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.
You have a string s consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
start=1
end=0
for i in range(1,n):
if s[i]==s[i-1]:
end+=1
end=min(end,start)
else:
start+=1
print(end+(start-end+1)//2)
``` | instruction | 0 | 68,351 | 0 | 136,702 |
Yes | output | 1 | 68,351 | 0 | 136,703 |
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 consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import deque
from math import ceil
def main():
t = int(input())
for _ in range (t):
n = int(input())
s = input()
prefix = [1]*n
revPrefix = [1]*n
for i in range (1, n):
if (s[i] == s[i-1]):
prefix[i] = 1+prefix[i-1]
for i in range (n-2, -1, -1):
if (s[i] == s[i+1]):
revPrefix[i] = 1+revPrefix[i+1]
modified = list()
bank = deque()
steps = 0
pos = 0
for i in range (n):
if prefix[i] == 1:
extras = max (0, revPrefix[i]-2)
if extras > 0:
bank.append ([pos, extras])
if revPrefix[i] > 1:
number = 2
else:
number = 1
for _ in range (number):
modified.append (int(s[i]))
pos += number
revPrefix = [1]*pos
for i in range (pos-2, -1, -1):
if (modified[i] == modified[i+1]):
revPrefix[i] = 1+revPrefix[i+1]
i = 0
while i<pos:
if len(bank) == 0:
steps += ceil((pos-i)/2)
break
steps += 1
if (revPrefix[i] > 1):
if (bank[0][0] == i):
_ = bank.popleft()
i += 2
else:
bank[0][1] -= 1
if (bank[0][1] == 0):
_ = bank.popleft()
i += 1
print (steps)
# 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 = lambda s: self.buffer.write(s.encode()) if self.writable else None
def read(self):
if self.buffer.tell():
return self.buffer.read().decode("ascii")
return os.read(self._fd, os.fstat(self._fd).st_size).decode("ascii")
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().decode("ascii")
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
def print(*args, sep=" ", end="\n", file=sys.stdout, flush=False):
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(end)
if flush:
file.flush()
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
sys.setrecursionlimit(10000)
if __name__ == "__main__":
main()
``` | instruction | 0 | 68,352 | 0 | 136,704 |
Yes | output | 1 | 68,352 | 0 | 136,705 |
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 consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineering College
'''
from os import path
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
# mod=1000000007
mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('a')
def solve():
for _ in range(ii()):
n = ii()
s = si()
a = []
c = 1
for i in range(1,n):
if s[i] != s[i-1]:
a.append(c)
c = 1
else:
c += 1
a.append(c)
n = len(a)
suf = [0]*n
x = n
for i in range(n-1,-1,-1):
suf[i] = x
if a[i] > 1:
x = i
ans = 0
x = suf[0]
while(i < n):
ans += 1
x = max(x,suf[i])
if a[i] > 1:
i += 1
continue
i += 1
if x >= n:
i += 1
continue
a[x] -= 1
if a[x] == 1:
x = suf[x]
print(ans)
if __name__ =="__main__":
if path.exists('input.txt'):
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output.txt','w')
else:
input=sys.stdin.readline
solve()
``` | instruction | 0 | 68,353 | 0 | 136,706 |
Yes | output | 1 | 68,353 | 0 | 136,707 |
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 consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def RLE(s):
cc=[]
ww=[]
pc=s[0]
w=0
for c in s:
if c==pc:w+=1
else:
cc.append(pc)
ww.append(w)
w=1
pc=c
cc.append(pc)
ww.append(w)
return cc,ww
for _ in range(II()):
n=II()
s=SI()
cc,ww=RLE(s)
ans=0
j=0
back=False
for i in range(len(ww)):
# print(ww)
w=ww[i]
if w==0:break
if back:
j-=1
ww[j]=0
else:
if j<i:j=i
while j<len(ww) and ww[j]==1:j+=1
if j==len(ww):
back=True
j-=1
ww[j]=0
else:ww[j]-=1
ans+=1
print(ans)
``` | instruction | 0 | 68,354 | 0 | 136,708 |
Yes | output | 1 | 68,354 | 0 | 136,709 |
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 consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")
t = ip()
for _ in range(t):
n = ip()
s = ips()
gap = []
ct0,ct1 = 0,0
for i in range(n):
if s[i] == '1':
ct1 += 1
if ct0 != 0:
gap.append(ct0)
ct0 = 0
else:
ct0 += 1
if ct1 != 0:
gap.append(ct1)
ct1 = 0
if ct0 != 0:
gap.append(ct0)
if ct1 != 0:
gap.append(ct1)
ans = 0
great = deque()
nn = len(gap)
for i in range(nn):
if gap[i]>1:
great.append([gap[i],i])
i = 0
while i<nn:
if len(great) == 0:
i += 2
ans += 1
else:
ele,pos = great[0]
if ele == 1:
great.popleft()
if len(great) == 0:
i += 2
ans += 1
else:
i += 1
ele -= 1
great[0] = list([ele,pos])
ans += 1
else:
if pos< i:
great.popleft()
if len(great) == 0:
i += 2
ans += 1
else:
i += 1
ele -= 1
great[0] = list([ele,pos])
ans += 1
elif pos == i:
great.popleft()
i += 1
ans +=1
else:
i += 1
ele -= 1
great[0] = list([ele,pos])
ans += 1
out(ans)
``` | instruction | 0 | 68,355 | 0 | 136,710 |
No | output | 1 | 68,355 | 0 | 136,711 |
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 consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
from math import inf,sqrt,floor,ceil
from collections import Counter,defaultdict,deque
from heapq import heappush as hpush,heappop as hpop,heapify as h
from operator import itemgetter
from itertools import product
from bisect import bisect_left,bisect_right
for _ in range(int(input())):
n=int(input())
s=input()
d=deque(s)
#print(d)
res=0
while len(d)>0:
d.popleft()
if len(d)>0:
k=d.popleft()
while len(d)>0 and k==d[0]:
d.popleft()
res+=1
print(res)
``` | instruction | 0 | 68,356 | 0 | 136,712 |
No | output | 1 | 68,356 | 0 | 136,713 |
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 consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
"""T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
T=int(input())
for _ in range(0,T):
n=int(input())
s=input()
c=1
many=0
for i in range(1,len(s)):
if(s[i]!=s[i-1]):
many+=(c-1)
c=1
else:
c+=1
many+=(c-1)
turn=0
i=0
ans=0
while(i<len(s)):
if(turn==1):
ptr=len(s)
for j in range(i,len(s)):
if(s[j]!=s[i]):
ptr=j
break
turn=0
i=ptr
ans+=1
else:
ct=0
ptr=len(s)
for j in range(i,len(s)):
if(s[j]==s[i]):
ct+=1
else:
ptr=j
break
if(ct>1):
many-=(ct-1)
i=ptr
ans+=1
else:
if(many>0):
many-=1
i=ptr
ans+=1
else:
turn=1
i=ptr
if(turn==1):
ans+=1
print(ans)
``` | instruction | 0 | 68,357 | 0 | 136,714 |
No | output | 1 | 68,357 | 0 | 136,715 |
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 consisting of n characters. Each character is either 0 or 1.
You can perform operations on the string. Each operation consists of two steps:
1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1);
2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix).
Note that both steps are mandatory in each operation, and their order cannot be changed.
For example, if you have a string s = 111010, the first operation can be one of the following:
1. select i = 1: we'll get 111010 → 11010 → 010;
2. select i = 2: we'll get 111010 → 11010 → 010;
3. select i = 3: we'll get 111010 → 11010 → 010;
4. select i = 4: we'll get 111010 → 11110 → 0;
5. select i = 5: we'll get 111010 → 11100 → 00;
6. select i = 6: we'll get 111010 → 11101 → 01.
You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.
The second line contains string s of n characters. Each character is either 0 or 1.
It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.
Output
For each test case, print a single integer — the maximum number of operations you can perform.
Example
Input
5
6
111010
1
0
1
1
2
11
6
101010
Output
3
1
1
1
3
Note
In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
Submitted Solution:
```
def solve(s,n):
ans=0
while(len(s)>0):
if(s.count(s[0])==len(s)):
return(ans+1)
if(len(s)==1):
return(ans+1)
count=0
start=s[0]
for i in range(1,len(s)):
if(s[i]==start):
count+=1
else:
break
if(count>1):
s=s[i:]
else:
start=s[1]
prev=len(s)
for i in range(2,len(s)):
if(s[i]==start):
s=s[1:i-1]+s[i:]
break
else:
start=s[i]
if(len(s)==prev):
s=s[2:]
ans+=1
return(ans)
t=int(input())
for _ in range(t):
n=int(input())
s=input()
print(solve(s,n))
``` | instruction | 0 | 68,358 | 0 | 136,716 |
No | output | 1 | 68,358 | 0 | 136,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,374 | 0 | 136,748 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
# from sys import stdin,stdout
# input=stdin.readline
import math
# t=int(input())
from collections import Counter
import bisect
for _ in range(int(input())):
n = int(input())
a = list(map(int,input()))
b = list(map(int,input()))
# print(a,b)
count = [0 for i in range(n)]
if a[0]:
count[0] = 1
else:
count[0] = -1
for i in range(1,n):
if a[i] == 0:
count[i] = count[i-1] - 1
else:
count[i] = count[i-1] + 1
c = 0
flag = 0
for i in range(n-1,-1,-1):
if c%2:
k = not a[i]
else:
k = a[i]
if k != b[i]:
if count[i] != 0:
flag = 1
break
else:
c += 1
if flag:
print('NO')
else:
print('YES')
``` | output | 1 | 68,374 | 0 | 136,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,375 | 0 | 136,750 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys,os,io
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = sys.stdin.readline
for _ in range (int(input())):
n = int(input())
a = input().strip()
b = input().strip()
ind = [0]*len(a)
zc = 0
oc = 0
for i in range (n):
if a[i]=='0':
zc+=1
else:
oc+=1
if zc==oc:
ind[i]=1
temp = 0
flag = 0
for i in range (n-1,-1,-1):
if temp==0:
if a[i]!=b[i]:
if ind[i]:
temp = 1
else:
flag = 1
break
if temp==1:
if a[i]==b[i]:
if ind[i]:
temp=0
else:
flag = 1
break
if flag:
print("NO")
else:
print("YES")
``` | output | 1 | 68,375 | 0 | 136,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,376 | 0 | 136,752 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
#region Header
#!/usr/bin/env python3
# from typing import *
import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq
def input():
return sys.stdin.readline()[:-1]
