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

Given a matrix of m*n order, the task is to find the maximum difference between two rows Rj and Ri such that i < j, i.e., we need to find maximum value of sum(Rj) – sum(Ri) such that row i is above row j.

Examples:

Input : mat[5][4] = {{-1, 2, 3, 4},
                     {5, 3, -2, 1},
                     {6, 7, 2, -3},
                     {2, 9, 1, 4},
                     {2, 1, -2, 0}}
Output: 9
// difference of R3 - R1 is maximum

A simple solution for this problem is to one by one select each row, compute sum of elements in it and take difference from sum of next rows in forward direction. Finally return the maximum difference. Time complexity for this approach will be O(n*m²).

An efficient solution solution for this problem is to first calculate the sum of all elements of each row and store them in an auxiliary array rowSum[] and then calculate maximum difference of two elements max(rowSum[j] – rowSum[i]) such that rowSum[i] < rowSum[j] in linear time. See this article. In this method, we need to keep track of 2 things: 

1. Maximum difference found so far (max_diff). 
2. Minimum number visited so far (min_element). 

Implementation:

C++

// C++ program to find maximum difference of sum of
// elements of two rows
#include<bits/stdc++.h>
#define MAX 100
using namespace std;

// Function to find maximum difference of sum of
// elements of two rows such that second row appears
// before first row.
int maxRowDiff(int mat[][MAX], int m, int n)
{
    // auxiliary array to store sum of all elements
    // of each row
    int rowSum[m];

    // calculate sum of each row and store it in
    // rowSum array
    for (int i=0; i<m; i++)
    {
        int sum = 0;
        for (int j=0; j<n; j++)
            sum += mat[i][j];
        rowSum[i] = sum;
    }

    // calculating maximum difference of two elements
    // such that rowSum[i]<rowsum[j]
    int max_diff = rowSum[1] - rowSum[0];
    int min_element = rowSum[0];
    for (int i=1; i<m; i++)
    {
        // if current difference is greater than
        // previous then update it
        if (rowSum[i] - min_element > max_diff)
            max_diff = rowSum[i] - min_element;

        // if new element is less than previous minimum
        // element then update it so that
        // we may get maximum difference in remaining array
        if (rowSum[i] < min_element)
            min_element = rowSum[i];
    }

    return max_diff;
}

// Driver program to run the case
int main()
{
    int m = 5, n = 4;
    int mat[][MAX] = {{-1, 2, 3, 4},
                     {5, 3, -2, 1},
                     {6, 7, 2, -3},
                     {2, 9, 1, 4},
                     {2, 1, -2, 0}};

    cout << maxRowDiff(mat, m, n);
    return 0;
}

C

// C program to find maximum difference of sum of
// elements of two rows
#include <stdio.h>

#define MAX 100

// Function to find maximum difference of sum of
// elements of two rows such that second row appears
// before first row.
int maxRowDiff(int mat[][MAX], int m, int n)
{
    // auxiliary array to store sum of all elements
    // of each row
    int rowSum[m];

    // calculate sum of each row and store it in
    // rowSum array
    for (int i=0; i<m; i++)
    {
        int sum = 0;
        for (int j=0; j<n; j++)
            sum += mat[i][j];
        rowSum[i] = sum;
    }

    // calculating maximum difference of two elements
    // such that rowSum[i]<rowsum[j]
    int max_diff = rowSum[1] - rowSum[0];
    int min_element = rowSum[0];
    for (int i=1; i<m; i++)
    {
        // if current difference is greater than
        // previous then update it
        if (rowSum[i] - min_element > max_diff)
            max_diff = rowSum[i] - min_element;

        // if new element is less than previous minimum
        // element then update it so that
        // we may get maximum difference in remaining array
        if (rowSum[i] < min_element)
            min_element = rowSum[i];
    }

    return max_diff;
}

// Driver program to run the case
int main()
{
    int m = 5, n = 4;
    int mat[][MAX] = {{-1, 2, 3, 4},
                     {5, 3, -2, 1},
                     {6, 7, 2, -3},
                     {2, 9, 1, 4},
                     {2, 1, -2, 0}};
    printf("%d",maxRowDiff(mat, m, n));

    return 0;
}

// This code is contributed by kothavvsaakash.

