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: Accepted)

#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define N 1000010
int a[N],b[N],ca=0,cb=0;
char s[100010];
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    scanf("%s",s+1);
    int flag1=0,flag0=0,ok=0;
    if(s[1]=='1')ok=1;
     int ans=0;
    for(int i=1;i<=n;i++)
        if(s[i]=='1')
        {
            flag0=0;
            if(!flag1)ca++,flag1=1;
            a[ca]++;
            ans=max(ans,a[ca]);
        }
        else
        {
            ans=max(ans,a[ca]);
            flag1=0;
            if(!flag0)cb++,flag0=1;
            b[cb]++;
        }
     for(int i=2;i<=ca;i++)a[i]+=a[i-1];
     for(int i=2;i<=cb;i++)b[i]+=b[i-1];

    if(k>=cb)ans=n;
    for(int i=1;i+k-1<=cb;i++)
    {
        int r=i+k-1;
        int lb=b[r]-b[i-1];
        int x,y;
        if(ok)
            x=max(1,i),y=min(ca,i+k);
        else
            x=max(1,i-1),y=min(ca,i+k-1);
        int la=a[y]-a[x-1];
        ans=max(ans,la+lb);
    }
    printf("%d\n",ans);
    return 0;
}