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Minimum flip required to make Binary Matrix symmetric

Last Updated : 26 Dec, 2024
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Given a Binary Matrix mat[][] of size n x n, consisting of 1s and 0s. The task is to find the minimum flips required to make the matrix symmetric along the main diagonal.

Examples : 

Input: mat[][] = [[0, 0, 1],
[1, 1, 1],
[1, 0, 0]];
Output: 2
Value of mat[1][0] is not equal to mat[0][1].
Value of mat[2][1] is not equal to mat[1][2].
So, two flip are required.


Input: mat[][] = [[1, 1, 1, 1, 0],
[0, 1, 0, 1, 1],
[ 1, 0, 0, 0, 1],
[0, 1, 0, 1, 0],
[0, 1, 0, 0, 1]]
Output: 3

Table of Content

[Naive Approach] Using Matrix Transpose - O(n^2) Time and O(n^2) Space

The idea is to find the transpose of the matrix and find minimum number of flip required to make transpose and original matrix equal. To find minimum flip, find the number of position where original matrix and transpose matrix are not same, say x. So, our answer will be x/2.

C++ 
// C++ Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal
#include <bits/stdc++.h>
using namespace std;

// Return the minimum flip required to make
// Binary Matrix symmetric along main diagonal.
int minimumflip(vector<vector<int>> &mat) {
  
  	int n = mat.size();
    vector<vector<int>> transpose(n, vector<int>(n));

    // finding the transpose of the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            transpose[i][j] = mat[j][i];
        }
    }

    // Finding the number of positions where
    // elements are not the same.
    int flip = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (transpose[i][j] != mat[i][j]) {
                flip++;
            }
        }
    }

    return flip / 2;
}

int main() {
  
    vector<vector<int>> mat = {{0, 0, 1}, {1, 1, 1}, {1, 0, 0}};
    cout << minimumflip(mat) << endl;
    return 0;
}
Java 
// Java Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal
import java.util.*;

class GfG {
    static int minimumflip(int[][] mat) {
      
      	int n = mat.length;
        int[][] transpose = new int[n][n];

        // finding the transpose of the matrix
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                transpose[i][j] = mat[j][i];
            }
        }

        // Finding the number of positions where
        // elements are not the same.
        int flip = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (transpose[i][j] != mat[i][j]) {
                    flip++;
                }
            }
        }

        return flip / 2;
    }

    public static void main(String[] args) {

        int[][] mat = {
            { 0, 0, 1 },
            { 1, 1, 1 },
            { 1, 0, 0 }
        };
        System.out.println(minimumflip(mat));
    }
}
Python 
# Python Program to find minimum flip required to make
# Binary Matrix symmetric along main diagonal

def minimumflip(mat):
    n = len(mat)
    transpose = [[0] * n for _ in range(n)]

    # Finding the transpose of the matrix
    for i in range(n):
        for j in range(n):
            transpose[i][j] = mat[j][i]

    # Finding the number of positions where elements 
    # are not the same.
    flip = 0
    for i in range(n):
        for j in range(n):
            if transpose[i][j] != mat[i][j]:
                flip += 1

    return flip // 2

if __name__ == "__main__":
    mat = [
        [0, 0, 1],
        [1, 1, 1],
        [1, 0, 0]
    ]
    print(minimumflip(mat))
C# 
// C# Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal
using System;

class GfG {
    static int minimumflip(int[, ] mat) {
        int n = mat.GetLength(0);
        int[, ] transpose = new int[n, n];

        // Calculate the transpose of the matrix
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                transpose[i, j] = mat[j, i];
            }
        }

        // Count the number of positions where mat[i, j] !=
        // transpose[i, j]
        int flip = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (mat[i, j] != transpose[i, j]) {
                    flip++;
                }
            }
        }

        return flip / 2;
    }

    static void Main(string[] args) {
        int[, ] mat
            = { { 0, 0, 1 }, { 1, 1, 1 }, { 1, 0, 0 } };

