info@jrdhub.com

Count of distinct pair sum in given Array

gfg_200x200-min.png
Posted by Jrd Hub
Uncategorized 0


Given an array arr[] of size N, the task is to find the total number of unique pair sums possible from the array elements.

Examples:

Input: arr[] = {6, 1, 4, 3}
Output: 5
Explanation: All pair possible are {6, 1}, {6, 4}, {6, 3}, {1, 4}, {1, 3}, {4, 3}. S
ums of these pairs are 7, 10, 9, 5, 4, 7. So unique sums 7, 10, 9, 5, 4. So answer is 5.

Input: arr[] = {8, 7, 6, 5, 4, 3, 2, 1}
Output: 13

 

Approach: This problem can be efficiently solved by using unordered_set. 

Calculate all possible sum of pairs and store them in an unordered set. This is done to store the store elements in an average time of O(1) with no value repeating. 

Algorithm:

  • Use nested loops to get all possible sums of elements of the array.
  • Store all the possible sums into an unordered_set.
  • Total possible unique sums would be equal to the size of unordered_set. So return the size of the unordered set.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>

using namespace std;

  

int count(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.size();

}

  

int main()

{

    int arr[] = { 6, 1, 4, 3 };

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

  

    

    cout << count(arr, N);

    return 0;

}

Java

import java.util.*;

  

class GFG {

    

    

    static int count(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.size();

    }

  

    

    public static void main(String args[])

    {

        int arr[] = { 6, 1, 4, 3 };

        int N = arr.length;

  

        

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

    }

}

Python3

  

def count(arr, n):

    s = set()

      

    

    for i in range(n-1):

        for j in range(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(count(arr, N))

Javascript

<script>

  

  

function count(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.size;

}

  

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

let N = arr.length;

  

document.write(count(arr, N));

  

</script>

Time complexity: O(N2)
Auxiliary Space: O(N)



Source link

Share:
Previous Post

Dedicated Server Clearance sale, high spec, Fully Managed ready from ardhosting.com
Next Post

'Dungeons & Dragons' movie trailer looks like a loud, dumb and hopefully fun time

Leave a Reply