# sys.setrecursionlimit(1000000)
#endregion
# _INPUT = """# paste here...
# """
# sys.stdin = io.StringIO(_INPUT)
YES = 'YES'
NO = 'NO'
def solve(N, A, B):
T_A0, T_A1, T_B0, T_B1 = [], [], [], []
n_A0, n_A1, n_B0, n_B1 = 0, 0, 0, 0
for i in range(N):
if A[i] == '0':
n_A0 += 1
else:
n_A1 += 1
T_A0.append(n_A0)
T_A1.append(n_A1)
if B[i] == '0':
n_B0 += 1
else:
n_B1 += 1
T_B0.append(n_B0)
T_B1.append(n_B1)
flag = True # True: same; False: different
for i in reversed(range(N)):
if flag and A[i] != B[i]:
flag = False
if T_A0[i] != T_A1[i] or T_B0[i] != T_B1[i]:
return False
elif (not flag) and A[i] == B[i]:
flag = True
if T_A0[i] != T_A1[i] or T_B0[i] != T_B1[i]:
return False
return True
def main():
T0 = int(input())
for _ in range(T0):
N = int(input())
A = input()
B = input()
ans = solve(N, A, B)
if ans:
print('YES')
else:
print('NO')
if __name__ == '__main__':
main()
``` | output | 1 | 68,376 | 0 | 136,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,377 | 0 | 136,754 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
# aadiupadhyay
import os.path
from math import gcd, floor, ceil
from collections import *
from bisect import *
import sys
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def solve():
n = inp()
a = st()
b = st()
a = list(map(int, a))
b = list(map(int, b))
pref = []
cur = [0, 0]
for i in a:
cur[i] += 1
pref.append(list(cur))
add = 0
# print(pref)
for i in range(n-1, -1, -1):
if (a[i]+add) % 2 == b[i]:
continue
if pref[i][0] != pref[i][1]:
pr('NO')
return
add += 1
pr('YES')
for _ in range(inp()):
solve()
``` | output | 1 | 68,377 | 0 | 136,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,378 | 0 | 136,756 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = input()
b = input()
a = a + ' '
b = b + ' '
count = prefix = x = 0
for j in range(n):
if a[j] == '1':
x = x+1
else:
x = x-1
count = x
if (a[j] == b[j]) != (a[j+1] == b[j+1]) and count!=0:
prefix = 1
break
if prefix == 1:
print("NO")
else:
print("YES")
``` | output | 1 | 68,378 | 0 | 136,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,379 | 0 | 136,758 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
"""
inp_start
6
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
1
1
0
inp_end
"""
tcs = int(input())
for tc in range(tcs):
n = int(input())
a = list(map(int, list(input())))
b = list(map(int, list(input())))
f = 0
a1 = a.count(1)
a0 = a.count(0)
for i in range(n-1,-1, -1):
if (a[i]^f)!=b[i]:
if a1!=a0:
print("NO")
break
f = 1-f
a1 -= a[i]==1
a0 -= a[i]==0
else:
print("YES")
``` | output | 1 | 68,379 | 0 | 136,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,380 | 0 | 136,760 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys
from collections import *
from heapq import *
import math
import bisect
def input():
return sys.stdin.readline()
for _ in range(int(input())):
n=int(input())
a=list(input())
b=list(input())
same=False
diff=False
ans=True
zero=0
one=0
v=-1
for i in a:
v+=1
if i=='0':
zero+=1
else:
one+=1
if i==b[v]:
same=True
else:
diff=True
if one==zero:
if same and diff:
ans=False
break
same=False
diff=False
if diff or not ans:
print("NO")
else:
print("YES")
``` | output | 1 | 68,380 | 0 | 136,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b. | instruction | 0 | 68,381 | 0 | 136,762 |
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
a = input()
b = input()
a = a + " "
b = b + " "
count = prefix = x = 0
for j in range(n):
if a[j] == '1':
x = x+1
else:
x = x-1
count = x
if (a[j] == b[j]) != (a[j+1] == b[j+1]) and count!=0:
prefix = 1
break
if prefix == 1:
print("NO")
else:
print("YES")
``` | output | 1 | 68,381 | 0 | 136,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a= input()
b= input()
box = [0]
nocount = 0
yescount = 0
res = True
for i in a:
if i == '0':
nocount += 1
else:
yescount += 1
if yescount == nocount:
box.append(yescount*2)
if box[-1]!=n:
for x in range(box[-1], n):
if a[x]!=b[x]:
res= False
break
box.reverse()
for i in range(len(box)-1):
s = a[box[i+1]]
f= b[box[i+1]]
if s==f:
for x in range(box[i+1], box[i]):
if a[x]!=b[x]:
res= False
break
if s!=f:
for x in range(box[i+1], box[i]):
if a[x]==b[x]:
res= False
break
if n%2 ==1:
if a[-1]!=b[-1]:
res= False
if res:
print("YES")
else:
print("NO")
``` | instruction | 0 | 68,382 | 0 | 136,764 |
Yes | output | 1 | 68,382 | 0 | 136,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
import math
from collections import defaultdict, Counter, deque
from heapq import heapify, heappush, heappop
def solve():
n = int(input())
a = list(input())
b = list(input())
pre = [0 for i in range(n)]
o = z = 0
for i in range(n):
if a[i] == '0':
z += 1
else:
o += 1
if o == z:
pre[i] = 1
inverted = 0
# print(pre)
for i in range(n - 1, -1, -1):
if inverted:
a[i] = '1' if a[i] == '0' else '0'
if a[i] != b[i]:
if pre[i]:
inverted ^= 1
else:
print('NO')
return
print('YES')
t = 1
t = int(input())
for _ in range(t):
solve()
``` | instruction | 0 | 68,383 | 0 | 136,766 |
Yes | output | 1 | 68,383 | 0 | 136,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
def is_inv(a,b):
#print(a,b)
one = False
two = False
for i in range(len(a)):
if a[i] == b[i]:
one = True
if two:
#print(a,b,"NE")
return False
if a[i] != b[i]:
two = True
if one:
#print(a,b,"NE")
return False
#print(a,b, "JO")
return True
for _ in range(int(input())):
n = int(input())
a = str(input())
b = str(input())
count0 = 0
count1 = 0
zoz = [0]
for i in range(n):
if a[i] == "1":
count1 += 1
if a[i] == "0":
count0 += 1
if count1 == count0:
zoz.append(i)
#print(zoz)
for i in range(1,len(zoz)):
#print(a[zoz[i-1]:zoz[i]+1], b[zoz[i-1]:zoz[i]+1])
if not is_inv(a[zoz[i-1]:zoz[i]+1], b[zoz[i-1]:zoz[i]+1]):
print("NO")
break
else:
zoz[i] += 1
else:
if a[zoz[-1]:] == b[zoz[-1]:]:
print("YES")
else:
print("NO")
``` | instruction | 0 | 68,384 | 0 | 136,768 |
Yes | output | 1 | 68,384 | 0 | 136,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
import math
from collections import defaultdict
from sys import stdin
#input=stdin.readline
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
flag = 0
zero,one = a.count('0'),a.count('1')
for i in range(n-1,-1,-1):
if int(a[i])^flag!=int(b[i]):
if zero!=one:
print("NO")
break
flag = not flag
zero-= a[i]=='0'
one-=a[i]=='1'
else:
print("YES")
``` | instruction | 0 | 68,385 | 0 | 136,770 |
Yes | output | 1 | 68,385 | 0 | 136,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
for i in range(int(input())):
n=int(input())
a=input()
b=input()
if a==b:
print("YES")
elif a.count("1")!=b.count("1") or a.count("0")!=b.count("0"):
print("NO")
else:
if "".join("0" if j=="1" else "1" for j in a)[::-1]==b:
print("YES")
else:
c=[]
z=""
l=0
for v,i in enumerate(a):
z+=i
if z.count("1")==z.count("0"):
o="".join("0" if j=="1" else "1" for j in z)
c+=[(z[l:],o[l:],l,v)]
l=v+1
if c==[]:
print("NO")
else:
for j in c:
v=b[j[2]:j[-1]+1]
tt=[j[0],j[1]]
if v not in tt:
print("NO")
break
else:
print("YES")
``` | instruction | 0 | 68,386 | 0 | 136,772 |
No | output | 1 | 68,386 | 0 | 136,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
def inverted_check(s1, s2, i, ind):
for j in range(i,ind+1):
if s1[j] == s2[j]:
return False
return True
def solve(s1, s2, n):
if s1.count('1') != s2.count('1'):
return "NO"
i = 0
while i < n:
if s1[i] != s2[i]:
ind = i+1
count= [0,0]
count[int(s1[i])] += 1
while ind < n :
count[int(s1[ind])] += 1
if count[0] == count[1]:
break
ind += 1
if count[0] == count[1] and inverted_check(s1, s2, i, ind):
i = ind+1
continue
else:
return "NO"
i += 1
return "YES"
local_ans = []
local_mode = False
def getArr():
return list(map(int, input().split()))
def getNums():
return map(int, input().split())
# local_mode = True
t = int(input())
for _ in range(t):
n = int(input())
s1 = input()
s2 = input()
if local_mode:
local_ans.append(solve(s1, s2, n))
else:
print(solve(s1, s2, n))
if local_mode:
def printAll(sol):
for val in sol:
print(val)
printAll(local_ans)
``` | instruction | 0 | 68,387 | 0 | 136,774 |
No | output | 1 | 68,387 | 0 | 136,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
def inv(s,k):
d={'0':'1','1':'0'}
l=list(s)
#print("L= ",l)
for i in range(k):
l[i]=d[l[i]]
return "".join(l)
for _ in range(int(input())):
n=int(input())
a=input()
b=input()
f=nigg=0
for i in range(n):
if a.count('0')!=b.count('0') or a.count("1")!=b.count('1'):
print('NO')
nigg=1
break
else:
c=[]
a0=a1=0
for i in range(n):
if i!=0:
if a0==a1:
f+=1
c.append(i-1)
if a[i]=='0':
a0+=1
else:
a1+=1
if a0==a1:
f+=1
c.append(n-1)
if f==0 and nigg==0:
if a==b:
print('YES')
else:
print('NO')
elif nigg==0:
pre=g=0
#print(c,f)
for i in range(f):
an=inv(a[pre:c[i]+1],c[i]+1-pre)
#print(an,a[pre:c[i]+1])
if an!=b[pre:c[i]+1] and a[pre:c[i]+1]!=b[pre:c[i]+1]:
print('NO')
g=1
break
pre=c[i]+1
if g==0:
print('YES')
``` | instruction | 0 | 68,388 | 0 | 136,776 |
No | output | 1 | 68,388 | 0 | 136,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.
