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++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the maximum // absolute difference between // any two elements of the array int 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 code int main()
{ int arr[] = { 2, 1, 5, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxAbsDiff(arr, n);
return 0;
} |
// Java implementation of the approach import 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));
}
} |
# python implementation of the approach # Function to return the maximum # absolute difference between # any two elements of the array def maxAbsDiff(arr, n):
# sorting the array
arr.sort()
# returning the difference between last and first element
return arr[n - 1 ] - arr[ 0 ]
#driver code arr = [ 2 , 1 , 5 , 3 ]
n = len (arr)
print (maxAbsDiff(arr, n))
|
// C# implementation of the approach using System;
class Program
{ // Function to return the maximum // absolute difference between // any two elements of the array static int maxAbsDiff( int [] arr, int n)
{ //Sorting the array Array.Sort(arr); //returning the difference between last and first element
return arr[n - 1] - arr[0];
} // Driver code static 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 implementation of the approach // Function to return the maximum // absolute difference between // any two elements of the array function maxAbsDiff(arr, n)
{ //Sorting the array
arr.sort();
//returning the difference between last and first element
return arr[n-1]-arr[0];
} // Driver code let arr = [ 2, 1, 5, 3 ]; let n = arr.length; console.log(maxAbsDiff(arr, n)); // The code is contributed by Arushi Jindal. |
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++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the maximum // absolute difference between // any two elements of the array 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 = min(minEle, arr[i]);
maxEle = max(maxEle, arr[i]);
}
return (maxEle - minEle);
} // Driver code int main()
{ int arr[] = { 2, 1, 5, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxAbsDiff(arr, n);
return 0;
} |
// Java implementation of the approach 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(String[] args)
{
int [] arr = { 2 , 1 , 5 , 3 };
int n = arr.length;
System.out.print(maxAbsDiff(arr, n));
}
} |
# Python3 implementation of the approach # Function to return the maximum # absolute difference between # any two elements of the array def 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 code arr = [ 2 , 1 , 5 , 3 ]
n = len (arr)
print (maxAbsDiff(arr, n))
# This code is contributed # by mohit kumar |
// C# implementation of the approach using 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 implementation of the approach // Function to return the maximum // absolute difference between // any two elements of the array function 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 |
<script> // JavaScript implementation of the approach // Function to return the maximum // absolute difference between // any two elements of the array function 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> |
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++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the maximum // absolute difference between // any two elements of the array int 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 code int main()
{ int arr[] = { -10, 4, -9, -5 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxAbsDiff(arr, n);
return 0;
} |
//Java program for the above approach import 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 |
# Python program for the above approach # Function to return the maximum # absolute difference between # any two elements of the array def 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 code arr = [ - 10 , 4 , - 9 , - 5 ]
n = len (arr)
print (maxAbsDiff(arr, n))
# This code is contributed by Shubham Singh |
// C# program for the above approach using 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 |
<script> // Javascript program for the above approach // Function to return the maximum // absolute difference between // any two elements of the array function 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 code var arr = [ -10, 4, -9, -5 ];
var n = arr.Length;
document.write(maxAbsDiff(arr, n)); // This code is contributed by Shubham Singh </script> |
14
Time Complexity : O(n)
Auxiliary Space: O(1)