message stringlengths 2 23.8k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Paul hates palindromes. He assumes that string s is tolerable if each its character is one of the first p letters of the English alphabet and s doesn't contain any palindrome contiguous substring of length 2 or more.
Paul has found a tolerable string s of length n. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.
Input
The first line contains two space-separated integers: n and p (1 ≤ n ≤ 1000; 1 ≤ p ≤ 26). The second line contains string s, consisting of n small English letters. It is guaranteed that the string is tolerable (according to the above definition).
Output
If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).
Examples
Input
3 3
cba
Output
NO
Input
3 4
cba
Output
cbd
Input
4 4
abcd
Output
abda
Note
String s is lexicographically larger (or simply larger) than string t with the same length, if there is number i, such that s1 = t1, ..., si = ti, si + 1 > ti + 1.
The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.
A palindrome is a string that reads the same forward or reversed.
Submitted Solution:
```
def nxt(c):
return chr(ord(c) + 1) if c < maxc else 'a'
n, p = map(int, input().split())
s = input()
if p == 2 and s != 'ab' and s != 'a':
print("NO")
else:
s = list("#%" + s)
maxc = chr(ord('a') + p - 1)
found = False
for i in range(-1, -n - 1, -1):
while s[i] < maxc:
s[i] = nxt(s[i])
while s[i] == s[i - 1] or s[i] == s[i - 2]:
if s[i] == maxc:
break
s[i] = nxt(s[i])
else:
found = True
for j in range(i + 1, 0):
s[j] = nxt(s[j - 1])
while s[i] == s[i - 1] or s[i] == s[i - 2]:
s[i] = nxt(s[i])
break
if found:
print(*s[2:], sep="")
break
else:
print("NO")
``` | instruction | 0 | 58,399 | 0 | 116,798 |
No | output | 1 | 58,399 | 0 | 116,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Paul hates palindromes. He assumes that string s is tolerable if each its character is one of the first p letters of the English alphabet and s doesn't contain any palindrome contiguous substring of length 2 or more.
Paul has found a tolerable string s of length n. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.
Input
The first line contains two space-separated integers: n and p (1 ≤ n ≤ 1000; 1 ≤ p ≤ 26). The second line contains string s, consisting of n small English letters. It is guaranteed that the string is tolerable (according to the above definition).
Output
If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).
Examples
Input
3 3
cba
Output
NO
Input
3 4
cba
Output
cbd
Input
4 4
abcd
Output
abda
Note
String s is lexicographically larger (or simply larger) than string t with the same length, if there is number i, such that s1 = t1, ..., si = ti, si + 1 > ti + 1.
The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.
A palindrome is a string that reads the same forward or reversed.
Submitted Solution:
```
import math
from os import curdir, startfile
import random
from queue import Queue
import time
import heapq
import sys
def main(arr,p):
p+=96
arr_c=[0]*(len(arr))
idx=None
for i in range(len(arr)-1,-1,-1):
e=ord(arr[i])+1
if e<=p:
a=None if i-1<0 else ord(arr[i-1])
b=None if i-2<0 else ord(arr[i-2])
found=True
while (a==e or b==e):
e+=1
if e>p:
found=False
break
if found:
idx=i
arr_c[idx]=chr(e)
letter=97
for i in range(idx+1,len(arr_c)):
arr_c[i]=chr(letter)
letter+=1
if letter==100:
letter=97
for i in range(0,idx):
arr_c[i]=arr[i]
return "".join(arr_c)
return "NO"
n,p=list(map(int,input().split()))
s=input()
print(main(s,p))
``` | instruction | 0 | 58,400 | 0 | 116,800 |
No | output | 1 | 58,400 | 0 | 116,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Paul hates palindromes. He assumes that string s is tolerable if each its character is one of the first p letters of the English alphabet and s doesn't contain any palindrome contiguous substring of length 2 or more.
Paul has found a tolerable string s of length n. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.
Input
The first line contains two space-separated integers: n and p (1 ≤ n ≤ 1000; 1 ≤ p ≤ 26). The second line contains string s, consisting of n small English letters. It is guaranteed that the string is tolerable (according to the above definition).
Output
If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).
Examples
Input
3 3
cba
Output
NO
Input
3 4
cba
Output
cbd
Input
4 4
abcd
Output
abda
Note
String s is lexicographically larger (or simply larger) than string t with the same length, if there is number i, such that s1 = t1, ..., si = ti, si + 1 > ti + 1.
The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.
A palindrome is a string that reads the same forward or reversed.
Submitted Solution:
```
def nxt(c):
return chr(ord(c) + 1) if c < maxc else 'a'
n, p = map(int, input().split())
s = input()
if p == 2 and s != 'ab' and s != 'a':
print("NO")
else:
s = list("#%" + s)
maxc = chr(ord('a') + p - 1)
found = False
for i in range(-1, -n - 1, -1):
while s[i] < maxc:
s[i] = nxt(s[i])
while s[i] == s[i - 1] or s[i] == s[i - 2]:
if s[i] == maxc:
break
s[i] = nxt(s[i])
else:
found = True
if i < -1:
j = i + 1
s[j] = 'a'
while s[j] == s[j - 1] or s[j] == s[j - 2]:
s[j] = nxt(s[j])
for j in range(i + 2, 0):
s[j] = nxt(s[j - 1])
while s[j] == s[j - 1] or s[j] == s[j - 2]:
s[j] = nxt(s[j])
break
if found:
print(*s[2:], sep="")
break
else:
print("NO")
``` | instruction | 0 | 58,401 | 0 | 116,802 |
No | output | 1 | 58,401 | 0 | 116,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a string s consisting of `a` and `b`. Snuke can perform the following two kinds of operation any number of times in any order:
* Choose an occurrence of `aa` as a substring, and replace it with `b`.
* Choose an occurrence of `bb` as a substring, and replace it with `a`.
How many strings s can be obtained by this sequence of operations? Find the count modulo 10^9 + 7.
Constraints
* 1 \leq |s| \leq 10^5
* s consists of `a` and `b`.
Input
Input is given from Standard Input in the following format:
s
Output
Print the number of strings s that can be obtained, modulo 10^9 + 7.