For example, suppose a=0111010000.
* In the first operation, we can select the prefix of length 8 since it has four 0's and four 1's: [01110100]00→ [10001011]00.
* In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→ [01]00101100.
* It is illegal to select the prefix of length 4 for the third operation, because it has three 0's and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?
Input
The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the strings a and b.
The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.
The sum of n across all test cases does not exceed 3⋅ 10^5.
Output
For each test case, output "YES" if it is possible to transform a into b, or "NO" if it is impossible. You can print each letter in any case (upper or lower).
Example
Input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
Output
YES
YES
NO
YES
NO
Note
The first test case is shown in the statement.
In the second test case, we transform a into b by using zero operations.
In the third test case, there is no legal operation, so it is impossible to transform a into b.
In the fourth test case, here is one such transformation:
* Select the length 2 prefix to get 100101010101.
* Select the length 12 prefix to get 011010101010.
* Select the length 8 prefix to get 100101011010.
* Select the length 4 prefix to get 011001011010.
* Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
z = [0]*n
o = [0]*n
if a[0]=='0':
z[0]+=1
elif a[0]=='1':
o[0] += 1
for i in range(1,n):
if a[i]=='0':
z[i] = z[i-1] + 1
o[i] = o[i-1]
else:
o[i] = o[i-1] + 1
z[i] = z[i-1]
flag = 0
x = ''
for i in range(n-1,-1,-1):
if a[i]!=b[i]:
if flag%2==0:
if z[i]==o[i]:
flag += 1
x += b[i]
else:
x += a[i]
else:
x += b[i]
else:
if flag%2==0:
x += b[i]
else:
if z[i]==o[i]:
flag += 1
x += b[i]
else:
x += str(int(not int(b[i])))
x = x[::-1]
print(x)
if x==b:
print('YES')
else:
print('NO')
``` | instruction | 0 | 68,389 | 0 | 136,778 |
No | output | 1 | 68,389 | 0 | 136,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | instruction | 0 | 68,550 | 0 | 137,100 |
Tags: divide and conquer, hashing, sortings, strings
Correct Solution:
```
def getmin(s):
ls = len(s)
if ls % 2 == 1:
return s
s1 = getmin(s[:ls//2])
s2 = getmin(s[ls//2:])
return s1 + s2 if s1 < s2 else s2 + s1
s1 = input()
s2 = input()
print("YES") if getmin(s1) == getmin(s2) else print("NO")
``` | output | 1 | 68,550 | 0 | 137,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:
1. They are equal.
2. If we split string a into two halves of the same size a1 and a2, and string b into two halves of the same size b1 and b2, then one of the following is correct:
1. a1 is equivalent to b1, and a2 is equivalent to b2
2. a1 is equivalent to b2, and a2 is equivalent to b1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.
Output
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Examples
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Note
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | instruction | 0 | 68,553 | 0 | 137,106 |
Tags: divide and conquer, hashing, sortings, strings
Correct Solution:
```
def smallest(s):
if len(s) % 2 == 1:
return s
s1 = smallest(s[:len(s)//2])
s2 = smallest(s[len(s)//2:])
if s1 < s2:
return s1 + s2
else:
return s2 + s1
a = input()
b = input()
if smallest(a) == smallest(b):
print('YES')
else:
print('NO')
``` | output | 1 | 68,553 | 0 | 137,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice wants to send an important message to Bob. Message a = (a1, ..., an) is a sequence of positive integers (characters).
To compress the message Alice wants to use binary Huffman coding. We recall that binary Huffman code, or binary prefix code is a function f, that maps each letter that appears in the string to some binary string (that is, string consisting of characters '0' and '1' only) such that for each pair of different characters ai and aj string f(ai) is not a prefix of f(aj) (and vice versa). The result of the encoding of the message a1, a2, ..., an is the concatenation of the encoding of each character, that is the string f(a1)f(a2)... f(an). Huffman codes are very useful, as the compressed message can be easily and uniquely decompressed, if the function f is given. Code is usually chosen in order to minimize the total length of the compressed message, i.e. the length of the string f(a1)f(a2)... f(an).
Because of security issues Alice doesn't want to send the whole message. Instead, she picks some substrings of the message and wants to send them separately. For each of the given substrings ali... ari she wants to know the minimum possible length of the Huffman coding. Help her solve this problem.
Input
The first line of the input contains the single integer n (1 ≤ n ≤ 100 000) — the length of the initial message. The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000) — characters of the message.
Next line contains the single integer q (1 ≤ q ≤ 100 000) — the number of queries.
Then follow q lines with queries descriptions. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the position of the left and right ends of the i-th substring respectively. Positions are numbered from 1. Substrings may overlap in any way. The same substring may appear in the input more than once.
Output
Print q lines. Each line should contain a single integer — the minimum possible length of the Huffman encoding of the substring ali... ari.
Example
Input
7
1 2 1 3 1 2 1
5
1 7
1 3
3 5
2 4
4 4
Output
10
3
3
5
0
Note
In the first query, one of the optimal ways to encode the substring is to map 1 to "0", 2 to "10" and 3 to "11".
Note that it is correct to map the letter to the empty substring (as in the fifth query from the sample).
Submitted Solution:
```
import time
class Profiler(object):
def __enter__(self):
self._startTime = time.time()
def __exit__(self, type, value, traceback):
print("Elapsed time: {:.3f} sec".format(time.time() - self._startTime))
input()
words = input().split()
num_iter = int(input())
def get_result(words):
counter = {}
for word in words:
if counter.get(word):
counter[word] += 1
else:
counter[word] = 0
d1 = [(k, v) for k, v in counter.items()]
len_d1 = len(d1)
if len_d1 < 2:
return ''
def get_min_item_and_list_data(list_data):
min_i = min(list_data, key=lambda i: i[1])
list_data.pop(list_data.index(min_i))
return min_i, list_data
for i in range(len_d1-1):
min_i_1, d1 = get_min_item_and_list_data(d1)
min_i_2, d1 = get_min_item_and_list_data(d1)
d1.append(([min_i_1[0], min_i_2[0]], min_i_1[1]+min_i_2[1]))
tree = d1[0][0]
table = {}
def generate_table_bin(tree, current_bin, index):
if isinstance(tree, list):
generate_table_bin(tree[0], current_bin+'0', 0), generate_table_bin(tree[1], current_bin+'1', 1)
else:
table[tree] = current_bin
generate_table_bin(tree[0], '0', 0)
generate_table_bin(tree[1], '1', 1)
r = ''
for l in words:
r += table[l]
return r
list_result_num = []
for i in range(num_iter):
from_slice, to_slice = [int(i) for i in input().split()]
list_result_num.append(len(get_result(words[from_slice-1:to_slice])))
for result in list_result_num:
print(result)
``` | instruction | 0 | 68,591 | 0 | 137,182 |
No | output | 1 | 68,591 | 0 | 137,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice wants to send an important message to Bob. Message a = (a1, ..., an) is a sequence of positive integers (characters).
To compress the message Alice wants to use binary Huffman coding. We recall that binary Huffman code, or binary prefix code is a function f, that maps each letter that appears in the string to some binary string (that is, string consisting of characters '0' and '1' only) such that for each pair of different characters ai and aj string f(ai) is not a prefix of f(aj) (and vice versa). The result of the encoding of the message a1, a2, ..., an is the concatenation of the encoding of each character, that is the string f(a1)f(a2)... f(an). Huffman codes are very useful, as the compressed message can be easily and uniquely decompressed, if the function f is given. Code is usually chosen in order to minimize the total length of the compressed message, i.e. the length of the string f(a1)f(a2)... f(an).
Because of security issues Alice doesn't want to send the whole message. Instead, she picks some substrings of the message and wants to send them separately. For each of the given substrings ali... ari she wants to know the minimum possible length of the Huffman coding. Help her solve this problem.
Input
The first line of the input contains the single integer n (1 ≤ n ≤ 100 000) — the length of the initial message. The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000) — characters of the message.
Next line contains the single integer q (1 ≤ q ≤ 100 000) — the number of queries.
Then follow q lines with queries descriptions. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the position of the left and right ends of the i-th substring respectively. Positions are numbered from 1. Substrings may overlap in any way. The same substring may appear in the input more than once.
Output
Print q lines. Each line should contain a single integer — the minimum possible length of the Huffman encoding of the substring ali... ari.
Example
Input
7
1 2 1 3 1 2 1
5
1 7
1 3
3 5
2 4
4 4
Output
10
3
3
5
0
Note
In the first query, one of the optimal ways to encode the substring is to map 1 to "0", 2 to "10" and 3 to "11".
Note that it is correct to map the letter to the empty substring (as in the fifth query from the sample).
Submitted Solution:
```
1
``` | instruction | 0 | 68,592 | 0 | 137,184 |
No | output | 1 | 68,592 | 0 | 137,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice wants to send an important message to Bob. Message a = (a1, ..., an) is a sequence of positive integers (characters).
