Given an N*N matrix. The task is to find the index of a row with the maximum sum. That is the row whose sum of elements is 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 : Row 3 has max sum 35
Input : mat[][] = {
{ 1, 2, 3 },
{ 4, 2, 1 },
{ 5, 6, 7 },
};
Output : Row 3 has max sum 18
The idea is to traverse the matrix row-wise and find the sum of elements in each row and check for every row if current sum is greater than the maximum sum obtained till the current row and update the maximum_sum accordingly.
Algorithm:
- Define a constant N as the number of rows and columns in the matrix.
- Define a function colMaxSum that takes a 2D array of integers mat of size N*N as its input.
- Initialize two variables idx and maxSum to -1 and INT_MIN respectively.
- Traverse the matrix row-wise.
- For each row, calculate the sum of all the elements in that row.
- If the sum of the current row is greater than the current maxSum, update maxSum to be the sum of the current row and set idx to be the index of the current row.
- Return a pair of integers, with the first element being the index of the row with the maximum sum (idx) and the second element being the maximum sum (maxSum).
Pseudocode:
1. N ? number of rows and columns in the matrix 2. function colMaxSum(mat[N][N]) 3. idx ? -1 4. maxSum ? INT_MIN 5. for i from 0 to N-1 6. sum ? 0 7. for j from 0 to N-1 8. sum ? sum + mat[i][j] 9. if sum > maxSum 10. maxSum ? sum 11. idx ? i 12. return pair(idx, maxSum) 13. end function
Below is the implementation of the above approach:
C++
// C++ program to find row 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 row with max sumpair<int, int> colMaxSum(int mat[N][N])
{ // Variable to store index of row
// with maximum
int idx = -1;
// Variable to store max sum
int maxSum = INT_MIN;
// Traverse matrix row wise
for (int i = 0; i < N; i++) {
int sum = 0;
// calculate sum of row
for (int j = 0; j < N; j++) {
sum += mat[i][j];
}
// 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 codeint 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 << "Row " << ans.first + 1 << " has max sum "
<< ans.second;
return 0;
} |
Java
// Java program to find row with// max sum in a matriximport java.util.ArrayList;
class MaxSum {
public static int N;
static ArrayList<Integer> colMaxSum(int mat[][])
{
// Variable to store index of row
// with maximum
int idx = -1;
// Variable to store maximum sum
int maxSum = Integer.MIN_VALUE;
// Traverse the matrix row wise
for (int i = 0; i < N; i++) {
int sum = 0;
for (int j = 0; j < N; j++) {
sum += mat[i][j];
}
// Update maxSum if it is less than
// current row sum
if (maxSum < sum) {
maxSum = sum;
// store index
idx = i;
}
}
// Arraylist to store values of index
// of maximum sum and the maximum sum together
ArrayList<Integer> res = new ArrayList<>();
res.add(idx);
res.add(maxSum);
return res;
}
// Driver code
public static void main(String[] args)
{
N = 5;
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 },
};
ArrayList<Integer> ans = colMaxSum(mat);
System.out.println("Row " + (ans.get(0) + 1)
+ " has max sum " + ans.get(1));
}
}// This code is contributed by Vivekkumar Singh |
Python3
# Python3 program to find row with# max sum in a matriximport sys
N = 5 # No of rows and column
# Function to find the row with max sumdef colMaxSum(mat):
# Variable to store index of row
# with maximum
idx = -1
# Variable to store max sum
maxSum = -sys.maxsize
# Traverse matrix row wise
for i in range(0, N):
sum = 0
# calculate sum of row
for j in range(0, N):
sum += mat[i][j]
# Update maxSum if it is less than
# current sum
if (sum > maxSum):
maxSum = sum
# store index
idx = i
res = [idx, maxSum]
# return result
return res
# Driver codemat = [[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 = colMaxSum(mat)
print("Row", ans[0] + 1, "has max sum", ans[1])
# This code is contributed by Sanjit_Prasad |
C#
// C# program to find row with// max sum in a matrixusing System;
using System.Collections.Generic;
public class MaxSum {
public static int N;
static List<int> colMaxSum(int[, ] mat)
{
// Variable to store index of row
// with maximum
int idx = -1;
// Variable to store maximum sum
int maxSum = int.MinValue;
// Traverse the matrix row wise
for (int i = 0; i < N; i++) {
int sum = 0;
for (int j = 0; j < N; j++) {
sum += mat[i, j];
}
// Update maxSum if it is less than
// current row sum
if (maxSum < sum) {
maxSum = sum;
// store index
idx = i;
}
}
// Arraylist to store values of index
// of maximum sum and the maximum sum together
List<int> res = new List<int>();
res.Add(idx);
res.Add(maxSum);
return res;
}
// Driver code
public static void Main(String[] args)
{
N = 5;
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 },
};
List<int> ans = colMaxSum(mat);
Console.WriteLine("Row " + (ans[0] + 1)
+ " has max sum " + ans[1]);
}
}// This code has been contributed by 29AjayKumar |
Javascript
<script>// JavaScript program to find row with// max sum in a matrixvar N;
function colMaxSum(mat)
{ // Variable to store index of row
// with maximum
var idx = -1;
// Variable to store maximum sum
var maxSum = -1000000000;
// Traverse the matrix row wise
for (var i = 0; i < N; i++)
{
var sum = 0;
for (var j = 0; j < N; j++)
{
sum += mat[i][j];
}
// Update maxSum if it is less than
// current row sum
if (maxSum < sum)
{
maxSum = sum;
// store index
idx = i;
}
}
// Arraylist to store values of index
// of maximum sum and the maximum sum together
var res = [];
res.push(idx);
res.push(maxSum);
return res;
}// Driver codeN = 5;var 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]];
var ans = colMaxSum(mat);
document.write("Row "+ (ans[0]+1)+ " has max sum "
+ ans[1]);</script> |
Output
Row 3 has max sum 35
Time complexity: O(N2)
Auxiliary space: O(1)
Example in c:
Approach:
Initialize a variable max_sum to zero and a variable max_row to -1.
