Given an array arr[] of N integers, the task is to find the maximum difference between any two elements of the array.
Examples:
Input: arr[] = {2, 1, 5, 3}
Output: 4
|5 – 1| = 4Input: arr[] = {-10, 4, -9, -5}
Output: 14
Naive Approach:- As the maximum difference will be in between smallest and the largest array so we will simply sort the array and get the maximum difference.
Below is the implementation for the same:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum// absolute difference between// any two elements of the arrayint maxAbsDiff(int arr[], int n){ //Sorting the array sort(arr,arr+n); //returning the difference between last and first element return arr[n-1]-arr[0];}// Driver codeint main(){ int arr[] = { 2, 1, 5, 3 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxAbsDiff(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{ // Function to return the maximum // absolute difference between // any two elements of the array static int maxAbsDiff(int arr[], int n) { // Sorting the array Arrays.sort(arr); // returning the difference between last and first element return arr[n-1] - arr[0]; } // Driver code public static void main(String args[]) { int arr[] = {2,1,5,3}; int n = arr.length; System.out.println(maxAbsDiff(arr, n)); }} |
Python3
# python implementation of the approach# Function to return the maximum# absolute difference between# any two elements of the arraydef maxAbsDiff(arr, n): # sorting the array arr.sort() # returning the difference between last and first element return arr[n - 1] - arr[0]#driver codearr = [2, 1, 5, 3]n = len(arr)print(maxAbsDiff(arr, n)) |
C#
// C# implementation of the approachusing System;class Program{// Function to return the maximum// absolute difference between// any two elements of the arraystatic int maxAbsDiff(int[] arr, int n){//Sorting the arrayArray.Sort(arr); //returning the difference between last and first element return arr[n - 1] - arr[0];}// Driver codestatic void Main(string[] args){ int[] arr = { 2, 1, 5, 3 }; int n = arr.Length; Console.WriteLine(maxAbsDiff(arr, n));}}//This code is contributed by shivamsharma215 |
Javascript
// Javascript implementation of the approach// Function to return the maximum// absolute difference between// any two elements of the arrayfunction maxAbsDiff(arr, n){ //Sorting the array arr.sort(); //returning the difference between last and first element return arr[n-1]-arr[0];}// Driver codelet arr = [ 2, 1, 5, 3 ];let n = arr.length;console.log(maxAbsDiff(arr, n));// The code is contributed by Arushi Jindal. |
Output
4
Time Complexity:- O(nlogn)
Auxiliary Space:- O(1)
Approach: The maximum absolute difference in the array will always be the absolute difference between the minimum and the maximum element from the array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum// absolute difference between// any two elements of the arrayint maxAbsDiff(int arr[], int n){ // To store the minimum and the maximum // elements from the array int minEle = arr[0]; int maxEle = arr[0]; for (int i = 1; i < n; i++) { minEle = min(minEle, arr[i]); maxEle = max(maxEle, arr[i]); } return (maxEle - minEle);}// Driver codeint main(){ int arr[] = { 2, 1, 5, 3 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxAbsDiff(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to return the maximum // absolute difference between // any two elements of the array static int maxAbsDiff(int arr[], int n) { // To store the minimum and the maximum // elements from the array int minEle = arr[0]; int maxEle = arr[0]; for (int i = 1; i < n; i++) { minEle = Math.min(minEle, arr[i]); maxEle = Math.max(maxEle, arr[i]); } return (maxEle - minEle); } // Driver code public static void main(String[] args) { int[] arr = { 2, 1, 5, 3 }; int n = arr.length; System.out.print(maxAbsDiff(arr, n)); }} |
Python3
# Python3 implementation of the approach# Function to return the maximum# absolute difference between# any two elements of the arraydef maxAbsDiff(arr, n): # To store the minimum and the maximum # elements from the array minEle = arr[0] maxEle = arr[0] for i in range(1, n): minEle = min(minEle, arr[i]) maxEle = max(maxEle, arr[i]) return (maxEle - minEle)# Driver codearr = [2, 1, 5, 3]n = len(arr)print(maxAbsDiff(arr, n))# This code is contributed# by mohit kumar |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the maximum // absolute difference between // any two elements of the array static int maxAbsDiff(int []arr, int n) { // To store the minimum and the maximum // elements from the array int minEle = arr[0]; int maxEle = arr[0]; for (int i = 1; i < n; i++) { minEle = Math.Min(minEle, arr[i]); maxEle = Math.Max(maxEle, arr[i]); } return (maxEle - minEle); } // Driver code public static void Main() { int[] arr = { 2, 1, 5, 3 }; int n = arr.Length; Console.WriteLine(maxAbsDiff(arr, n)); }}// This code is contributed by Ryuga |
PHP