To compress the message Alice wants to use binary Huffman coding. We recall that binary Huffman code, or binary prefix code is a function f, that maps each letter that appears in the string to some binary string (that is, string consisting of characters '0' and '1' only) such that for each pair of different characters ai and aj string f(ai) is not a prefix of f(aj) (and vice versa). The result of the encoding of the message a1, a2, ..., an is the concatenation of the encoding of each character, that is the string f(a1)f(a2)... f(an). Huffman codes are very useful, as the compressed message can be easily and uniquely decompressed, if the function f is given. Code is usually chosen in order to minimize the total length of the compressed message, i.e. the length of the string f(a1)f(a2)... f(an).
Because of security issues Alice doesn't want to send the whole message. Instead, she picks some substrings of the message and wants to send them separately. For each of the given substrings ali... ari she wants to know the minimum possible length of the Huffman coding. Help her solve this problem.
Input
The first line of the input contains the single integer n (1 ≤ n ≤ 100 000) — the length of the initial message. The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000) — characters of the message.
Next line contains the single integer q (1 ≤ q ≤ 100 000) — the number of queries.
Then follow q lines with queries descriptions. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the position of the left and right ends of the i-th substring respectively. Positions are numbered from 1. Substrings may overlap in any way. The same substring may appear in the input more than once.
Output
Print q lines. Each line should contain a single integer — the minimum possible length of the Huffman encoding of the substring ali... ari.
Example
Input
7
1 2 1 3 1 2 1
5
1 7
1 3
3 5
2 4
4 4
Output
10
3
3
5
0
Note
In the first query, one of the optimal ways to encode the substring is to map 1 to "0", 2 to "10" and 3 to "11".
Note that it is correct to map the letter to the empty substring (as in the fifth query from the sample).
Submitted Solution:
```
# http://stackoverflow.com/questions/11587044/how-can-i-create-a-tree-for-huffman-encoding-and-decoding
# https://docs.python.org/3/library/asyncio-queue.html#priorityqueue
import queue
class Item:
__slots__ = '_key' , '_value'
def __init__ (self, k, v=None):
self._key = k
self._value = v
def __lt__ (self, other):
return self._key < other._key #
class HuffmanNode(object):
def __init__(self,left=None,right=None,root=None):
self.left = left
self.right = right
self.root = root
def children(self):
return (self.left,self.right)
def preorder(self,f_dict,path=None):
if path is None:
path = []
if self.left is not None:
if isinstance(self.left, HuffmanNode):
self.left[1].preorder(f_dict, path+[0])
else:
#print(self.left,path+[0])
f_dict[self.left] = len(path+[0])
if self.right is not None:
if isinstance(self.right, HuffmanNode):
self.right[1].preorder(f_dict, path+[1])
else:
#print(self.right,path+[1])
f_dict[self.right] = len(path+[1])
'''
freq = [
(8.167, 'a'), (1.492, 'b'), (2.782, 'c'), (4.253, 'd'),
(12.702, 'e'),(2.228, 'f'), (2.015, 'g'), (6.094, 'h'),
(6.966, 'i'), (0.153, 'j'), (0.747, 'k'), (4.025, 'l'),
(2.406, 'm'), (6.749, 'n'), (7.507, 'o'), (1.929, 'p'),
(0.095, 'q'), (5.987, 'r'), (6.327, 's'), (9.056, 't'),
(2.758, 'u'), (1.037, 'v'), (2.365, 'w'), (0.150, 'x'),
(1.974, 'y'), (0.074, 'z') ]
'''
def encode(frequencies):
p = queue.PriorityQueue()
for item in frequencies:
p.put(Item(item[0],item[1]))
#invariant that order is ascending in the priority queue
#p.size() gives list of elements
while p.qsize() > 1:
#left,right = p.get(),p.get()
left = p.get()
right = p.get()
node = HuffmanNode(left,right)
#print(left[0]+right[0], node)
p.put( Item(right._key+left._key, node) )
return p.get()
#node = encode(freq)
#print(node[1].preorder())
####################################
# a solution for http://codeforces.com/problemset/problem/700/D
n = int(input())
a = input().split()
q = int(input())
for qi in range(q):
l,r = [int(i) for i in input().split()]
if l==r:
print(0); continue
new_a = a[l-1:r]
new_a_int = [int(i) for i in new_a]
f = (max(new_a_int)+1)*[0]
#print('len(f): ', len(f))
for e in new_a:
#print('index:', ord(e)-ord('0'))
f[int(e)] += 1
freq = []
for e in set(new_a):
freq.append( (f[int(e)], e) )
if len(freq)==1:
print(0); continue
#print(freq)
node = encode(freq)
f_dict = {}
#print(node[1].preorder(f_dict) )
node._value.preorder(f_dict)
cnt = 0
for key in f_dict:
#print( key, f_dict[key] )
cnt += key._key*f_dict[key]
print(cnt)
``` | instruction | 0 | 68,593 | 0 | 137,186 |
No | output | 1 | 68,593 | 0 | 137,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice wants to send an important message to Bob. Message a = (a1, ..., an) is a sequence of positive integers (characters).
To compress the message Alice wants to use binary Huffman coding. We recall that binary Huffman code, or binary prefix code is a function f, that maps each letter that appears in the string to some binary string (that is, string consisting of characters '0' and '1' only) such that for each pair of different characters ai and aj string f(ai) is not a prefix of f(aj) (and vice versa). The result of the encoding of the message a1, a2, ..., an is the concatenation of the encoding of each character, that is the string f(a1)f(a2)... f(an). Huffman codes are very useful, as the compressed message can be easily and uniquely decompressed, if the function f is given. Code is usually chosen in order to minimize the total length of the compressed message, i.e. the length of the string f(a1)f(a2)... f(an).
Because of security issues Alice doesn't want to send the whole message. Instead, she picks some substrings of the message and wants to send them separately. For each of the given substrings ali... ari she wants to know the minimum possible length of the Huffman coding. Help her solve this problem.
Input
The first line of the input contains the single integer n (1 ≤ n ≤ 100 000) — the length of the initial message. The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000) — characters of the message.
Next line contains the single integer q (1 ≤ q ≤ 100 000) — the number of queries.
Then follow q lines with queries descriptions. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n) — the position of the left and right ends of the i-th substring respectively. Positions are numbered from 1. Substrings may overlap in any way. The same substring may appear in the input more than once.
Output
Print q lines. Each line should contain a single integer — the minimum possible length of the Huffman encoding of the substring ali... ari.
Example
Input
7
1 2 1 3 1 2 1
5
1 7
1 3
3 5
2 4
4 4
Output
10
3
3
5
0
Note
In the first query, one of the optimal ways to encode the substring is to map 1 to "0", 2 to "10" and 3 to "11".
Note that it is correct to map the letter to the empty substring (as in the fifth query from the sample).
Submitted Solution:
```
from collections import Counter
input()
string = input().replace(' ', '')
num_iter = int(input())
def get_result(string):
if len(string) < 2:
return ''
d1 = sorted([(k, v) for k, v in Counter(string).items()], key=lambda i: i[1])
if len(d1) < 2:
return ''
def get_min_item_and_list_data(list_data):
min_i = min(list_data, key=lambda i: i[1])
list_data.pop(list_data.index(min_i))
return min_i, list_data
while len(d1) != 1:
min_i_1, d1 = get_min_item_and_list_data(d1)
min_i_2, d1 = get_min_item_and_list_data(d1)
d1.append(([min_i_1[0], min_i_2[0]], min_i_1[1]+min_i_2[1]))
tree = d1[0][0]
table = {}
def generate_table_bin(tree, current_bin, index):
if isinstance(tree, list):
generate_table_bin(tree[0], current_bin+'0', 0), generate_table_bin(tree[1], current_bin+'1', 1)
else:
table[tree] = current_bin
generate_table_bin(tree[0], '0', 0)
generate_table_bin(tree[1], '1', 1)
r = ''
for l in string:
r += table[l]
return r
list_result_num = []
for i in range(num_iter):
from_slice, to_slice = [int(i) for i in input().split()]
list_result_num.append(len(get_result(string[from_slice-1:to_slice])))
for result in list_result_num:
print(result)
``` | instruction | 0 | 68,594 | 0 | 137,188 |
No | output | 1 | 68,594 | 0 | 137,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bear Limak prepares problems for a programming competition. Of course, it would be unprofessional to mention the sponsor name in the statement. Limak takes it seriously and he is going to change some words. To make it still possible to read, he will try to modify each word as little as possible.
Limak has a string s that consists of uppercase English letters. In one move he can swap two adjacent letters of the string. For example, he can transform a string "ABBC" into "BABC" or "ABCB" in one move.
Limak wants to obtain a string without a substring "VK" (i.e. there should be no letter 'V' immediately followed by letter 'K'). It can be easily proved that it's possible for any initial string s.
What is the minimum possible number of moves Limak can do?
Input
The first line of the input contains an integer n (1 ≤ n ≤ 75) — the length of the string.
The second line contains a string s, consisting of uppercase English letters. The length of the string is equal to n.
Output
Print one integer, denoting the minimum possible number of moves Limak can do, in order to obtain a string without a substring "VK".
Examples
Input
4
VKVK
Output
3
Input
5
BVVKV
Output
2
Input
7
VVKEVKK
Output
3
Input
20
VKVKVVVKVOVKVQKKKVVK
Output
8
Input
5
LIMAK
Output
0
Note
In the first sample, the initial string is "VKVK". The minimum possible number of moves is 3. One optimal sequence of moves is:
1. Swap two last letters. The string becomes "VKKV".
2. Swap first two letters. The string becomes "KVKV".
3. Swap the second and the third letter. The string becomes "KKVV". Indeed, this string doesn't have a substring "VK".
In the second sample, there are two optimal sequences of moves. One is "BVVKV" → "VBVKV" → "VVBKV". The other is "BVVKV" → "BVKVV" → "BKVVV".
In the fifth sample, no swaps are necessary. | instruction | 0 | 68,611 | 0 | 137,222 |
Tags: dp
Correct Solution:
```
# http://codeforces.com/contest/771/problem/D
"""
DP-solution.
For each state (v, k, x, v_is_last_letter) we trial a step along the v, k and x
axes and check that
dp[future_state] = min(dp[future_state], dp[state] + cost_of_move)
Hence this implicitly reults in the one with least cost.
V, K, X are arrays that contain the number of occurences of v, k, x at the i'th
index of s.
