# Maximum value of |arr[0] β arr[1] - + |arr[1] β arr[2] - + β¦ +|arr[n β 2] β arr[n β 1] - when elements are from 1 to n in C++

In this problem, we are given an array of n integers of range [1,n]. Our task is to create a program that will find the maximum value of |arr[0] – arr[1] - + |arr[1] – arr[2] - + … +|arr[n – 2] – arr[n – 1].

Let’s take an example to understand the problem,

Input − array= {1, 2, 3}

Output − 3

Explanation

max sum is
|1-3|+|2-1| = 3

To solve this problem, a simple approach is to create all permutations from the array. And find the maximum value of all the values from permutation. A more effective method is to generalize all the maximum values for each value of n and then create a general formula.

So,

Maximum sum for (n = 1) = 0
Maximum sum for (n = 2) = 1
Maximum sum for (n = 3) = 3
Maximum sum for (n = 4) = 7
Maximum sum for (n = 5) = 11
So, the maximum value is 0, 1, 3, 7, 11…

The general formula is, ((n*n/2)-1)

## Example

Program to illustrate the working of our solution,

Live Demo

#include <iostream>
using namespace std;
int maxAbsVal(int n) {
if (n == 1)
return 0;
return ((n*n/2) - 1);
}
int main() {
int n = 4;
cout<<"The maximum sum of absolute difference is "<<maxAbsVal(n);
return 0;
}

## Output

The maximum sum of absolute difference is 7