Maximize the binary matrix by filpping submatrix once

Given a binary matrix of R rows and C columns. We are allowed flip to any size of sub matrix. Flipping means changing 1 to 0 and 0 to 1. The task is maximize the number of 1s in the matrix. Output the maximum number of 1s.

Examples:

Input : R = 3, C =3
mat[][] = { 0, 0, 1,
            0, 0, 1,
            1, 0, 1 }

Output : 8
Flip
0 0 1
0 0 1
1 0 1

to get

1 1 1
1 1 1
0 1 1

Input : R = 2, C = 3
mat[][] = { 0, 0, 0,
            0, 0, 0 }
Output : 6

Create a matrix ones[][] of R rows and C columns, which precomputes the number of ones in the submatrix from (0, 0) to (i, j) by 

// Common elements in ones[i-1][j] and 
// ones[i][j-1] are ones[i-1][j-1]
ones[i][j] = ones[i-1][j] + ones[i][j-1] - 
             ones[i-1][j-1] + (mat[i][j] == 1)

Since, we are allowed to flip sub matrix only once. We iterate over all possible submatrices of all possible sizes for each cell (i, j) to (i + k – 1, i + k – 1). We calculate the total number of ones after the digits are filliped in the chosen submatrix.



Total number of ones in the final matrix after flipping submatrix (i, j) to (i + k – 1) will be Ones in the whole matrix – Ones in the chosen submatrix + Zeroes in the chosen sub matrix. That comes out to be :-
ones[R][C] – cal(i, j, i + k, j + k – 1) + k*k – cal(i, j, i + k – 1, j + k – 1)
where cal(a, b, c, d) denotes the number of ones in square submatrix of length c – a.

Now cal(x1, y1, x2, y2) can be define by:
ones[x2][y2] – ones[x2][y1 – 1] – ones[x1 – 1][y2] + ones[x1 – 1][y1 – 1].

Below is the implementation of this approach:

C++

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// C++ program to find maximum number of ones after
// one flipping in Binary Matrix
#include <bits/stdc++.h>
#define R 3
#define C 3
using namespace std;
  
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
int cal(int ones[R + 1][C + 1], int x, int y, int k)
{
    return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1]
           - ones[x + k - 1][y - 1] + ones[x - 1][y - 1];
}
  
// Return maximum number of 1s after flipping a submatrix
int sol(int mat[R][C])
{
    int ans = 0;
  
    // Precomputing the number of 1s
    int ones[R + 1][C + 1] = {0};
    for (int i = 1; i <= R; i++)
        for (int j = 1; j <= C; j++)
            ones[i][j] = ones[i - 1][j] + ones[i][j - 1] -
                         ones[i - 1][j - 1] +
                         (mat[i - 1][j - 1] == 1);
  
    // Finding the maximum number of 1s after flipping
    for (int k = 1; k <= min(R, C); k++)
        for (int i = 1; i + k - 1 <= R; i++)
            for (int j = 1; j + k - 1 <= C; j++)
                ans = max(ans, (ones[R][C] + k * k -
                                2 * cal(ones, i, j, k)));
    return ans;
}
  
// Driver code
int main()
{
    int mat[R][C] = {{0, 0, 1},
        { 0, 0, 1},
        { 1, 0, 1 }
    };
  
    cout << sol(mat) << endl;
  
    return 0;
}

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Java














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// Java program to find maximum number of ones after  


// one flipping in Binary Matrix  


class GFG { 


  


static final int R = 3;  


static final int C = 3 ; 


  


  


// Return number of ones in square submatrix of size  


// k x k starting from (x, y)  


static int cal(int ones[][], int x, int y, int k)  


{  


    return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1]  


        - ones[x + k - 1][y - 1] + ones[x - 1][y - 1];  


}  


  


// Return maximum number of 1s after flipping a submatrix  


static int sol(int mat[][])  


{  


    int ans = 0;  


        int val =0; 


    // Precomputing the number of 1s  


    int ones[][] = new int[R + 1][C + 1];  


    for (int i = 1; i <= R; i++)  


        for (int j = 1; j <= C; j++) { 


                    if(mat[i - 1][j - 1] == 1) 


                        val=1; 


        ones[i][j] = ones[i - 1][j] + ones[i][j - 1] -  


                        ones[i - 1][j - 1] +  


                        (val);  


                } 


    // Finding the maximum number of 1s after flipping  


    for (int k = 1; k <= Math.min(R, C); k++)  


        for (int i = 1; i + k - 1 <= R; i++)  


            for (int j = 1; j + k - 1 <= C; j++)  


                ans = Math.max(ans, (ones[R][C] + k * k -  


                                2 * cal(ones, i, j, k)));  


    return ans;  


}  


  


// Driver code  


    static public void main(String[] args) { 


        int mat[][] = {{0, 0, 1},  


        { 0, 0, 1},  


        { 1, 0, 1 }  


    };  


  


       System.out.println(sol(mat));  


    } 


} 


  


// This code is contributed by Rajput-Ji 



















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Python3














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# Python 3 program to find maximum number of  


# ones after one flipping in Binary Matrix 


R = 3


C = 3


  


