Find zeroes to be flipped so that number of consecutive 1's is maximized 



Input:   arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1}
         m = 2
Output:  5 7
We are allowed to flip maximum 2 zeroes. If we flip
arr[5] and arr[7], we get 8 consecutive 1's which is
maximum possible under given constraints 

Input:   arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1}
         m = 1
Output:  7
We are allowed to flip maximum 1 zero. If we flip 
arr[7], we get 5 consecutive 1's which is maximum 
possible under given constraints.

Input:   arr[] = {0, 0, 0, 1}
         m = 4
Output:  0 1 2
Since m is more than number of zeroes, we can flip
all zeroes.

Steps:

It is based on sliding window method.
The window at any time should consists of only m zeroes.
If the zerocount is below m wr is incremented.
If the zerocount is more wl is incremented.

Code:

import java.util.*;

public class Solution {

 
    public static void main(String[] args) {
        int[] a = {1,1,0,1,1,0,0,1,1,1};
        int m = 2;
        int n = a.length;
        int zerocount = 0;
        int wr = 0, wl=0;
        int bestl = 0, bestr = 0, size = 0;
        int check = 0;
     
        while(wr<n){
         
           if(zerocount <= m){
               if(a[wr] == 0)
                   zerocount++;
               wr++;
           }

           if(zerocount > m){
               if(a[wl] == 0)
                   zerocount--;
               wl++;
           }
            if( wr - wl > size){
                size = wr - wl;
                bestl = wl;
                bestr = wr;
            }
         
        }
     
        for(int i=bestl; i<bestr; i++){
            if(a[i] == 0){
                System.out.print(i+" ");
            }
         
        }
     
 
    }
}

Output:
5 6

Illustration:




References:

 https://www.geeksforgeeks.org/window-sliding-technique/

https://www.geeksforgeeks.org/find-zeroes-to-be-flipped-so-that-number-of-consecutive-1s-is-maximized/

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