Maximum difference of sum of elements in two rows in a matrix in C

We are given a matrix and the task is to find the greatest difference between the sum of elements in two rows of a matrix. Suppose we have a matrix M[i,j] with i rows and j columns. Let the rows be R₀ to R_i-1. The difference will be calculated by subtracting the (sum of elements of R_y) - (sum of elements of R_x), where x<y.

Let's now understand what we have to do using an example −

Input

M[4][4] = {
   { 1,2,0,5 },
   {0,1,1,0},
   {7,2,3,2}
   {1,2,4,1}};

Output

Maximum difference here is : 12

Explanation − Here sum of elements of Row 2 is maximum that is 14 and sum of elements of Row 1 is minimum, that is 2. So the maximum difference is 14-2=12.

Syntax

int rowmaxd(int M[][MAX], int row, int col);

Approach

• Calculate the sum of elements for each row of the matrix
• Store these row sums in an array
• Find the maximum difference between any two row sums where the larger sum comes from a row with higher index
• Use the concept of finding maximum difference in an array where j > i

Example

#include <stdio.h>
#define MAX 100

// Function to calculate maximum difference between sum of elements of two rows
int rowmaxd(int M[][MAX], int row, int col) {
    // Array for storing sum of elements of each row
    int RSum[row];
    
    // Calculate sum of each row
    for(int i = 0; i < row; i++) {
        int sum = 0;
        for(int j = 0; j < col; j++)
            sum += M[i][j];
        RSum[i] = sum;
    }
    
    // Calculate max difference between two elements of RSum such that in RSum[j]-RSum[i], i<j
    int MD = RSum[1] - RSum[0];
    int MIN = RSum[0];
    
    for (int i = 1; i < row; i++) {
        // If this difference is more than MD, then update MD
        if(RSum[i] - MIN > MD)
            MD = RSum[i] - MIN;
        
        // If this value is even less than MIN, then update MIN
        if(RSum[i] < MIN)
            MIN = RSum[i];
    }
    return MD;
}

int main() {
    int r = 5, c = 4;
    int mat[][MAX] = {
        {-1, 2, 3, 4},
        {6, 3, 0, 1},
        {-1, 7, 8, -3},
        {3, 5, 1, 4},
        {2, 1, 1, 0}};
    
    printf("Maximum difference of sum of elements in two rows in a matrix is: %d<br>", rowmaxd(mat, r, c));
    return 0;
}

Output

Maximum difference of sum of elements in two rows in a matrix is: 5

How It Works

The algorithm works by:

• Step 1: Computing row sums: [8, 10, 11, 13, 4]
• Step 2: Finding maximum difference where later row sum minus earlier row sum gives maximum value
• Step 3: Tracking minimum seen so far and maximum difference simultaneously

Conclusion

This approach efficiently finds the maximum difference between row sums in O(n*m) time complexity where n is rows and m is columns. It uses the principle of finding maximum difference in an array with positional constraints.