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Maximum value of |arr[0] – arr[1]| + |arr[1] – arr[2]| + … +|arr[n – 2] – arr[n – 1]| when elements are from 1 to n

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Given an array arr[] of size n whose elements are from the range [1, n]. The task is to find maximum value of |arr[0] – arr[1]| + |arr[1] – arr[2]| + … +|arr[n – 2] – arr[n – 1]|. You can arrange the numbers in the array in any order.

Examples:

Input: arr[] = {1, 2, 3, 4} 
Output:
Arrange the array in this way for max value, arr[] = {3, 1, 4, 2} 
|3 – 1| + |1 – 4| + |4 – 2| = 2 + 3 + 2 = 7

Input: arr[] = {1, 2, 3} 
Output:
We arrange the array as {2, 1, 3} 

A Simple Approach is to generate all possible permutations. Compute the value for every permutation and find the maximum value.
Efficient Approach: 
The maximum sum with one element is 0. 
The maximum sum with two elements is 1 
The maximum sum with three elements is 3 (explained above) 
The maximum    sum with four elements is 7 (explained above)
It can be observed that for different values of n, a pattern for the maximum sum of absolute differences is 0, 1, 3, 7, 11, 17, 23, 31, 39, 49, ….. whose nth term is ((n * n / 2) – 1).

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
// required value
int maxValue(int n)
{
    if (n == 1)
        return 0;
 
    return ((n * n / 2) - 1);
}
 
// Driver code
int main()
{
    int n = 4;
    cout << maxValue(n);
 
    return 0;
}


Java




// Java implementation of the approach
class GFG {
 
    // Function to return the maximum
    // required value
    static int maxValue(int n)
    {
        if (n == 1)
            return 0;
 
        return ((n * n / 2) - 1);
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
        System.out.print(maxValue(n));
    }
}


Python




# Python3 implementation of the approach
 
# Function to return the maximum
# required value
def maxValue(n):
    if (n == 1):
        return 0
     
    return (( n * n // 2 ) - 1 )
 
# Driver code
n = 4
print(maxValue(n))


C#




// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the maximum
    // required value
    static int maxValue(int n)
    {
        if (n == 1)
            return 0;
 
        return ((n * n / 2) - 1);
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.WriteLine(maxValue(n));
    }
}


PHP




<?php
// Function to return the maximum
// required value
function maxValue($n)
{
    if ($n == 1)
        return 0;
 
    return (($n * $n / 2) - 1);
}
 
// Driver code
$n = 4;
echo maxValue($n);
?>


Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the maximum
// required value
function maxValue(n)
{
    if (n == 1)
        return 0;
    return (parseInt(n * n / 2) - 1);
}
 
// Driver code
var n = 4;
document.write(maxValue(n));
 
// This code is contributed by noob2000.
</script>


Output

7

Time Complexity: O(1)
Auxiliary Space: O(1)
 



Last Updated : 24 Jun, 2022
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