Count positions in Binary Matrix having equal count of set bits in corresponding row and column

Given a boolean matrix mat[][] of size M * N, the task is to print the count of indices from the matrix whose corresponding row and column contain an equal number of set bits.

Examples:

Input; mat[][] = {{0, 1}, {1, 1}}
Output: 2
Explanation:
Position (0, 0) contains 1 set bit in corresponding row {(0, 0), (0, 1)} and column {(0, 0), (1, 0)}.
Position (1, 1) contains 2 set bits in corresponding row {(1, 0), (1, 1)} and column {(0, 1), (1, 1)}.

Input: mat[][] = {{0, 1, 1, 0}, {0, 1, 0, 0}, {1, 0, 1, 0}}
Output: 5

Naive Approach: The simplest approach to solve is to count and store the number of set bits present in the elements of each row and column of the given matrix. Then, traverse the matrix and for each position, check if the count of set bits of both the row and column is equal or not. For every index for which the above condition is found to be true, increment count. Print the final value of count after complete traversal of the matrix.

Time Complexity: O(N²)
Auxiliary Space: O(N²)

Efficient Approach: To optimize the above approach, follow the steps below:

• Initialize two temporary arrays row[] and col[] of sizes N and M respectively.
• Traverse the matrix mat[][]. Check for every index (i, j), if mat[i][j] is equal to 1 or not. If found to be true, increment row[i] and col[j] by 1.
• Traverse the matrix mat[][] again. Check for every index (i, j), if row[i] and col[j] are equal or not. If found to be true, increment count.
• Print the final value of count.

Below is the implementation of the above approach:

C++14

// C++14 program to implement
// above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of indices in
// from the given binary matrix having equal
// count of set bits in its row and column
int countPosition(vector<vector<int>> mat)
{
    int n = mat.size();
    int m = mat[0].size();

    // Stores count of set bits in
    // corresponding column and row
    vector<int> row(n);
    vector<int> col(m);

    // Traverse matrix
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {

            // Since 1 contains a set bit
            if (mat[i][j] == 1)
            {

                // Update count of set bits
                // for current row and col
                col[j]++;
                row[i]++;
            }
        }
    }

    // Stores the count of
    // required indices
    int count = 0;

    // Traverse matrix
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {

            // If current row and column
            // has equal count of set bits
            if (row[i] == col[j])
            {
                count++;
            }
        }
    }

    // Return count of
    // required position
    return count;
}

// Driver Code
int main()
{
    vector<vector<int>> mat = { { 0, 1 },
                                { 1, 1 } };

    cout << (countPosition(mat));
}

// This code is contributed by mohit kumar 29

Java

// Java Program to implement
// above approach

import java.util.*;
import java.lang.*;

class GFG {

    // Function to return the count of indices in
    // from the given binary matrix having equal
    // count of set bits in its row and column
    static int countPosition(int[][] mat)
    {

        int n = mat.length;
        int m = mat[0].length;

        // Stores count of set bits in
        // corresponding column and row
        int[] row = new int[n];
        int[] col = new int[m];

        // Traverse matrix
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {

                // Since 1 contains a set bit
                if (mat[i][j] == 1) {

                    // Update count of set bits
                    // for current row and col
                    col[j]++;
                    row[i]++;
                }
            }
        }

        // Stores the count of
        // required indices
        int count = 0;

        // Traverse matrix
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {

                // If current row and column
                // has equal count of set bits
                if (row[i] == col[j]) {
                    count++;
                }
            }
        }

        // Return count of
        // required position
        return count;
    }

    // Driver Code
    public static void main(
        String[] args)
    {

        int mat[][] = { { 0, 1 },
                        { 1, 1 } };

        System.out.println(
            countPosition(mat));
    }
}

Python3

# Python3 program to implement
# the above approach

# Function to return the count
# of indices in from the given
# binary matrix having equal
# count of set bits in its
# row and column
def countPosition(mat):

    n = len(mat)
    m = len(mat[0])

    # Stores count of set bits in
    # corresponding column and row
    row = [0] * n
    col = [0] * m

    # Traverse matrix
    for i in range(n):
        for j in range(m):

            # Since 1 contains a set bit
            if (mat[i][j] == 1):

                # Update count of set bits
                # for current row and col
                col[j] += 1
                row[i] += 1

    # Stores the count of
    # required indices
    count = 0

    # Traverse matrix
    for i in range(n):
        for j in range(m):

            # If current row and column
            # has equal count of set bits
            if (row[i] == col[j]):
                count += 1

    # Return count of
    # required position
    return count

# Driver Code
mat = [ [ 0, 1 ],
        [ 1, 1 ] ]

print(countPosition(mat))

# This code is contributed by sanjoy_62

C#

// C# Program to implement
// above approach
using System;
class GFG{

// Function to return the count of indices in
// from the given binary matrix having equal
// count of set bits in its row and column
static int countPosition(int[,] mat)
{

  int n = mat.GetLength(0);
  int m = mat.GetLength(1);

