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

Related: Find zeroes to be flipped so that number of consecutive 1's is maximized
Find zeroes to be flipped so that number of consecutive 1's is maximized - GeeksforGeeks
Given a binary array and an integer m, find the position of zeroes flipping which creates maximum number of consecutive 1s in array.

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 

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] = 6 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.

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 << " ";

    }

}

Related: Find number of bits to be flipped to get maximum 1’s in array
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