Pair with maximum difference in a Matrix

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.

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++

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// 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;
}

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Java

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// 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..

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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 [tabby title = "C#"]

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// 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

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Output:

24


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Improved By : inderDuMCA, mohit kumar 29



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