Depend of distinct pair sum in given Array

#embody <bits/stdc++.h>

utilizing namespace std;

int rely(int arr[], int n)

{

    unordered_set<int> s;

    for (int i = 0; i < n - 1; i++) {

        for (int j = i + 1; j < n; j++) {

            s.insert(arr[i] + arr[j]);

        }

    }

    return s.dimension();

}

int major()

{

    int arr[] = { 6, 1, 4, 3 };

    int N = sizeof(arr) / sizeof(arr[0]);

    cout << rely(arr, N);

    return 0;

}

import java.util.*;

class GFG {

    static int rely(int arr[], int n)

    {

        Set<Integer> s = new HashSet<Integer>();

        for (int i = 0; i < n - 1; i++) {

            for (int j = i + 1; j < n; j++) {

                s.add(arr[i] + arr[j]);

            }

        }

        return s.dimension();

    }

    public static void major(String args[])

    {

        int arr[] = { 6, 1, 4, 3 };

        int N = arr.size;

        System.out.println(rely(arr, N));

    }

}

def rely(arr, n):

    s = set()

    for i in vary(n-1):

        for j in vary(i + 1, n):

            s.add(arr[i] + arr[j])

    return int(len(s))

if __name__ == '__main__':

    arr = [6, 1, 4, 3]

    N = len(arr)

    print(rely(arr, N))

<script>

operate rely(arr, n) {

    let s = new Set();

    for (let i = 0; i < n - 1; i++) {

        for (let j = i + 1; j < n; j++) {

            s.add(arr[i] + arr[j]);

        }

    }

    return s.dimension;

}

let arr = [6, 1, 4, 3];

let N = arr.size;

doc.write(rely(arr, N));

</script>