Java

// Java program to find maximum difference
// of sum of elements of two rows
import java.io.*;
class GFG {
static final int MAX = 100;

// Function to find maximum difference of sum
// of elements of two rows such that second
// row appears before first row.
static int maxRowDiff(int mat[][], int m, int n) {

    // auxiliary array to store sum
    // of all elements of each row
    int rowSum[] = new int[m];

    // calculate sum of each row and
    // store it in rowSum array
    for (int i = 0; i < m; i++) {
    int sum = 0;
    for (int j = 0; j < n; j++)
        sum += mat[i][j];
    rowSum[i] = sum;
    }

    // calculating maximum difference of two elements
    // such that rowSum[i]<rowsum[j]
    int max_diff = rowSum[1] - rowSum[0];
    int min_element = rowSum[0];
    for (int i = 1; i < m; i++) {

    // if current difference is greater than
    // previous then update it
    if (rowSum[i] - min_element > max_diff)
        max_diff = rowSum[i] - min_element;

    // if new element is less than previous
    // minimum element then update it so that
    // we may get maximum difference in remaining array
    if (rowSum[i] < min_element)
        min_element = rowSum[i];
    }

    return max_diff;
}

// Driver code
public static void main(String[] args) {
    int m = 5, n = 4;
    int mat[][] = {{-1, 2,  3, 4 },
                     {5,  3, -2, 1 },
                    {6,  7,  2, -3},
                   {2,  9,  1, 4 },
                   {2,  1, -2, 0}};

    System.out.print(maxRowDiff(mat, m, n));
}
}
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find maximum difference
# of sum of elements of two rows

# Function to find maximum difference of
# sum of elements of two rows such that
# second row appears before first row.
def maxRowDiff(mat, m, n):

    # auxiliary array to store sum of
    # all elements of each row
    rowSum = [0] * m

    # calculate sum of each row and
    # store it in rowSum array
    for i in range(0, m):
        sum = 0
        for j in range(0, n):
            sum += mat[i][j]
        rowSum[i] = sum

    # calculating maximum difference of
    # two elements such that
    # rowSum[i]<rowsum[j]
    max_diff = rowSum[1] - rowSum[0]
    min_element = rowSum[0]

    for i in range(1, m):

        # if current difference is greater
        # than previous then update it
        if (rowSum[i] - min_element > max_diff):
            max_diff = rowSum[i] - min_element

        # if new element is less than previous
        # minimum element then update it so
        # that we may get maximum difference
        # in remaining array
        if (rowSum[i] < min_element):
            min_element = rowSum[i]
    return max_diff

# Driver program to run the case
m = 5
n = 4
mat = [[-1, 2, 3, 4],
       [5, 3, -2, 1],
       [6, 7, 2, -3],
       [2, 9, 1, 4],
       [2, 1, -2, 0]]

print( maxRowDiff(mat, m, n))

# This code is contributed by Swetank Modi

C#

// C# program to find maximum difference
// of sum of elements of two rows
using System;

class GFG {

    // Function to find maximum difference
    // of sum of elements of two rows such
    // that second row appears before
    // first row.
    static int maxRowDiff(int [,] mat,
                              int m, int n)
    {

        // auxiliary array to store sum
        // of all elements of each row
        int [] rowSum = new int[m];

        // calculate sum of each row and
        // store it in rowSum array
        for (int i = 0; i < m; i++)
        {
            int sum = 0;

            for (int j = 0; j < n; j++)
                sum += mat[i,j];

            rowSum[i] = sum;
        }

        // calculating maximum difference
        // of two elements such that
        // rowSum[i] < rowsum[j]
        int max_diff = rowSum[1] - rowSum[0];
        int min_element = rowSum[0];

        for (int i = 1; i < m; i++)
        {

            // if current difference is
            // greater than previous then
            // update it
            if (rowSum[i] - min_element
                                  > max_diff)
                max_diff = rowSum[i]
                             - min_element;

