How can i approach this problem?

#1

please give me some hint

#2

use the approach of Sliding Window for the given array. use a window covering from index Left to index Right. Let the number of zeros inside the window be zeroCount. We maintain the window with at most m zeros inside.

  • While zeroCount is no more than m: expand the window to the right and update the count zeroCount.
  • While zeroCount exceeds m, shrink the window from left , update zeroCount;
  • Update the widest window along the way. The positions of output zeros are inside the best window.
#3

i am applying same approach but not passing all test cases

#4

@aa1

 import java.util.*;
public class Main {
  	public static long count(int arr[], int k){
	int n=arr.length;
	int left=0;
	int zero=0;
	long max_len=0;

	for(int right=0;right<n;right++){
		if(arr[right]==0){
			zero++;
		}
		while(zero>k){
			if(arr[left]==0)
				zero--;
			left++;
		}

		max_len=Math.max(max_len,right-left+1);
	}
	return max_len;
}
public static void main(String args[]) {
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	int k=sc.nextInt();
	int arr[]=new int[n];
	for(int i=0;i<n;i++){
		arr[i]=sc.nextInt();

	}
	System.out.println(count(arr,k));



}

}

#5

print the Array after flipping k 0s too in second line

#6

how can i know which zero need to flip??

#7

you are updating the window to find a way such that the positions of output zeros are inside the best window. and you print the count there .
just print this array after final change