Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>

using namespace std;

int main()
{
   int n, k, i, b[100000], m=1, j=0, p[100000], v, z;
   char a[110000];
   scanf("%d %d",&n,&k);
   getchar();
   scanf("%s",a);
   for(i=0;i<n-1;i++)
   {
       if(a[i]!=a[i+1])
       {
           b[j]=m;
           m=1;
           j++;
       }
       else
       {
           m++;
       }
   }
   if(a[n-1]!=a[n-2])
   {
        b[j]=m;
        m=1;
        j++;
   }
   memset(p,0,sizeof(b));
   for(i=0;i<j-2*k;i++)
   {
       if(a[0]=='0'&&i==0&&i+2*k==j-1)
       {
           if(b[j-1]>b[0])
           {
               for(v=i+1;v<=i+2*k;v++)
               {
                 p[i]=p[i]+b[v];
               }
           }
           else
           {
               for(v=i;v<i+2*k;v++)
               {
                p[i]=p[i]+b[v];
               }
           }
            break;
       }
       if(a[0]=='0'&&i==0)
       {
           for(v=i;v<i+2*k;v++)
           {
           p[i]=p[i]+b[v];
           }
           continue;
       }
       if(a[n-1]=='0'&&i==j-2*k-1)
       {
           for(v=i+1;v<i+2*k;v++)
           {
           p[i]=p[i]+b[v];
           }
           break;
       }
       for(v=i;v<i+2*k+1;v++)
       {
           p[i]=p[i]+b[v];
       }
   }
   z=p[0];
   for(i=1;p[i]!=0;i++)
   {
       if(z<p[i])
        z=p[i];
   }
   printf("%d\n",z);
   return 0;
}