Given a binary matrix, mat[][] of size M * N, the task is to count the number of 1s from the given binary matrix whose corresponding row and column consists of 0s only in the remaining indices.
Examples:
Input: mat[][] = {{1, 0, 0}, {0, 0, 1}, {0, 0, 0}}
Output: 2
Explanation:
The only two cells satisfying the conditions are (0, 0) and (1, 2).
Therefore, the count is 2.Input: mat[][] = {{1, 0}, {1, 1}}
Output: 0
Naive Approach: The simplest approach is to iterate over the matrix and check the given condition for all 1s present in the given matrix by traversing its corresponding row and column. Increase count of all 1s satisfying the condition. Finally, print the count as the required answer.
Time Complexity: O(M*N2)
Auxiliary Space: O(M + N)
Efficient Approach: The above approach can be optimized based on the idea that the sum of such rows and columns will be only 1. Follow the steps below to solve the problem:
- Initialize two arrays, rows[] and cols[], to store the sum of each row and each column of the matrix respectively.
- Initialize a variable, say cnt, to store the count of 1s satisfying given condition.
- Traverse the matrix for every mat[i][j] = 1, check if rows[i] and cols[j] is 1.If found to be true, then increment cnt.
- After completing the above steps, print the final value of count.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count required 1s// from the given matrixint numSpecial(vector<vector<int> >& mat){ // Stores the dimensions of the mat[][] int m = mat.size(), n = mat[0].size(); int rows[m]; int cols[n]; int i, j; // Calculate sum of rows for (i = 0; i < m; i++) { rows[i] = 0; for (j = 0; j < n; j++) rows[i] += mat[i][j]; } // Calculate sum of columns for (i = 0; i < n; i++) { cols[i] = 0; for (j = 0; j < m; j++) cols[i] += mat[j][i]; } // Stores required count of 1s int cnt = 0; for (i = 0; i < m; i++) { for (j = 0; j < n; j++) { // If current cell is 1 // and sum of row and column is 1 if (mat[i][j] == 1 && rows[i] == 1 && cols[j] == 1) // Increment count of 1s cnt++; } } // Return the final count return cnt;}// Driver Codeint main(){ // Given matrix vector<vector<int> > mat = { { 1, 0, 0 }, { 0, 0, 1 }, { 0, 0, 0 } }; // Function Call cout << numSpecial(mat) << endl; return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to count required 1s// from the given matrixstatic int numSpecial(int [][]mat){ // Stores the dimensions of the mat[][] int m = mat.length; int n = mat[0].length; int []rows = new int[m]; int []cols = new int[n]; int i, j; // Calculate sum of rows for(i = 0; i < m; i++) { rows[i] = 0; for(j = 0; j < n; j++) rows[i] += mat[i][j]; } // Calculate sum of columns for(i = 0; i < n; i++) { cols[i] = 0; for(j = 0; j < m; j++) cols[i] += mat[j][i]; } // Stores required count of 1s int cnt = 0; for(i = 0; i < m; i++) { for(j = 0; j < n; j++) { // If current cell is 1 // and sum of row and column is 1 if (mat[i][j] == 1 && rows[i] == 1 && cols[j] == 1) // Increment count of 1s cnt++; } } // Return the final count return cnt;} // Driver Codepublic static void main(String[] args){ // Given matrix int [][]mat = { { 1, 0, 0 }, { 0, 0, 1 }, { 0, 0, 0 } }; // Function call System.out.print(numSpecial(mat) + "\n");}} // This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach# Function to count required 1s# from the given matrixdef numSpecial(mat): # Stores the dimensions # of the mat m = len(mat) n = len(mat[0]) rows = [0] * m cols = [0] * n i, j = 0, 0 # Calculate sum of rows for i in range(m): rows[i] = 0 for j in range(n): rows[i] += mat[i][j] # Calculate sum of columns for i in range(n): cols[i] = 0 for j in range(m): cols[i] += mat[j][i] # Stores required count of 1s cnt = 0 for i in range(m): for j in range(n): # If current cell is 1 # and sum of row and column is 1 if (mat[i][j] == 1 and rows[i] == 1 and cols[j] == 1): # Increment count of 1s cnt += 1 # Return the final count return cnt# Driver Codeif __name__ == '__main__': # Given matrix mat = [ [ 1, 0, 0 ], [ 0, 0, 1 ], [ 0, 0, 0 ] ] # Function call print(numSpecial(mat))# This code is contributed by Amit Katiyar |
C#
// C# program for the above approachusing System;class GFG{ // Function to count required 1s// from the given matrixstatic int numSpecial(int [,]mat){ // Stores the dimensions of the [,]mat int m = mat.GetLength(0); int n = mat.GetLength(1); int []rows = new int[m]; int []cols = new int[n]; int i, j; // Calculate sum of rows for(i = 0; i < m; i++) { rows[i] = 0; for(j = 0; j < n; j++) rows[i] += mat[i, j]; } // Calculate sum of columns for(i = 0; i < n; i++) { cols[i] = 0; for(j = 0; j < m; j++) cols[i] += mat[j, i]; } // Stores required count of 1s int cnt = 0; for(i = 0; i < m; i++) { for(j = 0; j < n; j++) { // If current cell is 1 and // sum of row and column is 1 if (mat[i, j] == 1 && rows[i] == 1 && cols[j] == 1) // Increment count of 1s cnt++; } } // Return the readonly count return cnt;} // Driver Codepublic static void Main(String[] args){ // Given matrix int [,]mat = { { 1, 0, 0 }, { 0, 0, 1 }, { 0, 0, 0 } }; // Function call Console.Write(numSpecial(mat) + "\n");}} // This code is contributed by Amit Katiyar |
2
Time Complexity: O(M*N)
Auxiliary Space: O(M + N)
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