Examples
Input
aaaa
Output
6
Input
aabb
Output
5
Input
ababababa
Output
1
Input
babbabaaba
Output
35
Submitted Solution:
```
from collections import deque
s = input()
q = deque()
q.append(s)
c = []
c.append(s)
a = 1
p = 1000000007
L = len(s) - 1
while q:
x = q.popleft()
l = len(x)
if L == l:
L -= 1
c = []
for i in range(l-1):
if x[i] == "a" and x[i+1] == "a":
t = x[:i] + "b" + x[i+2:]
w = True
for j in c:
if t == j:
w = False
break
if w:
q.append(t)
c.append(t)
a += 1
elif x[i] == "b" and x[i+1] == "b":
t = x[:i] + "a" + x[i+2:]
w = True
for j in c:
if t == j:
w = False
break
if w:
q.append(t)
c.append(t)
a += 1
if a == p:
a = 0
print(a)
``` | instruction | 0 | 58,730 | 0 | 117,460 |
No | output | 1 | 58,730 | 0 | 117,461 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given two strings, find the longest of the strings contained in both strings and write a program that answers that length.
Here, the string s included in the string t means that s appears consecutively in t. An empty string, that is, a string of length 0, is included in any string. For example, the string ABRACADABRA contains the following strings: ABRA, RAC, D, ACADABRA, ABRACADABRA, the empty string, etc. On the other hand, the string ABRACADABRA does not contain the following strings: ABRC, RAA, BA , K etc.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of two lines, the first line is given the first string and the second line is given the second string. The strings consist of uppercase letters and each string is 1 in length. More than 4000 and less.
Of the scoring data, for 30% of the points, the length of each character string is 1 or more and 50 or less.
The end of input is indicated by EOF. The number of datasets does not exceed 10.
output
Outputs the length of the longest string contained in both of the two strings given for each dataset on one line.
Examples
Input
ABRACADABRA
ECADADABRBCRDARA
UPWJCIRUCAXIIRGL
SBQNYBSBZDFNEV
Output
5
0
Input
None
Output
None | instruction | 0 | 58,790 | 0 | 117,580 |
"Correct Solution:
```
def rolling_hash(S, base, MOD):
l = len(S)
h = [0]*(l + 1)
for i in range(l):
h[i+1] = (h[i] * base + ord(S[i])) % MOD
return h
C = open(0).read().split()
MOD = 358976445361682909
base = 31
for t in range(len(C)//2):
S = C[2*t]; T = C[2*t+1]
rhs = rolling_hash(S, base, MOD)
rht = rolling_hash(T, base, MOD)
L = len(S); M = len(T)
def solve(l):
v = pow(base, l, MOD)
hs = set()
for i in range(L-l+1):
hs.add((rhs[i+l] - rhs[i]*v) % MOD)
for i in range(M-l+1):
if (rht[i+l] - rht[i]*v) % MOD in hs:
return 1
return 0
left = 0; right = min(L, M)+1
while left+1 < right:
mid = (left + right) // 2
if solve(mid):
left = mid
else:
right = mid
print(left)
``` | output | 1 | 58,790 | 0 | 117,581 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given two strings, find the longest of the strings contained in both strings and write a program that answers that length.
Here, the string s included in the string t means that s appears consecutively in t. An empty string, that is, a string of length 0, is included in any string. For example, the string ABRACADABRA contains the following strings: ABRA, RAC, D, ACADABRA, ABRACADABRA, the empty string, etc. On the other hand, the string ABRACADABRA does not contain the following strings: ABRC, RAA, BA , K etc.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of two lines, the first line is given the first string and the second line is given the second string. The strings consist of uppercase letters and each string is 1 in length. More than 4000 and less.
Of the scoring data, for 30% of the points, the length of each character string is 1 or more and 50 or less.
The end of input is indicated by EOF. The number of datasets does not exceed 10.
output
Outputs the length of the longest string contained in both of the two strings given for each dataset on one line.
Examples
Input
ABRACADABRA
ECADADABRBCRDARA
UPWJCIRUCAXIIRGL
SBQNYBSBZDFNEV
Output
5
0
Input
None
Output
None | instruction | 0 | 58,791 | 0 | 117,582 |
"Correct Solution:
```
while True:
try:
s1 = input()
except EOFError:
break
s2 = input()
BASE = 27
MOD = 1000000007
def rolling_hash(s):
h_lst = [0]
for c in s:
h_lst.append((h_lst[-1] * BASE + ord(c)) % MOD)
return h_lst
h_lst1 = rolling_hash(s1)
h_lst2 = rolling_hash(s2)
length1 = len(s1)
length2 = len(s2)
pow_mem = [pow(BASE, i, MOD) for i in range(min(length1, length2) + 1)]
def check(center):
hash_set = {(h_lst1[i + center] - h_lst1[i] * pow_mem[center]) % MOD for i in range(length1 - center + 1)}
for i in range(length2 - center + 1):
h2 = (h_lst2[i + center] - h_lst2[i] * pow_mem[center]) % MOD
if h2 in hash_set:return True
return False
left = 0
right = min(length1, length2)
while right > left + 1:
center = (left + right) // 2
if check(center):
left = center
else:
right = center
print(right if check(right) else left)
``` | output | 1 | 58,791 | 0 | 117,583 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Given two strings, find the longest of the strings contained in both strings and write a program that answers that length.
Here, the string s included in the string t means that s appears consecutively in t. An empty string, that is, a string of length 0, is included in any string. For example, the string ABRACADABRA contains the following strings: ABRA, RAC, D, ACADABRA, ABRACADABRA, the empty string, etc. On the other hand, the string ABRACADABRA does not contain the following strings: ABRC, RAA, BA , K etc.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of two lines, the first line is given the first string and the second line is given the second string. The strings consist of uppercase letters and each string is 1 in length. More than 4000 and less.
Of the scoring data, for 30% of the points, the length of each character string is 1 or more and 50 or less.
The end of input is indicated by EOF. The number of datasets does not exceed 10.
output
Outputs the length of the longest string contained in both of the two strings given for each dataset on one line.
