.
Submitted Solution:
```
n = int(input())
s = input()
if 'B' in s:
s = s.split('W')
for i in range(len(s)):
if '' in s: s.remove('')
else: break
print(len(s))
for i in s:
print(len(i), end = ' ')
else: print(0)
``` | instruction | 0 | 17,291 | 18 | 34,582 |
Yes | output | 1 | 17,291 | 18 | 34,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia <https://en.wikipedia.org/wiki/Japanese_crossword>).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.
<image> The example of encrypting of a single row of japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the length of the row. The second line of the input contains a single string consisting of n characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output
The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Examples
Input
3
BBW
Output
1
2
Input
5
BWBWB
Output
3
1 1 1
Input
4
WWWW
Output
0
Input
4
BBBB
Output
1
4
Input
13
WBBBBWWBWBBBW
Output
3
4 1 3
Note
The last sample case correspond to the picture in the statement.
Submitted Solution:
```
input();s=input().replace('W',' ').split();print(len(s))
for i in s:
print(len(i),end=' ')
``` | instruction | 0 | 17,292 | 18 | 34,584 |
Yes | output | 1 | 17,292 | 18 | 34,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia <https://en.wikipedia.org/wiki/Japanese_crossword>).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.
<image> The example of encrypting of a single row of japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the length of the row. The second line of the input contains a single string consisting of n characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output
The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Examples
Input
3
BBW
Output
1
2
Input
5
BWBWB
Output
3
1 1 1
Input
4
WWWW
Output
0
Input
4
BBBB
Output
1
4
Input
13
WBBBBWWBWBBBW
Output
3
4 1 3
Note
The last sample case correspond to the picture in the statement.
Submitted Solution:
```
number = input()
string = [i for i in input().split("W") if i]
# print([[str(len(string))+"\n"+" ".join(map(str,list((len(i) for i in string))[0:-1])),str(len(string))+"\n"+" ".join(map(str,list((len(i) for i in string))))][string[0] ==""],0]["B" not in ("".join(string))])
print(len(string),"\n"+(" ".join(list(str(len(i)) for i in string))))
``` | instruction | 0 | 17,293 | 18 | 34,586 |
Yes | output | 1 | 17,293 | 18 | 34,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia <https://en.wikipedia.org/wiki/Japanese_crossword>).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.
<image> The example of encrypting of a single row of japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the length of the row. The second line of the input contains a single string consisting of n characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output
The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Examples
Input
3
BBW
Output
1
2
Input
5
BWBWB
Output
3
1 1 1
Input
4
WWWW
Output
0
Input
4
BBBB
Output
1
4
Input
13
WBBBBWWBWBBBW
Output
3
4 1 3
Note
The last sample case correspond to the picture in the statement.
Submitted Solution:
```
n = input()
letters = input()
B = letters.split('W')
length = []
for i in B:
if i is n:
length.apprend(len (i))
count += 1
print('1')
print('2')
``` | instruction | 0 | 17,294 | 18 | 34,588 |
No | output | 1 | 17,294 | 18 | 34,589 |
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