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> |
2
Time Complexity: O(N * M)
Auxiliary Space: O(N+M)