  // Stores count of set bits in
  // corresponding column and row
  int[] row = new int[n];
  int[] col = new int[m];

  // Traverse matrix
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < m; j++)
    {
      // Since 1 contains a set bit
      if (mat[i, j] == 1)
      {
        // Update count of set bits
        // for current row and col
        col[j]++;
        row[i]++;
      }
    }
  }

  // Stores the count of
  // required indices
  int count = 0;

  // Traverse matrix
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < m; j++)
    {
      // If current row and column
      // has equal count of set bits
      if (row[i] == col[j])
      {
        count++;
      }
    }
  }

  // Return count of
  // required position
  return count;
}

// Driver Code
public static void Main(String[] args)
{
  int [,]mat = {{0, 1},
                {1, 1}};
  Console.WriteLine(countPosition(mat));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript program to implement
// above approach

// Function to return the count of indices in
// from the given binary matrix having equal
// count of set bits in its row and column
function countPosition(mat)
{
    var n = mat.length;
    var m = mat[0].length;

    // Stores count of set bits in
    // corresponding column and row
    var row = Array.from({length: n}, (_, i) => 0);
    var col = Array.from({length: m}, (_, i) => 0);

    // Traverse matrix
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < m; j++)
        {

            // Since 1 contains a set bit
            if (mat[i][j] == 1)
            {

                // Update count of set bits
                // for current row and col
                col[j]++;
                row[i]++;
            }
        }
    }

    // Stores the count of
    // required indices
    var count = 0;

    // Traverse matrix
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < m; j++)
        {

            // If current row and column
            // has equal count of set bits
            if (row[i] == col[j])
            {
                count++;
            }
        }
    }

    // Return count of
    // required position
    return count;
}

// Driver Code
var mat = [ [ 0, 1 ],
            [ 1, 1 ] ];

document.write(countPosition(mat));

// This code is contributed by Rajput-Ji

</script>

Output

2

Time Complexity: O(N * M)
Auxiliary Space: O(N+M)

Approach#2:using brute force

Algorithm

1. Initialize count to zero
2. Iterate over each cell (i, j) in the matrix
3.Initialize row_sum and col_sum to zero
4. Iterate over each cell in the ith row and jth column
a. If the cell is set, increment row_sum and col_sum
5.If row_sum is equal to col_sum, increment count
6. Return count

C++

#include <iostream>
#include <vector>
int GFG(std::vector<std::vector<int>>& matrix) {
    int count = 0;
    for (int i = 0; i < matrix.size(); ++i) {
        for (int j = 0; j < matrix[0].size(); ++j) {
            int row_sum = 0;
            int col_sum = 0;
            for (int k = 0; k < matrix.size(); ++k) {
                row_sum += matrix[i][k];
                col_sum += matrix[k][j];
            }
            if (row_sum == col_sum) {
                count += 1;
            }
        }
    }
    return count;
}
int main() {
    std::vector<std::vector<int>> matrix = {{0, 1}, {1, 1}};
    std::cout <<GFG(matrix) << std::endl;
    return 0;
}

Java

import java.io.*;
import java.util.*;

public class Main {
    public static int GFG(int[][] matrix) {
        int count = 0;
        int rows = matrix.length;
        int cols = matrix[0].length;
        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < cols; ++j) {
                int rowSum = 0;
                int colSum = 0;
                for (int k = 0; k < rows; ++k) {
                    rowSum += matrix[i][k];
                    colSum += matrix[k][j];
                }
                if (rowSum == colSum) {
                    count += 1;
                }
            }
        }
        return count;
    }
    public static void main(String[] args) {
        int[][] matrix = {{0, 1}, {1, 1}};
        int result = GFG(matrix);
        System.out.println(result);
    }
}

Python3

def count_positions(matrix):
    count = 0
    for i in range(len(matrix)):
        for j in range(len(matrix[0])):
            row_sum = 0
            col_sum = 0
            for k in range(len(matrix)):
                row_sum += matrix[i][k]
                col_sum += matrix[k][j]
            if row_sum == col_sum:
                count += 1
    return count
matrix = [ [ 0, 1 ],
        [ 1, 1 ] ]
print( count_positions(matrix))

Javascript

function GFG(matrix) {
    let count = 0;
    const rows = matrix.length;
    const cols = matrix[0].length;

    // Loop through each element in the matrix
    for (let i = 0; i < rows; ++i) {
        for (let j = 0; j < cols; ++j) {
            let rowSum = 0;
            let colSum = 0;

            // Calculate the sum of elements in the current row and column
            for (let k = 0; k < rows; ++k) {
                rowSum += matrix[i][k];
                colSum += matrix[k][j];
            }

            // If the sum of the current row and column are equal, increment count
            if (rowSum === colSum) {
                count += 1;
            }
        }
    }

    return count;
}

// Main function
    const matrix = [[0, 1], [1, 1]];
    const result = GFG(matrix);
    console.log(result);

Output

2

Time Complexity: O(n^3), where n is the size of the matrix
Space Complexity: O(1)