Examples
Input
ABRACADABRA
ECADADABRBCRDARA
UPWJCIRUCAXIIRGL
SBQNYBSBZDFNEV
Output
5
0
Input
None
Output
None | instruction | 0 | 58,792 | 0 | 117,584 |
"Correct Solution:
```
while 1:
try:s=input()
except:break
t=input()
if len(s)<len(t):s,t=t,s
a=sum(min(s.count(i),t.count(i)) for i in set(s)&set(t))
b=0
for x in range(len(t)):
if len(t)-x<=b:break
for y in range(b+1,min(a+1,len(t)-x+1)):
if t[x:x+y] in s:b=y
else:break
print(b)
``` | output | 1 | 58,792 | 0 | 117,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Given two strings, find the longest of the strings contained in both strings and write a program that answers that length.
Here, the string s included in the string t means that s appears consecutively in t. An empty string, that is, a string of length 0, is included in any string. For example, the string ABRACADABRA contains the following strings: ABRA, RAC, D, ACADABRA, ABRACADABRA, the empty string, etc. On the other hand, the string ABRACADABRA does not contain the following strings: ABRC, RAA, BA , K etc.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of two lines, the first line is given the first string and the second line is given the second string. The strings consist of uppercase letters and each string is 1 in length. More than 4000 and less.
Of the scoring data, for 30% of the points, the length of each character string is 1 or more and 50 or less.
The end of input is indicated by EOF. The number of datasets does not exceed 10.
output
Outputs the length of the longest string contained in both of the two strings given for each dataset on one line.
Examples
Input
ABRACADABRA
ECADADABRBCRDARA
UPWJCIRUCAXIIRGL
SBQNYBSBZDFNEV
Output
5
0
Input
None
Output
None
Submitted Solution:
```
def solve():
while True:
try:
s1 = input()
s2 = input()
ans = 0
def search(x,y,count):
while x < len(s1) and y < len(s2) and s1[x] == s2[y]:
count += 1
x += 1
y += 1
return count
for i in range(len(s1)):
for j in range(len(s2)):
if s1[i] == s2[j]:
ans = max(ans,search(i,j,0))
print(ans)
except EOFError:
break
solve()
``` | instruction | 0 | 58,793 | 0 | 117,586 |
No | output | 1 | 58,793 | 0 | 117,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Given two strings, find the longest of the strings contained in both strings and write a program that answers that length.
Here, the string s included in the string t means that s appears consecutively in t. An empty string, that is, a string of length 0, is included in any string. For example, the string ABRACADABRA contains the following strings: ABRA, RAC, D, ACADABRA, ABRACADABRA, the empty string, etc. On the other hand, the string ABRACADABRA does not contain the following strings: ABRC, RAA, BA , K etc.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of two lines, the first line is given the first string and the second line is given the second string. The strings consist of uppercase letters and each string is 1 in length. More than 4000 and less.
Of the scoring data, for 30% of the points, the length of each character string is 1 or more and 50 or less.
The end of input is indicated by EOF. The number of datasets does not exceed 10.
output
Outputs the length of the longest string contained in both of the two strings given for each dataset on one line.
Examples
Input
ABRACADABRA
ECADADABRBCRDARA
UPWJCIRUCAXIIRGL
SBQNYBSBZDFNEV
Output
5
0
Input
None
Output
None
Submitted Solution:
```
while True:
try:
s = "." + input()
t = "." + input()
except:break
l = [[0 for _ in range(len(t))] for _ in range(len(s))]
for i in range(1, len(s)):
for j in range(1, len(t)):
if s[i] == t[j]:l[i][j] = l[i - 1][j - 1] + 1
print(max([max(i) for i in l]))
``` | instruction | 0 | 58,794 | 0 | 117,588 |
No | output | 1 | 58,794 | 0 | 117,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Given two strings, find the longest of the strings contained in both strings and write a program that answers that length.
Here, the string s included in the string t means that s appears consecutively in t. An empty string, that is, a string of length 0, is included in any string. For example, the string ABRACADABRA contains the following strings: ABRA, RAC, D, ACADABRA, ABRACADABRA, the empty string, etc. On the other hand, the string ABRACADABRA does not contain the following strings: ABRC, RAA, BA , K etc.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of two lines, the first line is given the first string and the second line is given the second string. The strings consist of uppercase letters and each string is 1 in length. More than 4000 and less.
Of the scoring data, for 30% of the points, the length of each character string is 1 or more and 50 or less.
The end of input is indicated by EOF. The number of datasets does not exceed 10.
output
Outputs the length of the longest string contained in both of the two strings given for each dataset on one line.
Examples
Input
ABRACADABRA
ECADADABRBCRDARA
UPWJCIRUCAXIIRGL
SBQNYBSBZDFNEV
Output
5
0
Input
None
Output
None
Submitted Solution:
```
while True:
try:
a, b = input().strip(), input().strip()
except:
break
n, m = len(a), len(b)
dp = [0] * m
tmp_max = 0
for i, ca in enumerate(a):
np = [0] * m
for j, cb in enumerate(b):
if ca == cb:
np[j] = dp[j - 1] + 1 if j else 1
tmp_max = max(tmp_max, max(np))
dp = np
print(tmp_max)
``` | instruction | 0 | 58,795 | 0 | 117,590 |
No | output | 1 | 58,795 | 0 | 117,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
Given two strings, find the longest of the strings contained in both strings and write a program that answers that length.
Here, the string s included in the string t means that s appears consecutively in t. An empty string, that is, a string of length 0, is included in any string. For example, the string ABRACADABRA contains the following strings: ABRA, RAC, D, ACADABRA, ABRACADABRA, the empty string, etc. On the other hand, the string ABRACADABRA does not contain the following strings: ABRC, RAA, BA , K etc.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of two lines, the first line is given the first string and the second line is given the second string. The strings consist of uppercase letters and each string is 1 in length. More than 4000 and less.
Of the scoring data, for 30% of the points, the length of each character string is 1 or more and 50 or less.
The end of input is indicated by EOF. The number of datasets does not exceed 10.
output
Outputs the length of the longest string contained in both of the two strings given for each dataset on one line.