"""
def cost_of_move(state, ss_ind):
"""
eg. ss = s[0:K.index(k+1)]
Note: ss includes the i+1'th occurence of letter I. We hence want
ss = s[0:ss_ind-1]
And then we cound the number of occurences of V, K, X in this substring.
However, we don't need ss now - this info is contained in lists V, K, X.
"""
curr_v, curr_k, curr_x = state
cost = sum([max(0, V[ss_ind-1] - curr_v), max(0, K[ss_ind-1] - curr_k),
max(0, X[ss_ind-1] - curr_x)])
return cost
if __name__ == "__main__":
n = int(input())
s = input()
V = [s[0:i].count('V') for i in range(n+1)]
K = [s[0:i].count('K') for i in range(n+1)]
X = [(i - V[i] - K[i]) for i in range(n+1)]
# Initialising
n_v, n_k, n_x = V[n], K[n], X[n]
dp = [[[[float('Inf') for vtype in range(2)] for x in range(n_x+1)]
for k in range(n_k+1)] for v in range(n_v+1)]
dp[0][0][0][0] = 0
for v in range(n_v + 1):
for k in range(n_k + 1):
for x in range(n_x + 1):
for vtype in range(2):
orig = dp[v][k][x][vtype]
if v < n_v:
dp[v+1][k][x][1] = min(dp[v+1][k][x][vtype],
orig + cost_of_move([v, k, x], V.index(v+1)))
if k < n_k and vtype == 0:
dp[v][k+1][x][0] = min(dp[v][k+1][x][0],
orig + cost_of_move([v, k, x], K.index(k+1)))
if x < n_x:
dp[v][k][x+1][0] = min(dp[v][k][x+1][0],
orig + cost_of_move([v, k, x], X.index(x+1)))
print(min(dp[n_v][n_k][n_x]))
``` | output | 1 | 68,611 | 0 | 137,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bear Limak prepares problems for a programming competition. Of course, it would be unprofessional to mention the sponsor name in the statement. Limak takes it seriously and he is going to change some words. To make it still possible to read, he will try to modify each word as little as possible.
Limak has a string s that consists of uppercase English letters. In one move he can swap two adjacent letters of the string. For example, he can transform a string "ABBC" into "BABC" or "ABCB" in one move.
Limak wants to obtain a string without a substring "VK" (i.e. there should be no letter 'V' immediately followed by letter 'K'). It can be easily proved that it's possible for any initial string s.
What is the minimum possible number of moves Limak can do?
Input
The first line of the input contains an integer n (1 ≤ n ≤ 75) — the length of the string.
The second line contains a string s, consisting of uppercase English letters. The length of the string is equal to n.
Output
Print one integer, denoting the minimum possible number of moves Limak can do, in order to obtain a string without a substring "VK".
Examples
Input
4
VKVK
Output
3
Input
5
BVVKV
Output
2
Input
7
VVKEVKK
Output
3
Input
20
VKVKVVVKVOVKVQKKKVVK
Output
8
Input
5
LIMAK
Output
0
Note
In the first sample, the initial string is "VKVK". The minimum possible number of moves is 3. One optimal sequence of moves is:
1. Swap two last letters. The string becomes "VKKV".
2. Swap first two letters. The string becomes "KVKV".
3. Swap the second and the third letter. The string becomes "KKVV". Indeed, this string doesn't have a substring "VK".
In the second sample, there are two optimal sequences of moves. One is "BVVKV" → "VBVKV" → "VVBKV". The other is "BVVKV" → "BVKVV" → "BKVVV".
In the fifth sample, no swaps are necessary.
Submitted Solution:
```
n = int(input())
s = list(input())
count = 0
count_right_K = 0
for i in range(n):
cur_count_right_K = 0
if i + 1 < n and s[i] == 'V' and s[i + 1] == 'K':
left = i
while left >= 0 and (s[left] == 'V'):
left -= 1
right = -1
if i + 2 < n:
right = i + 2
while right < n and (s[right] == 'V' or s[right] == 'K'):
if s[right] == 'K':
cur_count_right_K += 1
right += 1
if right != -1 and i - left > right - i - 1 and set(s[i + 2:]) != set(['V', 'K']) and \
set(s[i + 2:]) != set(['K']) and set(s[i + 2:]) != set(['V']):
count += right - i - 1
for j in range(i + 1, i + 1 + right - i - 1):
if j + 1 < n:
s[j], s[j + 1] = s[j + 1], s[j]
count_right_K += cur_count_right_K
else:
count += i - left
for j in range(i + 1, i + 1 - (i - left), -1):
if j - 1 >= 0:
s[j], s[j - 1] = s[j - 1], s[j]
# print(s, count - count_right_K)
print(count - count_right_K)
``` | instruction | 0 | 68,612 | 0 | 137,224 |
No | output | 1 | 68,612 | 0 | 137,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bear Limak prepares problems for a programming competition. Of course, it would be unprofessional to mention the sponsor name in the statement. Limak takes it seriously and he is going to change some words. To make it still possible to read, he will try to modify each word as little as possible.
Limak has a string s that consists of uppercase English letters. In one move he can swap two adjacent letters of the string. For example, he can transform a string "ABBC" into "BABC" or "ABCB" in one move.
Limak wants to obtain a string without a substring "VK" (i.e. there should be no letter 'V' immediately followed by letter 'K'). It can be easily proved that it's possible for any initial string s.
What is the minimum possible number of moves Limak can do?
Input
The first line of the input contains an integer n (1 ≤ n ≤ 75) — the length of the string.
The second line contains a string s, consisting of uppercase English letters. The length of the string is equal to n.
Output
Print one integer, denoting the minimum possible number of moves Limak can do, in order to obtain a string without a substring "VK".
Examples
Input
4
VKVK
Output
3
Input
5
BVVKV
Output
2
Input
7
VVKEVKK
Output
3
Input
20
VKVKVVVKVOVKVQKKKVVK
Output
8
Input
5
LIMAK
Output
0
Note
In the first sample, the initial string is "VKVK". The minimum possible number of moves is 3. One optimal sequence of moves is:
1. Swap two last letters. The string becomes "VKKV".
2. Swap first two letters. The string becomes "KVKV".
3. Swap the second and the third letter. The string becomes "KKVV". Indeed, this string doesn't have a substring "VK".
In the second sample, there are two optimal sequences of moves. One is "BVVKV" → "VBVKV" → "VVBKV". The other is "BVVKV" → "BVKVV" → "BKVVV".
In the fifth sample, no swaps are necessary.
Submitted Solution:
```
#This code is dedicated to Olya S.
n=int(input())
s=list(input()[::-1])
c=0
for i in range(n-1):
if s[i]=='K' and s[i+1]=='V':
s[i],s[i+1]='V','K'
c+=1
print(c)
``` | instruction | 0 | 68,613 | 0 | 137,226 |
No | output | 1 | 68,613 | 0 | 137,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bear Limak prepares problems for a programming competition. Of course, it would be unprofessional to mention the sponsor name in the statement. Limak takes it seriously and he is going to change some words. To make it still possible to read, he will try to modify each word as little as possible.
Limak has a string s that consists of uppercase English letters. In one move he can swap two adjacent letters of the string. For example, he can transform a string "ABBC" into "BABC" or "ABCB" in one move.
Limak wants to obtain a string without a substring "VK" (i.e. there should be no letter 'V' immediately followed by letter 'K'). It can be easily proved that it's possible for any initial string s.
What is the minimum possible number of moves Limak can do?
Input
The first line of the input contains an integer n (1 ≤ n ≤ 75) — the length of the string.
The second line contains a string s, consisting of uppercase English letters. The length of the string is equal to n.
Output
Print one integer, denoting the minimum possible number of moves Limak can do, in order to obtain a string without a substring "VK".
Examples
Input
4
VKVK
Output
3
Input
5
BVVKV
Output
2
Input
7
VVKEVKK
Output
3
Input
20
VKVKVVVKVOVKVQKKKVVK
Output
8
Input
5
LIMAK
Output
0
Note
In the first sample, the initial string is "VKVK". The minimum possible number of moves is 3. One optimal sequence of moves is:
1. Swap two last letters. The string becomes "VKKV".
2. Swap first two letters. The string becomes "KVKV".
3. Swap the second and the third letter. The string becomes "KKVV". Indeed, this string doesn't have a substring "VK".
In the second sample, there are two optimal sequences of moves. One is "BVVKV" → "VBVKV" → "VVBKV". The other is "BVVKV" → "BVKVV" → "BKVVV".
In the fifth sample, no swaps are necessary.
Submitted Solution:
```
n = int(input())
s = "".join(['O' if x not in ('V', 'K') else x for x in input()])
ar = s.split("O")
m = len(ar)
k = 0
for i, el in enumerate(ar):
for j, c in enumerate(el):
if c != "K": continue
l = el[:j].count("V")
if i != m-1:
l = min(l, el[j+1:].count("V") + 1)
k += l
print(k)
``` | instruction | 0 | 68,614 | 0 | 137,228 |
No | output | 1 | 68,614 | 0 | 137,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bear Limak prepares problems for a programming competition. Of course, it would be unprofessional to mention the sponsor name in the statement. Limak takes it seriously and he is going to change some words. To make it still possible to read, he will try to modify each word as little as possible.
Limak has a string s that consists of uppercase English letters. In one move he can swap two adjacent letters of the string. For example, he can transform a string "ABBC" into "BABC" or "ABCB" in one move.
Limak wants to obtain a string without a substring "VK" (i.e. there should be no letter 'V' immediately followed by letter 'K'). It can be easily proved that it's possible for any initial string s.
What is the minimum possible number of moves Limak can do?
Input
The first line of the input contains an integer n (1 ≤ n ≤ 75) — the length of the string.
The second line contains a string s, consisting of uppercase English letters. The length of the string is equal to n.
Output
Print one integer, denoting the minimum possible number of moves Limak can do, in order to obtain a string without a substring "VK".
