Given a binary matrix of **N** rows and **M** columns. The operation allowed on the matrix is to choose any index (x, y) and toggle all the elements between the rectangle having top-left as (0, 0) and bottom-right as (x-1, y-1). Toggling the element means changing 1 to 0 and 0 to 1. The task is to find minimum operations required to make set all the elements of the matrix i.e make all elements as 1.

Examples:

Input : mat[][] = 0 0 0 1 1 0 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 Output : 1 In one move, choose (3, 3) to make the whole matrix consisting of only 1s. Input : mat[][] = 0 011 1 0 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 Output : 3

The idea is to start from the end point (N – 1, M – 1) and traverse the matrix in reverse order. Whenever we encounter a cell which has a value of 0, flip it.

*Why traversing from end point ?*

Suppose there are 0 at (x, y) and (x + 1, y + 1) cell. You shouldn’t flip a cell (x + 1, y + 1) after cell (x, y) because after you flipped (x, y) to 1, in next move to flip (x + 1, y + 1) cell, you will flip again (x, y) to 0. So there is no benefit from the first move for flipping (x, y) cell.

Below is C++ implementation of this approach:

// C++ program to find minimum operations required // to set all the element of binary matrix #include <bits/stdc++.h> #define N 5 #define M 5 using namespace std; // Return minimum operation required to make all 1s. int minOperation(bool arr[N][M]) { int ans = 0; for (int i = N - 1; i >= 0; i--) { for (int j = M - 1; j >= 0; j--) { // check if this cell equals 0 if(arr[i][j] == 0) { // increase the number of moves ans++; // flip from this cell to the start point for (int k = 0; k <= i; k++) { for (int h = 0; h <= j; h++) { // flip the cell if (arr[k][h] == 1) arr[k][h] = 0; else arr[k][h] = 1; } } } } } return ans; } // Driven Program int main() { bool mat[N][M] = { 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }; cout << minOperation(mat) << endl; return 0; }

Output:

3

**Time Complexity: **O(N^{2} * M^{2}).

**Space Complexity: **O(N*M).

This article is contributed by **Anuj Chauhan**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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