Does array length plays any role while finding a repeated element in array list?

public static int findDuplicate(int[] arr) {
    
    int n = arr.length-2;
    int sum = 0;
    for(int i : arr){
        sum+=i;
    }
    
    return (sum - ((n*(n+1))/2));
}

Here is the code. This code must return any duplicate element present in the array. Say, if array is { 1, 2, 3, 4, 1 } then it must return 1. It works fine for the test cases but I just want to understand the role of arr.length-2 used here.

Sample input
5 size
1 2 3 4 1
Sample output :
1

>Solution :

Your solution is based on assumption that your array must contain

• all numbers from 0 till "some" N, like 0, 1, 2, 3, .. , N (in any order)
• and one extra element X.

How many elements are there?

• 0 ... N gives us N+1 elements
• X is 1 element.

So we have N+1 + 1 which is N+2 element.

In other words array has N+2 elements, which means N+2 = array.length. And that means

N = array.length - 2.

Now sum of all array elements is 0 + 1 + 2 + ... + N + X (their order doesn’t matter for calculating sum).

We can get rid of 0 since it doesn’t affect our sum so 0 + 1 + 2 + ... + N becomes 1 + 2 + ... + N which leaves us with

SUM = 1 + 2 + ... + N + X

and here we can also replace 1 + 2 + ... + N with N*(N+1)/2 based on formula on sum of arithmetic series which gives us

SUM = N*(N+1)/2 + X

Based on above the duplicate element X is X = SUM - N*(N+1)/2 where N = array.length-2.