Examples
Input
4
VKVK
Output
3
Input
5
BVVKV
Output
2
Input
7
VVKEVKK
Output
3
Input
20
VKVKVVVKVOVKVQKKKVVK
Output
8
Input
5
LIMAK
Output
0
Note
In the first sample, the initial string is "VKVK". The minimum possible number of moves is 3. One optimal sequence of moves is:
1. Swap two last letters. The string becomes "VKKV".
2. Swap first two letters. The string becomes "KVKV".
3. Swap the second and the third letter. The string becomes "KKVV". Indeed, this string doesn't have a substring "VK".
In the second sample, there are two optimal sequences of moves. One is "BVVKV" → "VBVKV" → "VVBKV". The other is "BVVKV" → "BVKVV" → "BKVVV".
In the fifth sample, no swaps are necessary.
Submitted Solution:
```
#This code is dedicated to Olya S.
n=int(input())
s=list(input()[::-1])
c=0
f=1
while f==1:
f=0
for i in range(n-1):
if s[i]=='K' and s[i+1]=='V':
s[i],s[i+1]='V','K'
c+=1
f=1
print(c)
``` | instruction | 0 | 68,615 | 0 | 137,230 |
No | output | 1 | 68,615 | 0 | 137,231 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,853 | 0 | 137,706 |
"Correct Solution:
```
def change(n):
x=n[0]
count=1
ans=""
for i in n[1:]:
if i==x:count+=1
else:
ans+=str(count)
ans+=x
x=i
count=1
ans+=str(count)
ans+=x
return(ans)
while 1:
n=int(input())
if n==0:break
word=input()
for i in range(n):
word=change(word)
print(word)
``` | output | 1 | 68,853 | 0 | 137,707 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,854 | 0 | 137,708 |
"Correct Solution:
```
while True:
n = int(input())
if not n:
break
ss = [s for s in input()[::-1]]
for i in range(n):
new = []
app = new.append
last = ss.pop()
count = 1
while ss:
a = ss.pop()
if a == last:
count += 1
else:
app(str(count))
app(last)
last = a
count = 1
app(str(count))
app(last)
new = "".join(new)
ss = [s for s in new[::-1]]
# print("".join(new))
print("".join(new))
``` | output | 1 | 68,854 | 0 | 137,709 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,855 | 0 | 137,710 |
"Correct Solution:
```
while True:
n = int(input())
if not n:
break
s = input().strip()
while n:
prev, cnt, new = s[0], 1, ''
for c in s[1:]:
if c == prev:
cnt += 1
else:
new += str(cnt) + prev
prev, cnt = c, 1
new += str(cnt) + prev
s = new
n -= 1
print(s)
``` | output | 1 | 68,855 | 0 | 137,711 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,856 | 0 | 137,712 |
"Correct Solution:
```
def runlen(a):
n = len(a)
result = []
count = 1
for i in range(n):
if i == n - 1 or a[i] != a[i + 1]:
result.append((count, a[i]))
count = 1
else:
count += 1
return result
def f(s):
rl = runlen(s)
result = ""
for (count, c) in rl:
result += str(count)
result += c
return result
def apply(f, n, x):
for _ in range(n):
x = f(x)
return x
while True:
n = int(input())
if n == 0:
break
s = input().strip()
print(apply(f, n, s))
``` | output | 1 | 68,856 | 0 | 137,713 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,857 | 0 | 137,714 |
"Correct Solution:
```
def compress(digits):
compressed_list = list()
for digit in digits:
if compressed_list == [] or compressed_list[-1][1] != digit:
compressed_list.append([1, digit])
else:
compressed_list[-1][0] += 1
return compressed_list
def decompress(compressed_list):
new_digits = ""
for pair in compressed_list:
new_digits += "{}{}".format(*pair)
return new_digits
while 1:
proc_num = int(input())
if proc_num == 0:
break
digits = input().strip()
for i in range(proc_num):
compressed_list = compress(digits)
digits = decompress(compressed_list)
print(digits)
``` | output | 1 | 68,857 | 0 | 137,715 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,858 | 0 | 137,716 |
"Correct Solution:
```
def main():
while True:
n = int(input())
if not n:
break
ss = [s for s in reversed(input())]
for i in range(n):
new = []
app = new.append
last = ss.pop()
count = 1
while ss:
a = ss.pop()
if a == last:
count += 1
else:
app(str(count))
app(last)
last = a
count = 1
app(str(count))
app(last)
new = "".join(new)
ss = [s for s in reversed(new)]
print(new)
main()
``` | output | 1 | 68,858 | 0 | 137,717 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,859 | 0 | 137,718 |
"Correct Solution:
```
while True:
n=int(input())
if n==0:break
s=input()
conv=""
for _ in range(n):
seq=1
pr=s[0]
for i in range(1,len(s)):
if pr==s[i]:seq+=1
else:
conv+=str(seq)+pr
pr=s[i]
seq=1
conv+=str(seq)+pr
s=conv
conv=""
print(s)
``` | output | 1 | 68,859 | 0 | 137,719 |
Provide a correct Python 3 solution for this coding contest problem.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221 | instruction | 0 | 68,860 | 0 | 137,720 |
"Correct Solution:
```
def main():
while True:
n = int(input())
if not n:
break
ss = [s for s in input()[::-1]]
for i in range(n):
new = []
ext = new.extend
last = ss.pop()
count = 1
while ss:
a = ss.pop()
if a == last:
count += 1
else:
ext([str(count),last])
last = a
count = 1
ext([str(count), last])
new = "".join(new)
ss = [s for s in new[::-1]]
print(new)
main()
``` | output | 1 | 68,860 | 0 | 137,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a string consisting of only numbers from 0 to 9, consider the operation of creating a new string from that string according to the following rules. Read the given string one character at a time from the left end. Go, if the same number a continues r, write the number r and the number a in this order without separating them with a space. Read to the right end of the given character string, and what is on the way to the end of the last writing Even if there are times of writing, all of them are counted as one operation. For the second and subsequent operations, the same operation is performed with the character string written out by the previous operation as the given character string. For example, "122244" If the character string "" is given, the character string obtained by one operation is "113224" and "44444444444" (11) because one 1, three 2, two 4s are given in order from the left end. In the case of 4), the obtained character string is “114”.
Create a program that outputs a string obtained by performing the above operation n times on a given string of 100 characters or less, where n ≤ 20.
The input data consists of two lines, the first line contains the number of operations n, and the second line contains the first character string.
Input example
---
Five
11
Output example
13112221
input
The input consists of multiple datasets. Input ends when n is 0. The number of datasets does not exceed 5.
output
For each data set, the character string that has been operated the specified number of times is output on one line.
Example
Input
5
11
5
11
0
Output
13112221
13112221
Submitted Solution:
```
while True:
n = int(input())
if not n:
break
ss = [s for s in input()[::-1]]
for i in range(n):
new = []
last = ss.pop()
count = 1
while ss:
a = ss.pop()
if a == last:
count += 1
else:
new.append(last)
new.append(str(count))
last = a
count = 1
else:
new.append(last)
new.append(str(count))
ss = [s for s in new[::-1]]
print("".join(ss))
``` | instruction | 0 | 68,867 | 0 | 137,734 |
No | output | 1 | 68,867 | 0 | 137,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long ago, when Petya was a schoolboy, he was very much interested in the Petr# language grammar. During one lesson Petya got interested in the following question: how many different continuous substrings starting with the sbegin and ending with the send (it is possible sbegin = send), the given string t has. Substrings are different if and only if their contents aren't equal, their positions of occurence don't matter. Petya wasn't quite good at math, that's why he couldn't count this number. Help him!
Input
The input file consists of three lines. The first line contains string t. The second and the third lines contain the sbegin and send identificators, correspondingly. All three lines are non-empty strings consisting of lowercase Latin letters. The length of each string doesn't exceed 2000 characters.
Output
Output the only number — the amount of different substrings of t that start with sbegin and end with send.
Examples
Input
round
ro
ou
Output
1
Input
codeforces
code
forca
Output
0
Input
abababab
a
b
Output
4
Input
aba
ab
ba
Output
1
Note
In the third sample there are four appropriate different substrings. They are: ab, abab, ababab, abababab.
In the fourth sample identificators intersect. | instruction | 0 | 68,982 | 0 | 137,964 |
Tags: brute force, data structures, hashing, strings
Correct Solution:
```
from functools import cmp_to_key
def calc_lcp(s, sa):
rank = [0 for _ in range(len(s))]
for i in range(len(s)):
rank[sa[i]] = i
lcp = [0 for _ in range(len(s) - 1)]
h = 0
for i in range(len(s)):
if rank[i] < len(s) - 1:
while max(i, sa[rank[i] + 1]) + h < len(s) and s[i + h] == s[sa[rank[i] + 1] + h]:
h += 1
lcp[rank[i]] = h
if h > 0:
h -= 1
return lcp, rank
def suffix_array(s):
s += chr(0)
sa = [i for i in range(len(s))]
rank = [ord(s[i]) for i in range(len(s))]
k = 1
def cmp(a, b):
if rank[a] != rank[b]:
return rank[a] - rank[b]
return rank[a + k] - rank[b + k]
while k < len(s):
sa.sort(key=cmp_to_key(cmp))
new_rank = [0 for _ in range(len(s))]
for i in range(1, len(s)):
new_rank[sa[i]] = new_rank[sa[i - 1]] if cmp(sa[i - 1], sa[i]) == 0 else new_rank[sa[i - 1]] + 1
k *= 2
rank = new_rank
return sa[1:]
def kmp(s, p):
pi = [0 for _ in range(len(p))]
k = 0
for i in range(1, len(p)):
while k > 0 and p[k] != p[i]:
k = pi[k - 1]
if p[k] == p[i]:
k += 1
pi[i] = k
k = 0
resp = []
for i in range(len(s)):
while k > 0 and p[k] != s[i]:
k = pi[k - 1]
if p[k] == s[i]:
k += 1
if k == len(p):
resp.append(i - len(p) + 1)
k = pi[k - 1]
return resp
def lower_bound(list, value):
left = 0
right = len(list)
while left < right:
mid = int((left + right) / 2)
if list[mid] < value:
left = mid + 1
else:
right = mid
return left
s = input()
start = input()
end = input()
indStart = kmp(s, start)
indEnd = kmp(s, end)
if len(indStart) == 0 or len(indEnd) == 0:
print(0)
else:
sa = suffix_array(s)
lcp, rank = calc_lcp(s, sa)
ind = rank[indStart[0]]
for st in indStart:
ind = min(ind, rank[st])
resp = len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, len(start) - len(end)))
while ind < len(lcp) and lcp[ind] >= len(start):
ind += 1
resp += len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, max(lcp[ind - 1] + 1, len(start)) - len(end)))
print(resp)
``` | output | 1 | 68,982 | 0 | 137,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long ago, when Petya was a schoolboy, he was very much interested in the Petr# language grammar. During one lesson Petya got interested in the following question: how many different continuous substrings starting with the sbegin and ending with the send (it is possible sbegin = send), the given string t has. Substrings are different if and only if their contents aren't equal, their positions of occurence don't matter. Petya wasn't quite good at math, that's why he couldn't count this number. Help him!