<?php// PHP implementation of the approach// Function to return the maximum// absolute difference between// any two elements of the arrayfunction maxAbsDiff($arr, $n){ // To store the minimum and the maximum // elements from the array $minEle = $arr[0]; $maxEle = $arr[0]; for ($i = 1; $i < $n; $i++) { $minEle = min($minEle, $arr[$i]); $maxEle = max($maxEle, $arr[$i]); } return ($maxEle - $minEle);}// Driver code$arr = array(2, 1, 5, 3);$n = sizeof($arr);echo maxAbsDiff($arr, $n);// This code is contributed// by Akanksha Rai |
Javascript
<script>// JavaScript implementation of the approach// Function to return the maximum// absolute difference between// any two elements of the arrayfunction maxAbsDiff(arr, n){ // To store the minimum and the maximum // elements from the array let minEle = arr[0]; let maxEle = arr[0]; for (let i = 1; i < n; i++) { minEle = Math.min(minEle, arr[i]); maxEle = Math.max(maxEle, arr[i]); } return (maxEle - minEle);}// Driver code let arr = [ 2, 1, 5, 3 ]; let n = arr.length; document.write(maxAbsDiff(arr, n));</script> |
Output
4
Time Complexity : O(n)
Auxiliary Space : O(1)
Another Approach ( Using STL): The maximum absolute difference in the array will always be the absolute difference between the minimum and the maximum element from the array. Below is the implementation of the above approach:
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum// absolute difference between// any two elements of the arrayint maxAbsDiff(int arr[], int n){ // To find the minimum and the maximum element // using stl int maxele = *max_element(arr, arr + n); int minele = *min_element(arr, arr + n); // make variable to store answer int ans = abs(maxele - minele); return ans;}// Driver codeint main(){ int arr[] = { -10, 4, -9, -5 }; int n = sizeof(arr) / sizeof(arr[0]); cout << maxAbsDiff(arr, n); return 0;} |
Java
//Java program for the above approachimport java.util.Arrays;class GFG { // Function to return the maximum // absolute difference between // any two elements of the array public static int maxAbsDiff(int arr[], int n) { // To find the minimum and the maximum element // using stl int maxele = Arrays.stream(arr).max().getAsInt(); int minele = Arrays.stream(arr).min().getAsInt(); // make variable to store answer int ans = Math.abs(maxele - minele); return ans; } // Driver code public static void main (String[] args) { int arr[] = { -10, 4, -9, -5 }; int n = arr.length; System.out.print(maxAbsDiff(arr, n)); }}// This code is contributed by Shubham Singh |
Python3
# Python program for the above approach# Function to return the maximum# absolute difference between# any two elements of the arraydef maxAbsDiff(arr, n): # To find the minimum and the maximum element maxele = max(arr) minele = min(arr) # make variable to store answer ans = abs(maxele - minele) return ans# Driver codearr = [ -10, 4, -9, -5 ]n = len(arr)print(maxAbsDiff(arr, n))# This code is contributed by Shubham Singh |
C#
// C# program for the above approachusing System;using System.Linq;public class GFG{ // Function to return the maximum // absolute difference between // any two elements of the array public static int maxAbsDiff(int[] arr, int n) { // To find the minimum and the maximum element int maxele = arr.Max(); int minele = arr.Min(); // make variable to store answer int ans = Math.Abs(maxele - minele); return ans; } // Driver code static public void Main () { int[] arr = { -10, 4, -9, -5 }; int n = arr.Length; Console.Write(maxAbsDiff(arr, n)); }}// This code is contributed by Shubham Singh |
Javascript
<script>// Javascript program for the above approach// Function to return the maximum// absolute difference between// any two elements of the arrayfunction maxAbsDiff(arr, n){ // To find the minimum and the maximum element var maxele = Math.max(...arr); var minele = Math.min(...arr); // make variable to store answer var ans = Math.abs(maxele - minele); return ans;}// Driver codevar arr = [ -10, 4, -9, -5 ];var n = arr.Length;document.write(maxAbsDiff(arr, n));// This code is contributed by Shubham Singh</script> |
Output
14
Time Complexity : O(n)
Auxiliary Space: O(1)