Traverse the matrix row by row:
a. Initialize a variable row_sum to zero.
b. Traverse the elements of the current row and add them to row_sum.
c. If row_sum is greater than max_sum, update max_sum to row_sum and max_row to the current row.
Return max_row.
C
#include <stdio.h>int main()
{ int mat[3][4] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
int m = 3;
int n = 4;
int max_sum = 0;
int max_row = -1;
// Traverse the matrix row by row and find the row with
// maximum sum
for (int i = 0; i < m; i++) {
int row_sum = 0;
for (int j = 0; j < n; j++) {
row_sum += mat[i][j];
}
if (row_sum > max_sum) {
max_sum = row_sum;
max_row = i;
}
}
printf("Row with maximum sum is: %d\n", max_row);
return 0;
} |
C++
#include <iostream>using namespace std;
int main()
{ int mat[3][4] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
int m = 3;
int n = 4;
int max_sum = 0;
int max_row = -1;
// Traverse the matrix row by row and find the row with
// maximum sum
for (int i = 0; i < m; i++) {
int row_sum = 0;
for (int j = 0; j < n; j++) {
row_sum += mat[i][j];
}
if (row_sum > max_sum) {
max_sum = row_sum;
max_row = i;
}
}
cout << "Row with maximum sum is: " << max_row << endl;
return 0;
} |
Java
class Main {
public static void main(String[] args)
{
int[][] mat = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 } };
int m = 3;
int n = 4;
int max_sum = 0;
int max_row = -1;
// Traverse the matrix row by row and find the row
// with maximum sum
for (int i = 0; i < m; i++) {
int row_sum = 0;
for (int j = 0; j < n; j++) {
row_sum += mat[i][j];
}
if (row_sum > max_sum) {
max_sum = row_sum;
max_row = i;
}
}
System.out.println("Row with maximum sum is: "
+ max_row);
}
} |
Python3
# Create a 3x4 matrixmat = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]]
m = 3
n = 4
# Initialize max_sum and max_row to 0max_sum = 0
max_row = -1
# Traverse the matrix row by row and find the row with maximum sumfor i in range(m):
row_sum = 0
for j in range(n):
row_sum += mat[i][j]
if row_sum > max_sum:
max_sum = row_sum
max_row = i
# Print the index of the row with maximum sumprint("Row with maximum sum is: ", max_row)
|
Javascript
// Create a 3x4 matrix let mat = [[1, 2, 3, 4], [5, 6, 7, 8],
[9, 10, 11, 12]];
let m = 3;let n = 4;// Initialize max_sum and max_row to 0 let max_sum = 0;let max_row = -1;// Traverse the matrix row by row and find the row with maximum sum for (let i = 0; i < m; i++) {
let row_sum = 0;
for (let j = 0; j < n; j++) {
row_sum += mat[i][j];
}
if (row_sum > max_sum) {
max_sum = row_sum;
max_row = i;
}
}// Print the index of the row with maximum sum console.log("Row with maximum sum is: ", max_row);
|
C#
using System;
class GFG {
public static void Main(string[] args)
{
int[][] mat
= new int[][] { new int[] { 1, 2, 3, 4 },
new int[] { 5, 6, 7, 8 },
new int[] { 9, 10, 11, 12 } };
int m = 3;
int n = 4;
// Initialize max_sum and max_row to 0
int max_sum = 0;
int max_row = -1;
// Traverse the matrix row by row and find the row
// with maximum sum
for (int i = 0; i < m; i++) {
int row_sum = 0;
for (int j = 0; j < n; j++) {
row_sum += mat[i][j];
}
if (row_sum > max_sum) {
max_sum = row_sum;
max_row = i;
}
}
// Print the index of the row with maximum sum
Console.WriteLine("Row with maximum sum is: "
+ max_row);
}
} |
Output
Row with maximum sum is: 2
time complexity of O(mn)
space complexity of O(n)