Input
The input file consists of three lines. The first line contains string t. The second and the third lines contain the sbegin and send identificators, correspondingly. All three lines are non-empty strings consisting of lowercase Latin letters. The length of each string doesn't exceed 2000 characters.
Output
Output the only number — the amount of different substrings of t that start with sbegin and end with send.
Examples
Input
round
ro
ou
Output
1
Input
codeforces
code
forca
Output
0
Input
abababab
a
b
Output
4
Input
aba
ab
ba
Output
1
Note
In the third sample there are four appropriate different substrings. They are: ab, abab, ababab, abababab.
In the fourth sample identificators intersect. | instruction | 0 | 68,983 | 0 | 137,966 |
Tags: brute force, data structures, hashing, strings
Correct Solution:
```
from functools import cmp_to_key
def calc_lcp(s, sa):
rank = [0 for _ in range(len(s))]
for i in range(len(s)):
rank[sa[i]] = i
lcp = [0 for _ in range(len(s) - 1)]
h = 0
for i in range(len(s)):
if rank[i] < len(s) - 1:
while max(i, sa[rank[i] + 1]) + h < len(s) and s[i + h] == s[sa[rank[i] + 1] + h]:
h += 1
lcp[rank[i]] = h
if h > 0:
h -= 1
return lcp, rank
def suffix_array(s):
sa = [i for i in range(len(s))]
rank = [ord(s[i]) for i in range(len(s))]
k = 1
while k < len(s):
key = [0 for _ in range(len(s))]
base = max(rank) + 2
for i in range(len(s)):
key[i] = rank[i] * base + (rank[i + k] + 1 if i + k < len(s) else 0)
sa.sort(key=(lambda i: key[i]))
rank[sa[0]] = 0
for i in range(1, len(s)):
rank[sa[i]] = rank[sa[i - 1]] if key[sa[i - 1]] == key[sa[i]] else i
k *= 2
# for i in sa:
# print(s[i:])
return sa
def kmp(s, p):
pi = [0 for _ in range(len(p))]
k = 0
for i in range(1, len(p)):
while k > 0 and p[k] != p[i]:
k = pi[k - 1]
if p[k] == p[i]:
k += 1
pi[i] = k
k = 0
resp = []
for i in range(len(s)):
while k > 0 and p[k] != s[i]:
k = pi[k - 1]
if p[k] == s[i]:
k += 1
if k == len(p):
resp.append(i - len(p) + 1)
k = pi[k - 1]
return resp
def lower_bound(list, value):
left = 0
right = len(list)
while left < right:
mid = int((left + right) / 2)
if list[mid] < value:
left = mid + 1
else:
right = mid
return left
s = input()
start = input()
end = input()
indStart = kmp(s, start)
indEnd = kmp(s, end)
if len(indStart) == 0 or len(indEnd) == 0:
print(0)
else:
sa = suffix_array(s)
lcp, rank = calc_lcp(s, sa)
ind = rank[indStart[0]]
for st in indStart:
ind = min(ind, rank[st])
resp = len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, len(start) - len(end)))
while ind < len(lcp) and lcp[ind] >= len(start):
ind += 1
resp += len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, max(lcp[ind - 1] + 1, len(start)) - len(end)))
print(resp)
# Made By Mostafa_Khaled
``` | output | 1 | 68,983 | 0 | 137,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long ago, when Petya was a schoolboy, he was very much interested in the Petr# language grammar. During one lesson Petya got interested in the following question: how many different continuous substrings starting with the sbegin and ending with the send (it is possible sbegin = send), the given string t has. Substrings are different if and only if their contents aren't equal, their positions of occurence don't matter. Petya wasn't quite good at math, that's why he couldn't count this number. Help him!
Input
The input file consists of three lines. The first line contains string t. The second and the third lines contain the sbegin and send identificators, correspondingly. All three lines are non-empty strings consisting of lowercase Latin letters. The length of each string doesn't exceed 2000 characters.
Output
Output the only number — the amount of different substrings of t that start with sbegin and end with send.
Examples
Input
round
ro
ou
Output
1
Input
codeforces
code
forca
Output
0
Input
abababab
a
b
Output
4
Input
aba
ab
ba
Output
1
Note
In the third sample there are four appropriate different substrings. They are: ab, abab, ababab, abababab.
In the fourth sample identificators intersect. | instruction | 0 | 68,984 | 0 | 137,968 |
Tags: brute force, data structures, hashing, strings
Correct Solution:
```
def calc_lcp(s, sa):
rank = [0 for _ in range(len(s))]
for i in range(len(s)):
rank[sa[i]] = i
lcp = [0 for _ in range(len(s) - 1)]
h = 0
for i in range(len(s)):
if rank[i] < len(s) - 1:
while s[i + h] == s[sa[rank[i] + 1] + h]:
h += 1
lcp[rank[i]] = h
if h > 0:
h -= 1
return lcp, rank
def suffix_array(s):
n = len(s)
na = max(n, 256)
sa = [0 for _ in range(n)]
top = [0 for _ in range(na)]
rank = [0 for _ in range(n)]
sa_new = [0 for _ in range(n)]
rank_new = [0 for _ in range(n)]
for i in range(n):
rank[i] = ord(s[i])
top[rank[i]] += 1
for i in range(1, na):
top[i] += top[i - 1]
for i in range(n):
top[rank[i]] -= 1
sa[top[rank[i]]] = i
k = 1
while k < n:
for i in range(n):
j = sa[i] - k
if j < 0:
j += n
sa_new[top[rank[j]]] = j
top[rank[j]] += 1
rank_new[sa_new[0]] = 0
top[0] = 0
cnt = 0
for i in range(1, n):
if rank[sa_new[i]] != rank[sa_new[i - 1]] or rank[sa_new[i] + k] != rank[sa_new[i - 1] + k]:
cnt += 1
top[cnt] = i
rank_new[sa_new[i]] = cnt
sa, sa_new = sa_new, sa
rank, rank_new = rank_new, rank
if cnt == n - 1:
break
k *= 2
return sa
def kmp(s, p):
pi = [0 for _ in range(len(p))]
k = 0
for i in range(1, len(p)):
while k > 0 and p[k] != p[i]:
k = pi[k - 1]
if p[k] == p[i]:
k += 1
pi[i] = k
k = 0
resp = []
for i in range(len(s)):
while k > 0 and p[k] != s[i]:
k = pi[k - 1]
if p[k] == s[i]:
k += 1
if k == len(p):
resp.append(i - len(p) + 1)
k = pi[k - 1]
return resp
def lower_bound(list, value):
left = 0
right = len(list)
while left < right:
mid = int((left + right) / 2)
if list[mid] < value:
left = mid + 1
else:
right = mid
return left
s = input()
start = input()
end = input()
indStart = kmp(s, start)
indEnd = kmp(s, end)
if len(indStart) == 0 or len(indEnd) == 0:
print(0)
else:
s += chr(0)
sa = suffix_array(s)
lcp, rank = calc_lcp(s, sa)
ind = rank[indStart[0]]
for st in indStart:
ind = min(ind, rank[st])
resp = len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, len(start) - len(end)))
while ind < len(lcp) and lcp[ind] >= len(start):
ind += 1
resp += len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, max(lcp[ind - 1] + 1, len(start)) - len(end)))
print(resp)
``` | output | 1 | 68,984 | 0 | 137,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long ago, when Petya was a schoolboy, he was very much interested in the Petr# language grammar. During one lesson Petya got interested in the following question: how many different continuous substrings starting with the sbegin and ending with the send (it is possible sbegin = send), the given string t has. Substrings are different if and only if their contents aren't equal, their positions of occurence don't matter. Petya wasn't quite good at math, that's why he couldn't count this number. Help him!
Input
The input file consists of three lines. The first line contains string t. The second and the third lines contain the sbegin and send identificators, correspondingly. All three lines are non-empty strings consisting of lowercase Latin letters. The length of each string doesn't exceed 2000 characters.
Output
Output the only number — the amount of different substrings of t that start with sbegin and end with send.
Examples
Input
round
ro
ou
Output
1
Input
codeforces
code
forca
Output
0
Input
abababab
a
b
Output
4
Input
aba
ab
ba
Output
1
Note
In the third sample there are four appropriate different substrings. They are: ab, abab, ababab, abababab.
