Find column with maximum sum in a Matrix
Given a N*N matrix. The task is to find the index of column with maximum sum. That is the column whose sum of elements are maximum.
Examples:
Input : mat[][] = { { 1, 2, 3, 4, 5 }, { 5, 3, 1, 4, 2 }, { 5, 6, 7, 8, 9 }, { 0, 6, 3, 4, 12 }, { 9, 7, 12, 4, 3 }, }; Output : Column 5 has max sum 31 Input : mat[][] = { { 1, 2, 3 }, { 4, 2, 1 }, { 5, 6, 7 }, }; Output : Column 3 has max sum 11
The idea is to traverse the matrix column-wise and find the sum of elements in each column and check for every column if current sum is greater than the maximum sum obtained till the current column and update the maximum_sum accordingly.
Below is the implementation of the above approach:
C++
// C++ program to find column with // max sum in a matrix #include <bits/stdc++.h> using namespace std; #define N 5 // No of rows and column // Function to find the column with max sum pair< int , int > colMaxSum( int mat[N][N]) { // Variable to store index of column // with maximum int idx = -1; // Variable to store max sum int maxSum = INT_MIN; // Traverse matrix column wise for ( int i = 0; i < N; i++) { int sum = 0; // calculate sum of column for ( int j = 0; j < N; j++) { sum += mat[j][i]; } // Update maxSum if it is less than // current sum if (sum > maxSum) { maxSum = sum; // store index idx = i; } } pair< int , int > res; res = make_pair(idx, maxSum); // return result return res; } // driver code int main() { int mat[N][N] = { { 1, 2, 3, 4, 5 }, { 5, 3, 1, 4, 2 }, { 5, 6, 7, 8, 9 }, { 0, 6, 3, 4, 12 }, { 9, 7, 12, 4, 3 }, }; pair< int , int > ans = colMaxSum(mat); cout << "Column " << ans.first + 1 << " has max sum " << ans.second; return 0; } |
Java
// Java program to find column // with max sum in a matrix import java.util.*; class GFG { // No of rows and column static final int N = 5 ; // structure for pair static class Pair { int first , second; Pair( int f, int s) { first = f; second = s; } } // Function to find the column // with max sum static Pair colMaxSum( int mat[][]) { // Variable to store index of // column with maximum int idx = - 1 ; // Variable to store max sum int maxSum = Integer.MIN_VALUE; // Traverse matrix column wise for ( int i = 0 ; i < N; i++) { int sum = 0 ; // calculate sum of column for ( int j = 0 ; j < N; j++) { sum += mat[j][i]; } // Update maxSum if it is // less than current sum if (sum > maxSum) { maxSum = sum; // store index idx = i; } } Pair res; res = new Pair(idx, maxSum); // return result return res; } // Driver code public static void main(String args[]) { int mat[][] = { { 1 , 2 , 3 , 4 , 5 }, { 5 , 3 , 1 , 4 , 2 }, { 5 , 6 , 7 , 8 , 9 }, { 0 , 6 , 3 , 4 , 12 }, { 9 , 7 , 12 , 4 , 3 }}; Pair ans = colMaxSum(mat); System.out.println( "Column " + ( int )(ans.first + 1 ) + " has max sum " + ans.second); } } // This code is contributed // by Arnab Kundu |
Python3
# Python3 program to find column with # max Sum in a matrix N = 5 # Function to find the column with max Sum def colMaxSum(mat): # Variable to store index of column # with maximum idx = - 1 # Variable to store max Sum maxSum = - 10 * * 9 # Traverse matrix column wise for i in range (N): Sum = 0 # calculate Sum of column for j in range (N): Sum + = mat[j][i] # Update maxSum if it is less # than current Sum if ( Sum > maxSum): maxSum = Sum # store index idx = i # return result return idx, maxSum # Driver Code mat = [[ 1 , 2 , 3 , 4 , 5 ], [ 5 , 3 , 1 , 4 , 2 ], [ 5 , 6 , 7 , 8 , 9 ], [ 0 , 6 , 3 , 4 , 12 ], [ 9 , 7 , 12 , 4 , 3 ]] ans, ans0 = colMaxSum(mat) print ( "Column" , ans + 1 , "has max Sum" , ans0) # This code is contributed by # Mohit kumar 29 |
C#
// C# program to find column // with max sum in a matrix using System; class GFG { // No of rows and column static readonly int N = 5; // structure for pair public class Pair { public int first , second; public Pair( int f, int s) { first = f; second = s; } } // Function to find the column // with max sum static Pair colMaxSum( int [,]mat) { // Variable to store index of // column with maximum int idx = -1; // Variable to store max sum int maxSum = int .MinValue; // Traverse matrix column wise for ( int i = 0; i < N; i++) { int sum = 0; // calculate sum of column for ( int j = 0; j < N; j++) { sum += mat[j, i]; } // Update maxSum if it is // less than current sum if (sum > maxSum) { maxSum = sum; // store index idx = i; } } Pair res; res = new Pair(idx, maxSum); // return result return res; } // Driver code public static void Main(String []args) { int [,]mat = { { 1, 2, 3, 4, 5 }, { 5, 3, 1, 4, 2 }, { 5, 6, 7, 8, 9 }, { 0, 6, 3, 4, 12 }, { 9, 7, 12, 4, 3 }}; Pair ans = colMaxSum(mat); Console.WriteLine( "Column " + ( int )(ans.first + 1) + " has max sum " + ans.second); } } // This code has been contributed by 29AjayKumar |
Javascript
<script> // Javascript program to find column // with max sum in a matrix // No of rows and column let N = 5; // Function to find the column // with max sum function colMaxSum(mat) { // Variable to store index of // column with maximum let idx = -1; // Variable to store max sum let maxSum = Number.MIN_VALUE; // Traverse matrix column wise for (let i = 0; i < N; i++) { let sum = 0; // calculate sum of column for (let j = 0; j < N; j++) { sum += mat[j][i]; } // Update maxSum if it is // less than current sum if (sum > maxSum) { maxSum = sum; // store index idx = i; } } let res; res = [idx, maxSum]; // return result return res; } // Driver code let mat = [[ 1, 2, 3, 4, 5 ], [ 5, 3, 1, 4, 2 ], [ 5, 6, 7, 8, 9 ], [ 0, 6, 3, 4, 12 ], [ 9, 7, 12, 4, 3 ]]; let ans = colMaxSum(mat); document.write( "Column " + (ans[0] + 1) + " has max sum " + ans[1]); // This code is contributed by unknown2108 </script> |
Output
Column 5 has max sum 31
Complexity Analysis
- Time Complexity: O(N*N) where N is the size of each row or column in the matrix.
- Auxiliary Space : O(1)