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

Given a binary array and an integer m, find the position of zeroes flipping which creates maximum number of consecutive 1’s in array.

Examples:

```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.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A Simple Solution is to consider every subarray by running two loops. For every subarray, count number of zeroes in it. Return the maximum size subarray with m or less zeroes. Time Complexity of this solution is O(n2).

A Better Solution is to use auxiliary space to solve the problem in O(n) time.

For all positions of 0’s calculate left[] and right[] which defines the number of consecutive 1’s to the left of i and right of i respectively.

For example, for arr[] = {1, 1, 0, 1, 1, 0, 0, 1, 1, 1} and m = 1, left[2] = 2 and right[2] = 2, left[5] = 2 and right[5] = 0, left[6] = 0 and right[6] = 3.

left[] and right[] can be filled in O(n) time by traversing array once and keeping track of last seen 1 and last seen 0. While filling left[] and right[], we also store indexes of all zeroes in a third array say zeroes[]. For above example, this third array stores {2, 5, 6}

Now traverse zeroes[] and for all consecutive m entries in this array, compute the sum of 1s that can be produced. This step can be done in O(n) using left[] and right[].

An Efficient Solution can solve the problem in O(n) time and O(1) space. The idea is to use Sliding Window for the given array. The solution is taken from here.
Let us use a window covering from index wL to index wR. Let the number of zeros inside the window be zeroCount. We maintain the window with at most m zeros inside.

The main steps are:
– While zeroCount is no more than m: expand the window to the right (wR++) and update the count zeroCount.
– While zeroCount exceeds m, shrink the window from left (wL++), update zeroCount;
– Update the widest window along the way. The positions of output zeros are inside the best window.

Below is C++ and Java implementation of the idea.

## C++

```// C++ program to find positions of zeroes flipping which
// produces maximum number of xonsecutive 1's
#include<bits/stdc++.h>
using namespace std;

// m is maximum of number zeroes allowed to flip
// n is size of array
void findZeroes(int arr[], int n, int m)
{
// Left and right indexes of current window
int wL = 0, wR = 0;

// Left index and size of the widest window
int bestL = 0, bestWindow = 0;

// Count of zeroes in current window
int zeroCount = 0;

// While right boundary of current window doesn't cross
// right end
while (wR < n)
{
// If zero count of current window is less than m,
// widen the window toward right
if (zeroCount <= m)
{
if (arr[wR] == 0)
zeroCount++;
wR++;
}

// If zero count of current window is more than m,
// reduce the window from left
if (zeroCount > m)
{
if (arr[wL] == 0)
zeroCount--;
wL++;
}

// Updqate widest window if this window size is more
if (wR-wL > bestWindow)
{
bestWindow = wR-wL;
bestL = wL;
}
}

// Print positions of zeroes in the widest window
for (int i=0; i<bestWindow; i++)
{
if (arr[bestL+i] == 0)
cout << bestL+i << " ";
}
}

// Driver program
int main()
{
int arr[] = {1, 0, 0, 1, 1, 0, 1, 0, 1, 1};
int m = 2;
int n =  sizeof(arr)/sizeof(arr[0]);
cout << "Indexes of zeroes to be flipped are ";
findZeroes(arr, n, m);
return 0;
}
```

## Java

```//Java to find positions of zeroes flipping which
// produces maximum number of consecutive 1's
class Test
{
static int arr[] = new int[]{1, 0, 0, 1, 1, 0, 1, 0, 1, 1};

// m is maximum of number zeroes allowed to flip
static void findZeroes(int m)
{
// Left and right indexes of current window
int wL = 0, wR = 0;

// Left index and size of the widest window
int bestL = 0, bestWindow = 0;

// Count of zeroes in current window
int zeroCount = 0;

// While right boundary of current window doesn't cross
// right end
while (wR < arr.length)
{
// If zero count of current window is less than m,
// widen the window toward right
if (zeroCount <= m)
{
if (arr[wR] == 0)
zeroCount++;
wR++;
}

// If zero count of current window is more than m,
// reduce the window from left
if (zeroCount > m)
{
if (arr[wL] == 0)
zeroCount--;
wL++;
}

// Update widest window if this window size is more
if (wR-wL > bestWindow)
{
bestWindow = wR-wL;
bestL = wL;
}
}

// Print positions of zeroes in the widest window
for (int i=0; i<bestWindow; i++)
{
if (arr[bestL+i] == 0)
System.out.print(bestL+i + " ");
}
}

// Driver method to test the above function
public static void main(String[] args)
{
int m = 2;
System.out.println("Indexes of zeroes to be flipped are ");

findZeroes(m);
}
}
```

Output:
`Indexes of zeroes to be flipped are 5 7`

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.7 Average Difficulty : 3.7/5.0
Based on 139 vote(s)