In the fourth sample identificators intersect. | instruction | 0 | 68,985 | 0 | 137,970 |
Tags: brute force, data structures, hashing, strings
Correct Solution:
```
def calc_lcp(s, sa):
rank = [0 for _ in range(len(s))]
for i in range(len(s)):
rank[sa[i]] = i
lcp = [0 for _ in range(len(s) - 1)]
h = 0
for i in range(len(s)):
if rank[i] < len(s) - 1:
while max(i, sa[rank[i] + 1]) + h < len(s) and s[i + h] == s[sa[rank[i] + 1] + h]:
h += 1
lcp[rank[i]] = h
if h > 0:
h -= 1
return lcp, rank
def suffix_array(s):
s += '\0'
sa = [0 for _ in range(len(s))]
cnt256 = [0 for _ in range(256)]
for c in s:
cnt256[ord(c)] += 1
for i in range(1, 256):
cnt256[i] += cnt256[i - 1]
for i in range(len(s) - 1, -1, -1):
cnt256[ord(s[i])] -= 1
sa[cnt256[ord(s[i])]] = i
rank = 0
ranks = [0 for _ in range(len(s))]
for i in range(1, len(s)):
if s[sa[i - 1]] != s[sa[i]]:
rank += 1
ranks[sa[i]] = rank
k = 1
while k < len(s):
sa_new = [0 for _ in range(len(s))]
rank_new = [0 for _ in range(len(s))]
for i in range(len(s)):
sa_new[i] = sa[i] - k
if sa_new[i] < 0:
sa_new[i] += len(s)
cnt = [0 for _ in range(len(s))]
for i in range(len(s)):
cnt[ranks[i]] += 1
for i in range(1, len(s)):
cnt[i] += cnt[i - 1]
for i in range(len(s) - 1, -1, -1):
cnt[ranks[sa_new[i]]] -= 1
sa[cnt[ranks[sa_new[i]]]] = sa_new[i]
rank = 0
for i in range(1, len(s)):
if ranks[sa[i - 1]] != ranks[sa[i]] or ranks[sa[i - 1] + k] != ranks[sa[i] + k]:
rank += 1
rank_new[sa[i]] = rank
ranks = rank_new
k *= 2
return sa[1:]
def kmp(s, p):
pi = [0 for _ in range(len(p))]
k = 0
for i in range(1, len(p)):
while k > 0 and p[k] != p[i]:
k = pi[k - 1]
if p[k] == p[i]:
k += 1
pi[i] = k
k = 0
resp = []
for i in range(len(s)):
while k > 0 and p[k] != s[i]:
k = pi[k - 1]
if p[k] == s[i]:
k += 1
if k == len(p):
resp.append(i - len(p) + 1)
k = pi[k - 1]
return resp
def lower_bound(list, value):
left = 0
right = len(list)
while left < right:
mid = int((left + right) / 2)
if list[mid] < value:
left = mid + 1
else:
right = mid
return left
s = input()
start = input()
end = input()
indStart = kmp(s, start)
indEnd = kmp(s, end)
if len(indStart) == 0 or len(indEnd) == 0:
print(0)
else:
sa = suffix_array(s)
lcp, rank = calc_lcp(s, sa)
ind = rank[indStart[0]]
for st in indStart:
ind = min(ind, rank[st])
resp = len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, len(start) - len(end)))
while ind < len(lcp) and lcp[ind] >= len(start):
ind += 1
resp += len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, max(lcp[ind - 1] + 1, len(start)) - len(end)))
print(resp)
``` | output | 1 | 68,985 | 0 | 137,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long ago, when Petya was a schoolboy, he was very much interested in the Petr# language grammar. During one lesson Petya got interested in the following question: how many different continuous substrings starting with the sbegin and ending with the send (it is possible sbegin = send), the given string t has. Substrings are different if and only if their contents aren't equal, their positions of occurence don't matter. Petya wasn't quite good at math, that's why he couldn't count this number. Help him!
Input
The input file consists of three lines. The first line contains string t. The second and the third lines contain the sbegin and send identificators, correspondingly. All three lines are non-empty strings consisting of lowercase Latin letters. The length of each string doesn't exceed 2000 characters.
Output
Output the only number — the amount of different substrings of t that start with sbegin and end with send.
Examples
Input
round
ro
ou
Output
1
Input
codeforces
code
forca
Output
0
Input
abababab
a
b
Output
4
Input
aba
ab
ba
Output
1
Note
In the third sample there are four appropriate different substrings. They are: ab, abab, ababab, abababab.
In the fourth sample identificators intersect. | instruction | 0 | 68,986 | 0 | 137,972 |
Tags: brute force, data structures, hashing, strings
Correct Solution:
```
from functools import cmp_to_key
def calc_lcp(s, sa):
rank = [0 for _ in range(len(s))]
for i in range(len(s)):
rank[sa[i]] = i
lcp = [0 for _ in range(len(s) - 1)]
h = 0
for i in range(len(s)):
if rank[i] < len(s) - 1:
while max(i, sa[rank[i] + 1]) + h < len(s) and s[i + h] == s[sa[rank[i] + 1] + h]:
h += 1
lcp[rank[i]] = h
if h > 0:
h -= 1
return lcp, rank
def suffix_array(s):
rank = [ord(d) for d in s]
k = 1
while k < len(s):
suffixes = [((rank[i], rank[i + k] if i + k < len(s) else -1), i) for i in range(len(s))]
suffixes.sort()
cnt = 0
rank[suffixes[0][1]] = 0
for i in range(1, len(s)):
if suffixes[i][0] != suffixes[i - 1][0]:
cnt += 1
rank[suffixes[i][1]] = cnt
if cnt == len(s) - 1:
break
k *= 2
sa = [0 for _ in s]
for i in range(len(s)):
sa[rank[i]] = i
# print(sa)
return sa
def kmp(s, p):
pi = [0 for _ in range(len(p))]
k = 0
for i in range(1, len(p)):
while k > 0 and p[k] != p[i]:
k = pi[k - 1]
if p[k] == p[i]:
k += 1
pi[i] = k
k = 0
resp = []
for i in range(len(s)):
while k > 0 and p[k] != s[i]:
k = pi[k - 1]
if p[k] == s[i]:
k += 1
if k == len(p):
resp.append(i - len(p) + 1)
k = pi[k - 1]
return resp
def lower_bound(list, value):
left = 0
right = len(list)
while left < right:
mid = int((left + right) / 2)
if list[mid] < value:
left = mid + 1
else:
right = mid
return left
s = input()
start = input()
end = input()
indStart = kmp(s, start)
indEnd = kmp(s, end)
if len(indStart) == 0 or len(indEnd) == 0:
print(0)
else:
sa = suffix_array(s)
lcp, rank = calc_lcp(s, sa)
ind = rank[indStart[0]]
for st in indStart:
ind = min(ind, rank[st])
resp = len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, len(start) - len(end)))
while ind < len(lcp) and lcp[ind] >= len(start):
ind += 1
resp += len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, max(lcp[ind - 1] + 1, len(start)) - len(end)))
print(resp)
``` | output | 1 | 68,986 | 0 | 137,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Long ago, when Petya was a schoolboy, he was very much interested in the Petr# language grammar. During one lesson Petya got interested in the following question: how many different continuous substrings starting with the sbegin and ending with the send (it is possible sbegin = send), the given string t has. Substrings are different if and only if their contents aren't equal, their positions of occurence don't matter. Petya wasn't quite good at math, that's why he couldn't count this number. Help him!
Input
The input file consists of three lines. The first line contains string t. The second and the third lines contain the sbegin and send identificators, correspondingly. All three lines are non-empty strings consisting of lowercase Latin letters. The length of each string doesn't exceed 2000 characters.
Output
Output the only number — the amount of different substrings of t that start with sbegin and end with send.
Examples
Input
round
ro
ou
Output
1
Input
codeforces
code
forca
Output
0
Input
abababab
a
b
Output
4
Input
aba
ab
ba
Output
1
Note
In the third sample there are four appropriate different substrings. They are: ab, abab, ababab, abababab.
In the fourth sample identificators intersect. | instruction | 0 | 68,987 | 0 | 137,974 |
Tags: brute force, data structures, hashing, strings
Correct Solution:
```
def calc_lcp(s, sa):
rank = [0 for _ in range(len(s))]
for i in range(len(s)):
rank[sa[i]] = i
lcp = [0 for _ in range(len(s) - 1)]
h = 0
for i in range(len(s)):
if rank[i] < len(s) - 1:
while max(i, sa[rank[i] + 1]) + h < len(s) and s[i + h] == s[sa[rank[i] + 1] + h]:
h += 1
lcp[rank[i]] = h
if h > 0:
h -= 1
return lcp, rank
def suffix_array(s):
def countinSort(array, key):
max_val = max(key)
cnt = [0 for _ in range(max_val + 1)]
for i in key:
cnt[i] += 1
for i in range(1, len(cnt)):
cnt[i] += cnt[i - 1]
resp = [0 for _ in array]
for i in range(len(array) - 1, -1, -1):
cnt[key[array[i]]] -= 1
resp[cnt[key[array[i]]]] = array[i]
return resp
sa = [i for i in range(len(s))]
ranks = [ord(c) for c in s]
k = 1
while k < len(s):
sa = countinSort(sa, [ranks[i + k] if i + k < len(s) else -1 for i in range(len(s))])
sa = countinSort(sa, ranks)
rank_new = [0 for _ in range(len(s))]
for i in range(1, len(s)):
if ranks[sa[i - 1]] == ranks[sa[i]] and sa[i] + k < len(s) and sa[i - 1] + k < len(s) and ranks[sa[i - 1] + k] == ranks[sa[i] + k]:
rank_new[sa[i]] = rank_new[sa[i - 1]]
else:
rank_new[sa[i]] = i
ranks = rank_new
k *= 2
return sa
def kmp(s, p):
pi = [0 for _ in range(len(p))]
k = 0
for i in range(1, len(p)):
while k > 0 and p[k] != p[i]:
k = pi[k - 1]
if p[k] == p[i]:
k += 1
pi[i] = k
k = 0
resp = []
for i in range(len(s)):
while k > 0 and p[k] != s[i]:
k = pi[k - 1]
if p[k] == s[i]:
k += 1
if k == len(p):
resp.append(i - len(p) + 1)
k = pi[k - 1]
return resp
def lower_bound(list, value):
left = 0
right = len(list)
while left < right:
mid = int((left + right) / 2)
if list[mid] < value:
left = mid + 1
else:
right = mid
return left
s = input()
start = input()
end = input()
indStart = kmp(s, start)
indEnd = kmp(s, end)
if len(indStart) == 0 or len(indEnd) == 0:
print(0)
else:
sa = suffix_array(s)
lcp, rank = calc_lcp(s, sa)
ind = rank[indStart[0]]
for st in indStart:
ind = min(ind, rank[st])
resp = len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, len(start) - len(end)))
while ind < len(lcp) and lcp[ind] >= len(start):
ind += 1
resp += len(indEnd) - lower_bound(indEnd, sa[ind] + max(0, max(lcp[ind - 1] + 1, len(start)) - len(end)))
print(resp)
``` | output | 1 | 68,987 | 0 | 137,975 |
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