Examples
Input
ABRACADABRA
ECADADABRBCRDARA
UPWJCIRUCAXIIRGL
SBQNYBSBZDFNEV
Output
5
0
Input
None
Output
None
Submitted Solution:
```
# coding: utf-8
while 1:
try:
a = input()
except:
break
z = input()
if len(a) > len(z):
b = a
a = z
else:
b = z
ans = 0
flg = 0
for i in range(len(a) + 1, 0, -1):
if flg == 1:
break
num = len(a) + 1 - i
if num > len(a):
break
for ii in range(num):
target = a[ii: i + ii]
if target in b:
ans = len(target)
flg = 1
print(ans)
``` | instruction | 0 | 58,796 | 0 | 117,592 |
No | output | 1 | 58,796 | 0 | 117,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,027 | 0 | 118,054 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
import sys
import math
from collections import defaultdict,Counter
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdout=open("CP2/output.txt",'w')
# sys.stdin=open("CP2/input.txt",'r')
# m=pow(10,9)+7
t=int(input())
for i in range(t):
n=int(input())
a=list(input())
b=list(input())
start=a[0]
curr=a[n-1]
c=0
l=[]
for j in range(n-1,0,-1):
if a[j]!=b[j]:
a1=[]
# a1=a[n-1:j:-1]
if a[0]==b[j]:
# c+=2
l+=[1,j+1]
a[0]=str(int(a[0])^1)
else:
# c+=1
l+=[j+1]
for k in range(j,-1,-1):
a1.append(str(int(a[k])^1))
a1+=a[j+1:]
a=a1[:]
# print(a)
# c+=1
# if c&1:
# start=b[j]
# curr=a[1]^1
if a[0]!=b[0]:
a[0]=str(int(a[0])^1)
l.append(1)
# print(a)
print(len(l),*l)
``` | output | 1 | 59,027 | 0 | 118,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,028 | 0 | 118,056 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
def funny(a,l):
m=[]
for i in range(a):
l[i]=1-l[i]
m.append(l[i])
m.reverse()
l=m[:]+l[a:]
return l
t=int(input())
final=[]
for i in range(t):
n=int(input())
a=[int(x) for x in input()]
b=[int(x) for x in input()]
k=[]
juju=[]
while(a!=b):
count=0
for j in range(len(a)):
if(a[len(a)-j-1]==b[len(b)-j-1]):
count+=1
else:
break
count=len(a)-count
if(a[0]==a[count-1]):
p=count
else:
p=1
for i in range(count-1):
if(a[0]==a[count-1-i]):
p=count-i
break
a=funny(p,a)
juju.append(p)
final.append([len(juju),*juju])
for i in range(t):
print(*final[i])
``` | output | 1 | 59,028 | 0 | 118,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,029 | 0 | 118,058 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
from sys import stdin
input = stdin.readline
for _ in range(int(input())):
n = int(input())
a = input().strip()
b = input().strip()
bi = n-1
f = 0
l, r = 0, n-1
ans = []
#temp = list(a)
while l!=r:
#print(l,r)
if (f==0 and a[l]==b[bi]) or (f!=0 and a[l]!=b[bi]):
#temp[0] = '0' if temp[0]=='1' else '1'
ans.append(1)
#for i in range(0, abs(l-r)+1):
# temp[i] = '0' if temp[i]=='1' else '1'
#print(temp,'mid')
#temp[0: abs(l-r)+1] = temp[0: abs(l-r)+1][::-1]
ans.append(abs(l-r)+1)
if l<r:
l += 1
else:
l -= 1
l, r = r, l
f ^= 1
bi-=1
#print(temp)
if (f==0 and a[l]!=b[0]) or (f and a[l]==b[0]):
#temp[0] = '0' if temp[0]=='1' else '1'
ans.append(1)
print(len(ans),end=' ')
print(*ans)
#print(temp)
``` | output | 1 | 59,029 | 0 | 118,059 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,030 | 0 | 118,060 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
import sys
t=int(input())
for _ in range(t):
n=int(input())
tempa=input()
tempb=input()
a=[]
for i in tempa:
a.append(int(i))
b=[]
for i in tempb:
b.append(int(i))
count=0
anslist=[]
for i in range(n-1,-1,-1):
if(a[i]!=b[i]):
if(a[0]!=b[i]):
anslist.append(i+1)
for j in range(i+1):
if(a[j]==0):
a[j]=1
else:
a[j]=0
a[0:i+1] = reversed(a[0:i+1])
count+=1
else:
if(a[0]==0):
a[0]=1
else:
a[0]=0
anslist.append(1)
for j in range(i+1):
if(a[j]==0):
a[j]=1
else:
a[j]=0
anslist.append(i+1)
a[0:i+1] = reversed(a[0:i+1])
count+=2
print(count,end=" ")
for i in anslist:
sys.stdout.write(str(i)+" ")
print()
``` | output | 1 | 59,030 | 0 | 118,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,031 | 0 | 118,062 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
z=input
mod = 10**9 + 7
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
def lcd(xnum1,xnum2):
return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if x%i==0:
return 0
return 1
################################################################################
"""
n=int(z())
for _ in range(int(z())):
x=int(z())
l=list(map(int,z().split()))
n=int(z())
l=sorted(list(map(int,z().split())))[::-1]
a,b=map(int,z().split())
l=set(map(int,z().split()))
led=(6,2,5,5,4,5,6,3,7,6)
vowel={'a':0,'e':0,'i':0,'o':0,'u':0}
color4=["G", "GB", "YGB", "YGBI", "OYGBI" ,"OYGBIV",'ROYGBIV' ]
"""
###########################---START-CODING---###############################################
num=int(z())
for _ in range(num):
n=int(z())
l1=list(z())
l2=list(z())
c=0
lp=[]
j=0
ans=0
lp=[]
for i in l1:
if i!=l2[j]:
ans=j
lp.append(ans+1)
ans=1
lp.append(ans)
lp.append(j+1)
c+=1
j+=1
print(c*3,*lp)
``` | output | 1 | 59,031 | 0 | 118,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,032 | 0 | 118,064 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
def flip(s, k):
pre = ''.join(
reversed([
'1' if _ == '0' else '0'