            // if new element is less than
            // previous minimum element then
            // update it so that we may get
            // maximum difference in
            // remaining array
            if (rowSum[i] < min_element)
                min_element = rowSum[i];
        }

        return max_diff;
    }

    // Driver code
    public static void Main()
    {
        int m = 5, n = 4;
        int [,] mat = { {-1, 2, 3, 4 },
                        {5, 3, -2, 1 },
                        {6, 7, 2, -3},
                        {2, 9, 1, 4 },
                        {2, 1, -2, 0} };

        Console.Write(maxRowDiff(mat, m, n));
    }
}

// This code is contributed by KRV.

PHP

<?php
// PHP program to find maximum
// difference of sum of
// elements of two rows
$MAX = 100;

// Function to find maximum
// difference of sum of
// elements of two rows such
// that second row appears
// before first row.
function maxRowDiff($mat, $m, $n)
{
    global $MAX;
    // auxiliary array to store
    // sum of all elements
    // of each row
    $rowSum = array();

    // calculate sum of each
    // row and store it in
    // rowSum array
    for ($i = 0; $i < $m; $i++)
    {
        $sum = 0;
        for ($j = 0; $j < $n; $j++)
            $sum += $mat[$i][$j];
        $rowSum[$i] = $sum;
    }

    // calculating maximum
    // difference of two
    // elements such that
    // rowSum[i]<rowsum[j]
    $max_diff = $rowSum[1] - $rowSum[0];
    $min_element = $rowSum[0];
    for ($i = 1; $i < $m; $i++)
    {
        // if current difference
        // is greater than
        // previous then update it
        if ($rowSum[$i] - $min_element > $max_diff)
            $max_diff = $rowSum[$i] - $min_element;

        // if new element is less
        // than previous minimum
        // element then update it
        // so that we may get maximum
        // difference in remaining array
        if ($rowSum[$i] < $min_element)
            $min_element = $rowSum[$i];
    }

    return $max_diff;
}

// Driver Code
$m = 5;
$n = 4;
$mat = array(array(-1, 2, 3, 4),
             array(5, 3, -2, 1),
             array(6, 7, 2, -3),
             array(2, 9, 1, 4),
             array(2, 1, -2, 0));

echo maxRowDiff($mat, $m, $n);

// This code is contributed by ajit
?>

Javascript

<script>

// Javascript program to find maximum difference
// of sum of elements of two rows

// Function to find maximum difference
// of sum of elements of two rows such
// that second row appears before
// first row.
function maxRowDiff(mat, m, n)
{

    // Auxiliary array to store sum
    // of all elements of each row
    let rowSum = new Array(m);

    // Calculate sum of each row and
    // store it in rowSum array
    for(let i = 0; i < m; i++)
    {
        let sum = 0;

        for(let j = 0; j < n; j++)
            sum += mat[i][j];

        rowSum[i] = sum;
    }

    // Calculating maximum difference
    // of two elements such that
    // rowSum[i] < rowsum[j]
    let max_diff = rowSum[1] - rowSum[0];
    let min_element = rowSum[0];

    for(let i = 1; i < m; i++)
    {

        // If current difference is
        // greater than previous then
        // update it
        if (rowSum[i] - min_element > max_diff)
            max_diff = rowSum[i] - min_element;

        // If new element is less than
        // previous minimum element then
        // update it so that we may get
        // maximum difference in
        // remaining array
        if (rowSum[i] < min_element)
            min_element = rowSum[i];
    }
    return max_diff;
}

// Driver code
let m = 5, n = 4;
let mat = [ [ -1, 2, 3, 4 ],
            [ 5, 3, -2, 1 ],
            [ 6, 7, 2, -3 ],
            [ 2, 9, 1, 4  ],
            [ 2, 1, -2, 0 ] ];

document.write(maxRowDiff(mat, m, n));

// This code is contributed by divyesh072019

</script>

Output

9

Time Complexity : O(m*n) 
Auxiliary Space : O(m)