Given a matrix mat[][] of dimensions N * M, the task is to count the number of rows whose sum exceeds the sum of the remaining elements of the Matrix.
Examples:
Input: mat[][] = {{1, 5}, {2, 3}}
Output: 1
Explanation:
Sum of elements of the first row {1, 5} exceeds the sum of the remaining matrix {2, 3}.Input: mat[][] = {{2, -1, 5}, {-3, 0, -2}, {5, 1, 2}}
Output: 2
Approach: To solve the problem, the idea is to pre-calculate the total sum of the elements of the given matrix and for every row, check if the sum of elements of the current row is greater than the sum of the elements of the rest of the matrix, i.e., currSum > (totalSum – currSum). For every row satisfying the afore-mentioned condition, increase count. Finally, print the value of count obtained after the complete traversal of the matrix.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include<bits/stdc++.h>using namespace std;#define N 3#define M 3// Function to count the number of rows// whose sum exceeds the sum of// elements of the remaining matrix void countRows(int mat[M][N]){ // To store the result int count = 0; // Stores the total sum of // the matrix elements int totalSum = 0; // Calculate the total sum for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { totalSum += mat[i][j]; } } // Traverse to check for each row for (int i = 0; i < N; i++) { // Stores the sum of elements // of the current row int currSum = 0; // Calculate the sum of elements // of the current row for (int j = 0; j < M; j++) { currSum += mat[i][j]; } // If sum of current row exceeds // the sum of rest of the matrix if (currSum > totalSum - currSum) // Increase count count++; } // Print the result cout << count;}// Driver Codeint main(){ // Given matrix int mat[N][M] = {{2, -1, 5}, {-3, 0, -2}, {5, 1, 2}}; // Function Call countRows(mat);}// This code is contributed by Rajput-Ji |
Java
// Java program to implement// the above approachimport java.io.*;class GFG { // Function to count the number of rows // whose sum exceeds the sum of // elements of the remaining matrix static void countRows(int[][] mat) { // Stores the matrix dimensions int n = mat.length; int m = mat[0].length; // To store the result int count = 0; // Stores the total sum of // the matrix elements int totalSum = 0; // Calculate the total sum for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { totalSum += mat[i][j]; } } // Traverse to check for each row for (int i = 0; i < n; i++) { // Stores the sum of elements // of the current row int currSum = 0; // Calculate the sum of elements // of the current row for (int j = 0; j < m; j++) { currSum += mat[i][j]; } // If sum of current row exceeds // the sum of rest of the matrix if (currSum > totalSum - currSum) // Increase count count++; } // Print the result System.out.println(count); } // Driver Code public static void main(String[] args) { // Given matrix int[][] mat = { { 2, -1, 5 }, { -3, 0, -2 }, { 5, 1, 2 } }; // Function Call countRows(mat); }} |
Python3
# Python3 program to implement# the above approach# Function to count the number of rows# whose sum exceeds the sum of elements# of the remaining matrixdef countRows(mat): # Stores the matrix dimensions n = len(mat) m = len(mat[0]) # To store the result count = 0 # Stores the total sum of # the matrix elements totalSum = 0 # Calculate the total sum for i in range(n): for j in range(m): totalSum += mat[i][j] # Traverse to check for each row for i in range(n): # Stores the sum of elements # of the current row currSum = 0 # Calculate the sum of elements # of the current row for j in range(m): currSum += mat[i][j] # If sum of current row exceeds # the sum of rest of the matrix if (currSum > totalSum - currSum): # Increase count count += 1 # Print the result print(count)# Driver Codeif __name__ == '__main__': # Given matrix mat = [ [ 2, -1, 5 ], [ -3, 0, -2 ], [ 5, 1, 2 ] ] # Function call countRows(mat)# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to count the number of rows// whose sum exceeds the sum of// elements of the remaining matrixstatic void countRows(int[,] mat){ // Stores the matrix dimensions int n = mat.GetLength(0); int m = mat.GetLength(1); // To store the result int count = 0; // Stores the total sum of // the matrix elements int totalSum = 0; // Calculate the total sum for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { totalSum += mat[i, j]; } } // Traverse to check for each row for (int i = 0; i < n; i++) { // Stores the sum of elements // of the current row int currSum = 0; // Calculate the sum of elements // of the current row for (int j = 0; j < m; j++) { currSum += mat[i, j]; } // If sum of current row exceeds // the sum of rest of the matrix if (currSum > totalSum - currSum) // Increase count count++; } // Print the result Console.WriteLine(count);}// Driver Codepublic static void Main(String[] args){ // Given matrix int[,] mat = {{2, -1, 5}, {-3, 0, -2}, {5, 1, 2}}; // Function Call countRows(mat);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to implement// the above approachvar N = 3var M = 3// Function to count the number of rows// whose sum exceeds the sum of// elements of the remaining matrixfunction countRows( mat){ // To store the result var count = 0; // Stores the total sum of // the matrix elements var totalSum = 0; // Calculate the total sum for (var i = 0; i < N; i++) { for (var j = 0; j < M; j++) { totalSum += mat[i][j]; } } // Traverse to check for each row for (var i = 0; i < N; i++) { // Stores the sum of elements // of the current row var currSum = 0; // Calculate the sum of elements // of the current row for (var j = 0; j < M; j++) { currSum += mat[i][j]; } // If sum of current row exceeds // the sum of rest of the matrix if (currSum > totalSum - currSum) // Increase count count++; } // Print the result document.write( count);}// Driver Code// Given matrixvar mat = [[2, -1, 5], [-3, 0, -2], [5, 1, 2]];// Function CallcountRows(mat);// This code is contributed by itsok.</script> |
Output
2
Time Complexity: O(N * M)
Auxiliary Space: O(1)
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