for _ in s[:k]])
)
pos = s[k:]
return pre+pos
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
ret = []
for i in reversed(list(range(n))):
while a[i] != b[i]:
if a[0] == b[i]:
p = 1
else:
p = i+1
ret.append(p)
a = flip(a, p)
# print(a)
# print(b)
print(len(ret), *ret)
``` | output | 1 | 59,032 | 0 | 118,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,033 | 0 | 118,066 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for tt in range(t):
n = int(input())
a1 = input().strip("\n")
b1 = input().strip("\n")
ans = []
a = []
b = []
for i in a1:
a.append(int(i))
for i in b1:
b.append(int(i))
for i in range(n-1,-1,-1):
if a[0] == b[i]:
a[0] ^= 1
ans.append(1)
ans.append(i+1)
temp = a[0:i+1]
temp = temp[::-1]
length = len(temp)
for j in range(length):
temp[j] ^= 1
for j in range(i+1):
a[j] = temp[j]
print(len(ans),end=" ")
for i in ans:
print(i,end=" ")
print()
``` | output | 1 | 59,033 | 0 | 118,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 59,034 | 0 | 118,068 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
from sys import stdin, gettrace
if gettrace():
inputi = input
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def solve():
n = int(input())
aa = list(input())
bb = list(input())
res = []
for i in range(n-1, 0, -1):
if aa[i] != bb[i]:
if aa[0] == aa[i]:
res += [i+1]
else:
res += [1, i+1]
aa[0] = '0' if aa[0] == '1' else '1'
an = []
for j in range(i, -1, -1):
an.append('1' if aa[j]=='0' else '0')
aa[:i+1] = an
if aa[0] != bb[0]:
res.append(1)
aa[0] = '0' if aa[0] == '1' else '1'
# print(''.join(aa))
print(len(res), ' '.join(map(str, res)))
def main():
t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
``` | output | 1 | 59,034 | 0 | 118,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def flip(arr_a, arr_b):
ops = 0
prefixlis = []
lis_a = arr_a.copy()
lis_b = arr_b.copy()
while arr_a != arr_b:
same = 0
for i in range(-1, -len(arr_a)-1, -1):
if arr_a[i] == arr_b[i]:
same += 1
else:
break
# print("same =", same)
if same >= 1:
arr_a = arr_a[:len(arr_a)-same]
arr_b = arr_b[:len(arr_b)-same]
if arr_a == []:
# print("solution found")
break
# print("arr_a =", arr_a)
# print("arr_b =", arr_b)
if arr_a[0] == arr_b[-1]:
prefixlis.append(1)
prefixlis.append(len(arr_a))
operation(arr_a, 1)
operation(arr_a, len(arr_a))
ops += 2
if arr_a == arr_b:
break
else:
prefixlis.append(len(arr_a))
operation(arr_a, len(arr_a))
ops += 1
if arr_a == arr_b:
break
print(ops, end = " ")
if ops == 0:
return
for i in range(ops):
print(prefixlis[i], end = " ")
def operation(arr_a, prefix):
for i in range(prefix):
if arr_a[i] == "0":
arr_a[i] = "1"
else:
arr_a[i] = "0"
arr_a[:prefix] = arr_a[:prefix][::-1]
# print("arr_a =", arr_a)
t = int(input())
for i in range(t):
# ab = input().strip().split()
n = int(input())
# arr = list(map(int, input().strip().split()))
a = input()
b = input()
arr_a = []
for element in a:
arr_a.append(element)
arr_b = []
for element in b:
arr_b.append(element)
# b = int(ab[1])
# arr_a = list(map(int, input().strip().split()))
# arr_b = list(map(int, input().strip().split()))
if len(arr_a) == 1:
if arr_a == arr_b:
print(0)
else:
print(1, 1)
else:
flip(arr_a, arr_b)
print()
"""
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
"""
"""
1
5
01011
11100
"""
``` | instruction | 0 | 59,035 | 0 | 118,070 |
Yes | output | 1 | 59,035 | 0 | 118,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s); sys.stdout.write('\n')
def wi(n): sys.stdout.write(str(n)); sys.stdout.write('\n')
def wia(a, sep=' '): sys.stdout.write(sep.join([str(x) for x in a])); sys.stdout.write('\n')
def solve(n, a, b):
res = []
for i in range(n-1, -1, -1):
if a[i] == b[i]:
continue
if a[0] == b[i]:
a[0] = 1 - a[0]
res.append(1)
if i % 2 == 0:
a[i // 2] = 1 - a[i // 2]
for j in range((i + 1) // 2):
a[j], a[i - j] = 1 - a[i - j], 1 - a[j]
res.append(i + 1)
return [len(res)] + res
def main():
for _ in range(ri()):
n = ri()
a = [int(si) for si in rs()]
b = [int(si) for si in rs()]
wia(solve(n, a, b))
if __name__ == '__main__':
main()
``` | instruction | 0 | 59,036 | 0 | 118,072 |
Yes | output | 1 | 59,036 | 0 | 118,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
a=int(input())
for i in range(a):
s=int(input())
a1=input()
a2=input()
ans1=[]
ans2=[]
for i in range(len(a1)):
ans1.append(int(a1[i]))
for i in range(len(a2)):
ans2.append(int(a2[i]))
fin=[]
for i in range(len(a2)-1,-1,-1):
if(ans2[i]==ans1[i]):
continue;
else:
if(ans1[0]!=ans2[i]):
fin.append(i+1)
temp=[]
for j in range(i,-1,-1):
temp.append(1-ans1[j])
for j in range(i+1):
ans1[j]=temp[j]
else:
fin.append(1)
ans1[0]=1-ans1[0]
fin.append(i+1)
temp=[]
for j in range(i,-1,-1):
temp.append(1-ans1[j])
for j in range(i+1):
ans1[j]=temp[j]
print(len(fin),end=" ")
print(*fin)
``` | instruction | 0 | 59,037 | 0 | 118,074 |
Yes | output | 1 | 59,037 | 0 | 118,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from __future__ import division, print_function
import sys
from atexit import register
