Maximum difference of sum of elements in two rows in a matrix
Last Updated :
23 Jul, 2025
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:
- Maximum difference found so far (max_diff).
- 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 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 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 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# 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 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
?>
<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>
Time Complexity : O(m*n)
Auxiliary Space : O(m)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem