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++
// 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 matrixint 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 Codeint 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 matriximport java.io.*;class GFG { static int N= 4; // Rowsstatic int M = 4; // Columns// Function to find pair with maximum// difference in a matrixstatic 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 matrixN = 4 # RowsM = 4 # Columns# Function to find pair with maximum# difference in a matrixdef 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# matrixmat = [[ 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 matrixusing System;class GFG {static int N = 4; // Rowsstatic int M = 4; // Columns// Function to find pair with // maximum difference in a matrixstatic 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 Codepublic 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 matrixlet N= 4; // Rowslet M = 4; // Columns// Function to find pair with maximum// difference in a matrixfunction 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 Codelet 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
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