if sys.version_info[0] < 3:
from io import BytesIO as stream
else:
from io import StringIO as stream
sys.stdin = stream(sys.stdin.read())
input = lambda: sys.stdin.readline().rstrip('\r\n')
sys.stdout = stream()
register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
def read_int():
return int(input())
def read_ints():
return list(map(int, input().split(' ')))
def flip(a):
return 0 if a == 1 else 1
def solve():
"""
a1 = 11 reverse whole
01
a0
"""
n = read_int()
a = list(map(int, input()))
b = list(map(int, input()))
end = n-1
operations = []
while end > -1:
if a[end] != b[end]:
if a[0] == a[end]: # just flip
flipped = list(map(flip, a[:end+1]))
flipped.reverse()
a[:end+1] = flipped
operations.append(end+1)
else:
a[0] = flip(a[0]) # flip first
operations.append(1)
flipped = list(map(flip, a[:end+1]))
flipped.reverse()
a[:end+1] = flipped
operations.append(end+1)
a[end] = b[end]
end -= 1
print(len(operations), ' '.join(map(str, operations)))
if __name__ == '__main__':
T = read_int()
for _ in range(T):
solve()
``` | instruction | 0 | 59,038 | 0 | 118,076 |
Yes | output | 1 | 59,038 | 0 | 118,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
# ------------------------------
# f = open('../input.txt')
# sys.stdin = f
def main():
for _ in range(N()):
n = N()
a = input()
b = input()
res = []
for i in range(n-1, -1, -1):
if a[0]==b[i]:
res.append(1)
res.append(i+1)
else:
res.append(i+1)
print(len(res), *res)
if __name__ == "__main__":
main()
``` | instruction | 0 | 59,039 | 0 | 118,078 |
No | output | 1 | 59,039 | 0 | 118,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def prefix_flip(index , arr):
s = []
for i in range(index-1 , -1 , -1):
s.append(str((int(arr[i])^1)))
s.append(arr[index:])
return s
t = int(input())
for _ in range(t):
n = int(input())
a = list(input())
b = list(input())
answer = []
for i in range(n-1 , -1 , -1):
if b[i] == a[i]:
continue
elif b[i] == a[0]:
answer.append(1)
answer.append(i+1)
a[0] = str(int(a[0])^1)
a = prefix_flip(i , a)
else:
answer.append(i+1)
a = prefix_flip(i , a)
if a[0] != b[0]:
answer.append(1)
print(len(answer) , *answer)
``` | instruction | 0 | 59,040 | 0 | 118,080 |
No | output | 1 | 59,040 | 0 | 118,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import sys, os.path
from collections import deque
from fractions import Fraction as f
def IO():
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
else:
input=sys.stdin.readline
print=sys.stdout.write
def nextInt():
return int(input())
def nextTuple():
return [int(a) for a in input().split()]
def nextArray():
return list(map(int, input().split()))
def nextLine():
return input()
def nextStringArray():
return list(input().split())
IO()
'''code starts here'''
t=nextInt()
for _ in range(t):
sb=[]
n=nextInt()
a=list(nextLine())
b=list(nextLine())
ptr=n-1
for i in range(n):
if b[ptr]==a[0]:
sb.append(str(ptr+1))
sb.append(str(ptr + 1))
else:
sb.append(str(ptr+1))
for i in range(ptr+1):
if a[i]=='0':
a[i]='1'
else:
a[i]='0'
a=a[0:ptr+1][::-1]+a[ptr+1:]
ptr-=1
print(len(sb),end=" ")
print(" ".join(sb),end="")
print()
``` | instruction | 0 | 59,041 | 0 | 118,082 |
No | output | 1 | 59,041 | 0 | 118,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
testcases=int(input())
for j in range(testcases):
n=int(input())
vala=[int(s) for s in input()]
valb=[int(s) for s in input()]
outy=[]
for s in range(n-1,-1,-1):
if vala[s]==valb[s]:
continue
else:
#2 checks
if vala[n-1-s]!=valb[s]:
for b in range(s+1):
vala[b]+=1
vala[b]=vala[b]%2
outy.append(s+1)
else:
for b in range(n-s):
vala[b]+=1
vala[b]=vala[b]%2
outy.append(n-s)
for b in range(s+1):
vala[b]+=1
vala[b]=vala[b]%2
outy.append(s+1)
if len(outy)>0:
outy=[len(outy)]+outy
print(*outy)
else:
print(0)
``` | instruction | 0 | 59,042 | 0 | 118,084 |
No | output | 1 | 59,042 | 0 | 118,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levko loves strings of length n, consisting of lowercase English letters, very much. He has one such string s. For each string t of length n, Levko defines its beauty relative to s as the number of pairs of indexes i, j (1 ≤ i ≤ j ≤ n), such that substring t[i..j] is lexicographically larger than substring s[i..j].
The boy wondered how many strings t are there, such that their beauty relative to s equals exactly k. Help him, find the remainder after division this number by 1000000007 (109 + 7).
A substring s[i..j] of string s = s1s2... sn is string sisi + 1... sj.
String x = x1x2... xp is lexicographically larger than string y = y1y2... yp, if there is such number r (r < p), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1. The string characters are compared by their ASCII codes.
Input
The first line contains two integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 2000).
The second line contains a non-empty string s of length n. String s consists only of lowercase English letters.
Output
Print a single number — the answer to the problem modulo 1000000007 (109 + 7).