# Return number of ones in square submatrix  


# of size k x k starting from (x, y) 


def cal(ones, x, y, k): 


    return (ones[x + k - 1][y + k - 1] - 


            ones[x - 1][y + k - 1] - 


            ones[x + k - 1][y - 1] + 


            ones[x - 1][y - 1]) 


  


# Return maximum number of 1s after  


# flipping a submatrix 


def sol(mat): 


    ans = 0


  


    # Precomputing the number of 1s 


    ones = [[0 for i in range(C + 1)]  


               for i in range(R + 1)] 


    for i in range(1, R + 1, 1): 


        for j in range(1, C + 1, 1): 


            ones[i][j] = (ones[i - 1][j] + ones[i][j - 1] - 


                          ones[i - 1][j - 1] + 


                         (mat[i - 1][j - 1] == 1)) 


  


    # Finding the maximum number of 1s 


    # after flipping 


    for k in range(1, min(R, C) + 1, 1): 


        for i in range(1, R - k + 2, 1): 


            for j in range(1, C - k + 2, 1): 


                ans = max(ans, (ones[R][C] + k * k - 2 * 


                            cal(ones, i, j, k))) 


    return ans 


  


# Driver code 


if __name__ == '__main__': 


    mat = [[0, 0, 1], 


           [0, 0, 1], 


           [1, 0, 1]] 


  


    print(sol(mat)) 


  


# This code is contributed by 


# Sahil_Shelangia 



















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C#














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// C# program to find maximum number of ones after  


// one flipping in Binary Matrix  


using System;     


  


public class GFG {  


  


    static readonly int R = 3;  


    static readonly int C = 3 ;  


  


  


    // Return number of ones in square submatrix of size  


    // k x k starting from (x, y)  


    static int cal(int [,]ones, int x, int y, int k)  


    {  


        return ones[x + k - 1,y + k - 1] - ones[x - 1,y + k - 1]  


            - ones[x + k - 1,y - 1] + ones[x - 1,y - 1];  


    }  


  


    // Return maximum number of 1s after flipping a submatrix  


    static int sol(int [,]mat)  


    {  


        int ans = 0;  


            int val =0;  


        // Precomputing the number of 1s  


        int [,]ones = new int[R + 1,C + 1];  


        for (int i = 1; i <= R; i++)  


            for (int j = 1; j <= C; j++) {  


                        if(mat[i - 1,j - 1] == 1)  


                            val=1;  


            ones[i,j] = ones[i - 1,j] + ones[i,j - 1] -  


                            ones[i - 1,j - 1] +  


                            (val);  


                    }  


        // Finding the maximum number of 1s after flipping  


        for (int k = 1; k <= Math.Min(R, C); k++)  


            for (int i = 1; i + k - 1 <= R; i++)  


                for (int j = 1; j + k - 1 <= C; j++)  


                    ans = Math.Max(ans, (ones[R,C] + k * k -  


                                    2 * cal(ones, i, j, k)));  


        return ans;  


    }  


  


    // Driver code  


    static public void Main() {  


        int [,]mat = {{0, 0, 1},  


        { 0, 0, 1},  


        { 1, 0, 1 }  


    };  


  


    Console.WriteLine(sol(mat));  


    }  


}  


// This code is contributed by 29AjayKumar 



















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PHP














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<?php 


// PHP program to find maximum number of ones after 


// one flipping in Binary Matrix 


  


$R = 3; 


$C = 3; 


  


// Return number of ones in square submatrix of size 


// k x k starting from (x, y) 


function cal($ones, $x, $y, $k) 


{ 


    return $ones[$x + $k - 1][$y + $k - 1] - 


            $ones[$x - 1][$y + $k - 1] - 


            $ones[$x + $k - 1][$y - 1] +  


            $ones[$x - 1][$y - 1]; 


} 


  


// Return maximum number of 1s after flipping a submatrix 


function sol($mat) 


{ 


    global $C,$R; 


    $ans = 0; 


  


    // Precomputing the number of 1s 


    $ones=array_fill(0,$R + 1,array_fill(0,$C + 1,0)); 


    for ($i = 1; $i <= $R; $i++) 


        for ($j = 1; $j <= $C; $j++) 


            $ones[$i][$j] = $ones[$i - 1][$j] + $ones[$i][$j - 1] - 


                        $ones[$i - 1][$j - 1] + 


                        (int)($mat[$i - 1][$j - 1] == 1); 


  


    // Finding the maximum number of 1s after flipping 


    for ($k = 1; $k <= min($R, $C); $k++) 


        for ($i = 1; $i + $k - 1 <= $R; $i++) 


            for ($j = 1; $j + $k - 1 <= $C; $j++) 


                $ans = max($ans, ($ones[$R][$C] + $k * $k - 


                                2 * cal($ones, $i, $j, $k))); 


    return $ans; 


} 


  


    // Driver code 


    $mat = array(array(0, 0, 1), 


        array( 0, 0, 1), 


        array( 1, 0, 1 )); 


  


    echo sol($mat); 


  


// This code is contributed by mits 


?> 



















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Output:


8



Time Complexity: O(R*C*min(R, C))


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.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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