Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Pair with maximum difference in a Matrix

  • Difficulty Level : Easy
  • Last Updated : 28 May, 2021

Given a NxM matrix with N rows and M columns of positive integers. The task is to find the pair with the maximum difference in the given matrix.
Note: Pairs at positions (a, b) and (b, a) are considered equivalent.
Examples
 

Input : mat[N][M] = {{1, 2, 3, 4},
                 {25, 6, 7, 8},
                 {9, 10, 11, 12},
                 {13, 14, 15, 16}}
Output : 24
Pair (25, 1) has the maximum difference

Input : mat[N][M] = {{1, 2, 3},
                 {4, 6, 7},
                 {9, 10, 5}}
Output : 9
Pair (10, 1) has the maximum difference.

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

The idea is to observe that the elements contributing to the pair with maximum difference are the maximum and minimum elements in the matrix. So, find the maximum and minimum elements in the matrix and return the difference between them.
Below is the implementation of the above approach: 
 

C++




// C++ program to find with maximum
// difference in a matrix
 
#include <bits/stdc++.h>
using namespace std;
 
#define N 4 // Rows
#define M 4 // Columns
 
// Function to find pair with maximum
// difference in a matrix
int maxDifferencePair(int mat[N][M])
{
    int maxElement = INT_MIN; // max
    int minElement = INT_MAX; // min
 
    // Traverse the matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            // Find max element
            if (mat[i][j] > maxElement) {
                maxElement = mat[i][j];
            }
 
            // Find min element
            if (mat[i][j] < minElement) {
                minElement = mat[i][j];
            }
        }
    }
 
    return abs(maxElement - minElement);
}
 
// Driver Code
int main()
{
    // matrix
    int mat[N][M] = { { 1, 2, 3, 4 },
                      { 25, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 } };
 
    cout << maxDifferencePair(mat) << endl;
 
    return 0;
}

Java




// Java program to find with maximum
// difference in a matrix
 
import java.io.*;
 
class GFG {
  
static int N= 4; // Rows
static int  M = 4; // Columns
 
// Function to find pair with maximum
// difference in a matrix
static int maxDifferencePair(int mat[][])
{
    int maxElement = Integer.MIN_VALUE; // max
    int minElement = Integer.MAX_VALUE; // min
 
    // Traverse the matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            // Find max element
            if (mat[i][j] > maxElement) {
                maxElement = mat[i][j];
            }
 
            // Find min element
            if (mat[i][j] < minElement) {
                minElement = mat[i][j];
            }
        }
    }
 
    return Math.abs(maxElement - minElement);
}
 
// Driver Code
 
    public static void main (String[] args) {
        // matrix
    int mat[][] = { { 1, 2, 3, 4 },
                    { 25, 6, 7, 8 },
                    { 9, 10, 11, 12 },
                    { 13, 14, 15, 16 } };
 
    System.out.println( maxDifferencePair(mat));
    }
}
 
// This code is contributed by inder_verma..

Python3




# Python3 program to find with maximum
# difference in a matrix
 
N = 4 # Rows
M = 4 # Columns
 
# Function to find pair with maximum
# difference in a matrix
def maxDifferencePair(mat):
 
    maxElement = -10**9 # max
    minElement = 10**9 # min
 
    # Traverse the matrix
    for i in range(N):
        for j in range(M):
             
            # Find max element
            if (mat[i][j] > maxElement):
                maxElement = mat[i][j]
 
            # Find min element
            if (mat[i][j] < minElement):
                minElement = mat[i][j]
 
    return abs(maxElement - minElement)
 
# Driver Code
 
# matrix
mat = [[ 1, 2, 3, 4 ],
       [ 25, 6, 7, 8 ],
       [ 9, 10, 11, 12 ],
       [ 13, 14, 15, 16]]
 
print(maxDifferencePair(mat))
 
# This code is contributed
# by mohit kumar

C#




// C# program to find with maximum
// difference in a matrix
using System;
 
class GFG
{
static int N = 4; // Rows
static int M = 4; // Columns
 
// Function to find pair with
// maximum difference in a matrix
static int maxDifferencePair(int [,]mat)
{
    int maxElement = int.MinValue; // max
    int minElement = int.MaxValue; // min
 
    // Traverse the matrix
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            // Find max element
            if (mat[i, j] > maxElement)
            {
                maxElement = mat[i, j];
            }
 
            // Find min element
            if (mat[i, j] < minElement)
            {
                minElement = mat[i, j];
            }
        }
    }
 
    return Math.Abs(maxElement -
                    minElement);
}
 
// Driver Code
public static void Main ()
{
    // matrix
    int [,]mat = {{ 1, 2, 3, 4 },
                  { 25, 6, 7, 8 },
                  { 9, 10, 11, 12 },
                  { 13, 14, 15, 16 }};
     
    Console.WriteLine( maxDifferencePair(mat));
}
}
 
// This code is contributed
// by inder_verma

Javascript




<script>
 
// JavaScript program to find with maximum
// difference in a matrix
 
let N= 4; // Rows
let M = 4; // Columns
 
// Function to find pair with maximum
// difference in a matrix
function maxDifferencePair(mat)
{
    let maxElement = Number.MIN_VALUE; // max
    let minElement = Number.MAX_VALUE; // min
   
    // Traverse the matrix
    for (let i = 0; i < N; i++) {
        for (let j = 0; j < M; j++) {
            // Find max element
            if (mat[i][j] > maxElement) {
                maxElement = mat[i][j];
            }
   
            // Find min element
            if (mat[i][j] < minElement) {
                minElement = mat[i][j];
            }
        }
    }
   
    return Math.abs(maxElement - minElement);
}
 
// Driver Code
let mat = [[ 1, 2, 3, 4 ],
       [ 25, 6, 7, 8 ],
       [ 9, 10, 11, 12 ],
       [ 13, 14, 15, 16]];
        
document.write( maxDifferencePair(mat));
 
 
// This code is contributed by unknown2108
 
</script>
Output: 
24

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!