        Console.WriteLine(minimumflip(mat));
    }
}
JavaScript 
// JavaScript Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal

function minimumflip(mat) {
	
    const n = mat.length;
    let transpose = Array.from({ length: n }, () => Array(n).fill(0));

    // finding the transpose of the matrix
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            transpose[i][j] = mat[j][i];
        }
    }

    // Finding the number of positions where
    // elements are not the same.
    let flip = 0;
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            if (transpose[i][j] !== mat[i][j]) {
                flip++;
            }
        }
    }

    return Math.floor(flip / 2);
}

//driver code
const mat = [
    [0, 0, 1],
    [1, 1, 1],
    [1, 0, 0]
];
console.log(minimumflip(mat));

Output
2

[Expected Approach] Comparing Lower and Upper Triangular Matrix - O(n^2) Time and O(1) Space

The idea is to find minimum flips required to make upper triangle of matrix equals to lower triangle of the matrix. To do so, we run two nested loop, outer loop from i = 0 to n i.e., for each row of the matrix and the inner loop from j = 0 to i, and check whether mat[i][j] is equal to mat[j][i]. Count of number of instances where they are not equal will be the minimum flips required to make matrix symmetric along main diagonal.

C++ 
// C++ Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal
#include <bits/stdc++.h>
using namespace std;

// Return the minimum flip required to make
// Binary Matrix symmetric along main diagonal.
int minimumflip(vector<vector<int>>& mat) {
  
  	int n = mat.size();
    int flip = 0;

    // Compare elements above and below the main diagonal
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) { 
            if (mat[i][j] != mat[j][i]) {
                flip++;
            }
        }
    }

    return flip;
}

int main() {
 
    vector<vector<int>> mat = {
        { 0, 0, 1 },
        { 1, 1, 1 },
        { 1, 0, 0 }
    };
    cout << minimumflip(mat) << endl;
    return 0;
}
Java 
// Java Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal
import java.util.*;

class GfG {
    static int minimumflip(int[][] mat) {
      
      	int n = mat.length;
        int flip = 0;

        // Compare elements above and below the main diagonal
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (mat[i][j] != mat[j][i]) {
                    flip++;
                }
            }
        }

        return flip;
    }

    public static void main(String[] args) {

        int[][] mat = {
            { 0, 0, 1 },
            { 1, 1, 1 },
            { 1, 0, 0 }
        };
        System.out.println(minimumflip(mat));
    }
}
Python 
# Python Program to find minimum flip required to make
# Binary Matrix symmetric along main diagonal

def minimumflip(mat):
    n = len(mat)
    flip = 0

    # Compare elements above and below the main diagonal
    for i in range(n):
        for j in range(i):
            if mat[i][j] != mat[j][i]:
                flip += 1

    return flip

if __name__ == "__main__":

    mat = [
        [0, 0, 1],
        [1, 1, 1],
        [1, 0, 0]
    ]
    print(minimumflip(mat))
C# 
// C# Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal
using System;

class GfG {
    static int minimumflip(int[,] mat) {
      
      	int n = mat.GetLength(0);
        int flip = 0;

        // Compare elements above and below the main diagonal
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (mat[i, j] != mat[j, i]) {
                    flip++;
                }
            }
        }

        return flip;
    }

    static void Main(string[] args) {
  
        int[,] mat = {
            { 0, 0, 1 },
            { 1, 1, 1 },
            { 1, 0, 0 }
        };
        Console.WriteLine(minimumflip(mat));
    }
}
JavaScript 
// JavaScript Program to find minimum flip required to make
// Binary Matrix symmetric along main diagonal

function minimumflip(mat) {
    const n = mat.length;
    let flip = 0;

    // Compare elements above and below the main diagonal
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < i; j++) {
            if (mat[i][j] !== mat[j][i]) {
                flip++;
            }
        }
    }

    return flip;
}

// Driver code
const mat = [
    [0, 0, 1],
    [1, 1, 1],
    [1, 0, 0]
];
console.log(minimumflip(mat));

Output
2

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