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 columnint 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 Codeint 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 approachimport 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 columndef 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 Codemat = [ [ 0, 1 ], [ 1, 1 ] ] print(countPosition(mat))# This code is contributed by sanjoy_62 |
C#
// C# Program to implement// above approachusing 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 columnstatic 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 Codepublic 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 columnfunction 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 Codevar 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 countmatrix = [ [ 0, 1 ], [ 1, 1 ] ]print( count_positions(matrix)) |
C#
using System;class GFG { // Function to count elements in the matrix // where row sum is equal to column sum static int CountElements(int[][] matrix) { int count = 0; int rowCount = matrix.Length; int colCount = matrix[0].Length; for (int i = 0; i < rowCount; i++) { for (int j = 0; j < colCount; j++) { int rowSum = 0; int colSum = 0; for (int k = 0; k < rowCount; k++) { rowSum += matrix[i][k]; colSum += matrix[k][j]; } if (rowSum == colSum) { count += 1; } } } return count; } static void Main() { int[][] matrix = new int[][] { new int[] { 0, 1 }, new int[] { 1, 1 } }; int result = CountElements(matrix); Console.WriteLine(result); }} |
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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