Examples
Input
2 2
yz
Output
26
Input
2 3
yx
Output
2
Input
4 7
abcd
Output
21962
Submitted Solution:
```
print (5)
``` | instruction | 0 | 59,180 | 0 | 118,360 |
No | output | 1 | 59,180 | 0 | 118,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,764 | 0 | 119,528 |
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
for i in range(n):
if(s[i]=='>'):
break
for j in range(n):
if(s[n-1-j]=='<'):
break
print(min(i,j))
``` | output | 1 | 59,764 | 0 | 119,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,765 | 0 | 119,530 |
Tags: implementation, strings
Correct Solution:
```
t = int(input())
while t:
t -= 1
n = int(input())
s = input()
if s[0] == '<' and s[n - 1] == '>':
c1 = 0
for i in s:
if i == '<': c1 += 1
else: break
c2 = 0
for i in range(n - 1, -1, -1):
if s[i] == '>': c2 += 1
else: break
print(min(c1, c2))
else: print(0)
``` | output | 1 | 59,765 | 0 | 119,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,766 | 0 | 119,532 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
Tests = []
for i in range(2*n):
if (i % 2 == 0):
input()
continue
else:
Tests.append(input())
for i in range(n):
max_left = 10**6
max_right = 10**6
for j in range(len(Tests[i])):
if Tests[i][j] == '>':
max_left = j
break
for j in range(len(Tests[i])):
if (Tests[i][::-1][j]) == '<':
max_right = j
break
print(min(max_left, max_right))
``` | output | 1 | 59,766 | 0 | 119,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,767 | 0 | 119,534 |
Tags: implementation, strings
Correct Solution:
```
Q = int(input())
for _ in range(Q):
n = int(input())
s = input()
ans = 1000
for i, v in enumerate(s):
if v == '>':
break
ans = i
for i, v in enumerate(s[::-1]):
if v == '<':
break
ans = min(ans, i)
print(ans)
``` | output | 1 | 59,767 | 0 | 119,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,768 | 0 | 119,536 |
Tags: implementation, strings
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
s=input()
l1=len(s)
if s[0]=='<' and s[-1]=='>':
c1=0
i=0
while s[i]!='>':
i=i+1
c1=c1+1
c2=0
i=n-1
while s[i]!='<':
i=i-1
c2=c2+1
print(min(c1,c2))
else:
print(0)
``` | output | 1 | 59,768 | 0 | 119,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,769 | 0 | 119,538 |
Tags: implementation, strings
Correct Solution:
```
import math
t = int(input())
for i in range(t):
n = int(input())
s = input()
first = 0
second = 0
for i in range(n):
if s[i] == '>':
first = i
break
for i in range(n - 1, -1, -1):
if s[i] == '<':
second = n - i - 1
break
print(min(first, second))
``` | output | 1 | 59,769 | 0 | 119,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,770 | 0 | 119,540 |
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
s = input()
ans = n - 1
for i in range(n):
if(s[i] == '>' or s[n-i-1] == '<'):
ans = i
break
print(ans)
``` | output | 1 | 59,770 | 0 | 119,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <. | instruction | 0 | 59,771 | 0 | 119,542 |
Tags: implementation, strings
Correct Solution:
```
T = int(input())
for _ in range(T):
N = int(input())
S = input()
before = 0
for c in S:
if c == '<':
before += 1
else:
break
if before == N:
print(0)
continue
after = 0
for i in range(N-1, -1, -1):
if S[i] == '>':
after += 1
else:
break
print(min(before, after))
``` | output | 1 | 59,771 | 0 | 119,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
n=int(input())
for i in range(n):
x=int(input())
s=input()
s1=s
s2=s
count=0
if len(s)==1:
print(0)
else:
while(s2[0]!=">") and (s1[len(s1)-1]!="<"):
s1=s1[0:len(s1)-1]
s2=s2[1:len(s2)]
count+=1
print(count)
``` | instruction | 0 | 59,772 | 0 | 119,544 |
Yes | output | 1 | 59,772 | 0 | 119,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=input()
count1=1
for i in range(0,n-1):
if(a[i]==a[i+1]):
count1=count1+1
else:
break
count2=1
for i in range(n-1,-1,-1):
if(a[i]==a[i-1]):
count2=count2+1
else:
break
if(count1==count2 and count1==n):
print('0')
if(a[0]=='<' and a[n-1]=='>'):
print(min(count1,count2))
else:
print('0')
``` | instruction | 0 | 59,773 | 0 | 119,546 |
Yes | output | 1 | 59,773 | 0 | 119,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
s = input()
if s[0] == '<' and s[n-1] == '>':
left = 0
while(s[left] == '<'):
left+= 1
s = s[::-1]
right = 0
while (s[right] == '>'):
right += 1
print(min(left,right))
else:
print(0)
``` | instruction | 0 | 59,774 | 0 | 119,548 |
Yes | output | 1 | 59,774 | 0 | 119,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
digl=['0','1','2','3','4','5','6','7','8','9']
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
mod = 10 ** 9 + 7
t=ii()
for _ in range(t):
n=ii()
s=si()
x1=x2=0
if '>' in s:
x1=s.index('>')
if '<' in s:
x2=s.rindex('<')
print(min(x1,n-x2-1))
``` | instruction | 0 | 59,775 | 0 | 119,550 |
Yes | output | 1 | 59,775 | 0 | 119,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
R = lambda a: list(map(a,input().split()))
t=R(int)[0]
for i in range(0,t):
n=R(int)[0]
s=R(str)[0]
#l = list(s)
if(s[0]=='<') & (s[n-1]=='>'):
print("1")
else:
print("0")
``` | instruction | 0 | 59,776 | 0 | 119,552 |
No | output | 1 | 59,776 | 0 | 119,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
"""609C"""
# import math
# import sys
def main():
n = int(input())
string = str(input())
cnt = 0
for i in range(n-1):
if string[i]=='<' and string[i+1]=='>':
cnt+=1
print(cnt)
return
# main()
def test():
t = int(input())
while t:
main()
t-=1
test()
``` | instruction | 0 | 59,777 | 0 | 119,554 |
No | output | 1 | 59,777 | 0 | 119,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
t = int(input())
ans = []
for _ in range(t):
n = int(input())
s = input()
count = 0
if s[0] == '<':
for i in range(1,n):
if s[i]== '>':
count+=1
ans.append(count)
break
else:
ans.append(0)
print(*ans)
``` | instruction | 0 | 59,778 | 0 | 119,556 |
No | output | 1 | 59,778 | 0 | 119,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
Submitted Solution:
```
'''input
3
2
<>
3
><<
1
>
'''
from sys import stdin, stdout
import math
# main starts
t = int(stdin.readline().strip())
for _ in range(t):
n = int(stdin.readline().strip())
string = list(stdin.readline().strip())
mystack = []
for i in range(n):
#print(mystack)
if len(mystack) == 0:
mystack.append(string[i])
continue
if string[i] == '<':
while len(mystack) > 0:
mystack.pop()
mystack.append(string[i])
else:
if mystack[-1] == '>':
pass
else:
mystack.append(string[i])
print(len(mystack) - 1)
``` | instruction | 0 | 59,779 | 0 | 119,558 |
No | output | 1 | 59,779 | 0 | 119,559 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:
* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.
Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.
Constraints
* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
N K
S
Output
Print the lexicographically smallest possible string that can be the string S after the K operations.
Examples
Input
5 1
bacba
Output
aabca
Input
10 2
bbaabbbaab
Output
aaaabbaabb | instruction | 0 | 60,431 | 0 | 120,862 |
"Correct Solution:
```
n,k=map(int,input().split())
s=input()
u=s+s[::-1]
t=min(u[i:i+n]for i in range(n+1))
h=t[0]
i=min(i for i in range(n)if t[i]!=h)
j=i<<min(k-1,13)
print((h*j+t[i:i+n-j])[:n])
``` | output | 1 | 60,431 | 0 | 120,863 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:
* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.
Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.
Constraints
* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
N K
S
Output
Print the lexicographically smallest possible string that can be the string S after the K operations.
Examples
Input
5 1
bacba
Output
aabca
Input
10 2
bbaabbbaab
Output
aaaabbaabb | instruction | 0 | 60,432 | 0 | 120,864 |
"Correct Solution:
```
n,k=map(int,input().split())
s=list(input())
l=0
c=0
for i in range(n):
if s[i]==min(s):
c+=1
l=max(l,c)
else:
c=0
import sys
input=sys.stdin.readline
s.reverse()
if (2**(k-1))*l>n:
print(min(s)*n)
else:
for i in range(min(k,13)):
t=list(reversed(s))
u=t+s
uu=u[:n]
for j in range(1,n+1):
x=u[j:j+n]
if uu>x:
uu=x
s=uu
print("".join(s))
``` | output | 1 | 60,432 | 0 | 120,865 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:
* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.
Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.
Constraints
* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
N K
S
Output
Print the lexicographically smallest possible string that can be the string S after the K operations.
Examples
Input
5 1
bacba
Output
aabca
Input
10 2
bbaabbbaab
Output
aaaabbaabb | instruction | 0 | 60,433 | 0 | 120,866 |
"Correct Solution:
```
n, k=map(int, input().split())
s=input()
u=s+s[::-1]
ans=min(u[i:i+n] for i in range(n+1))
c=ans[0]
i=min(i for i in range(n) if c!=ans[i])
h=(i<<min(k-1, 13))
ret=(h*c+ans[i:i+n-h])[:n]
print(ret)
``` | output | 1 | 60,433 | 0 | 120,867 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:
* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.
Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.
Constraints
* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
N K
S
Output
Print the lexicographically smallest possible string that can be the string S after the K operations.
Examples
Input
5 1
bacba
Output
aabca
Input
10 2
bbaabbbaab
Output
aaaabbaabb | instruction | 0 | 60,434 | 0 | 120,868 |
"Correct Solution:
```
n,k=map(int,input().split())
s=input()
r=range(n)
u=min((s+s[::-1])[i:]for i in r)
i=min(i for i in r if u[i]!=u[0])
print((u[0]*min(i<<k-1,n)+u[i:])[:n])
``` | output | 1 | 60,434 | 0 | 120,869 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:
* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.
Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.
Constraints
* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
N K
S
Output
Print the lexicographically smallest possible string that can be the string S after the K operations.
Examples
Input
5 1
bacba
Output
aabca
Input
10 2
bbaabbbaab
Output
aaaabbaabb | instruction | 0 | 60,435 | 0 | 120,870 |
"Correct Solution:
```
n,k=map(int,input().split())
s=input()
d=s+s[::-1]
d=d[::-1]
r=d[:n]
e=d[:n]
for i in range(n):
e=e[1:]+d[n+i]
r=min(r,e)
t=1
qo=r[0]
for i in range(1,n):
if qo==r[i]:
t+=1
else:
break
r=r[::-1]
for i in range(k-1):
p=r+r[n-t:]
p=p[::-1]
p=p[:n]
if r==p:
break
else:
r=p[::-1]
t+=t
print(r[::-1])
``` | output | 1 | 60,435 | 0 | 120,871 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a string S of length N consisting of lowercase English letters. On this string, he will perform the following operation K times:
* Let T be the string obtained by reversing S, and U be the string obtained by concatenating S and T in this order.
* Let S' be some contiguous substring of U with length N, and replace S with S'.
Among the strings that can be the string S after the K operations, find the lexicographically smallest possible one.
Constraints
* 1 \leq N \leq 5000
* 1 \leq K \leq 10^9
* |S|=N
* S consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
N K
S
Output
Print the lexicographically smallest possible string that can be the string S after the K operations.
Examples
Input
5 1
bacba
Output
aabca
Input
10 2
bbaabbbaab
Output
aaaabbaabb | instruction | 0 | 60,436 | 0 | 120,872 |
"Correct Solution:
```
n,k=map(int,input().split())
s=input()
slist=list(s)
num=0
for i in range(17):
if 2**num>=n:
num+=1
break
num+=1
if k>=num:
slist.sort()
ans=slist[0]*n
print(ans)
else:
for i in range(n):
slist.append(slist[n-1-i])
s1=[]
for i in range(n+1):
s1.append(slist[i:i+n])
s1.sort()
sstr=s1[0]
ss2=""
for i in range(n):
ss2+=sstr[i]
s3=sstr[0]
num10=1
for i in range(n-1):
if s3!=sstr[i+1]:
break
else:
num10+=1
s2=""
for i in range(n):
s2+=sstr[-1-i]
slist.sort()
num=min((2**(k-1))*num10,n)
num2=n-num
ans=slist[0]*num
ans+=ss2[num10:num10+num2]
print(ans)
``` | output | 1 | 60,436 | 0 | 120,873 |
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