Find the sum of all possible pairs in an array of N elements
Given an array arr[] of N integers, the task is to find the sum of all the pairs possible from the given array. Note that,
- (arr[i], arr[i]) is also considered as a valid pair.
- (arr[i], arr[j]) and (arr[j], arr[i]) are considered as two different pairs.
Examples:
Input: arr[] = {1, 2}
Output: 12
All valid pairs are (1, 1), (1, 2), (2, 1) and (2, 2).
1 + 1 + 1 + 2 + 2 + 1 + 2 + 2 = 12
Input: arr[] = {1, 2, 3, 1, 4}
Output: 110
Naive approach: Find all the possible pairs and calculate the sum of the elements of each pair.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int sumPairs( int arr[], int n)
{
int sum = 0;
for ( int i = 0; i < n; i++) {
for ( int j = 0; j < n; j++) {
sum += (arr[i] + arr[j]);
}
}
return sum;
}
int main()
{
int arr[] = { 1, 2, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << sumPairs(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int sumPairs( int arr[], int n)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
{
for ( int j = 0 ; j < n; j++)
{
sum += (arr[i] + arr[j]);
}
}
return sum;
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 };
int n = arr.length;
System.out.println(sumPairs(arr, n));
}
}
|
Python3
def summPairs(arr, n):
summ = 0
for i in range (n):
for j in range (n):
summ + = (arr[i] + arr[j])
return summ
arr = [ 1 , 2 , 3 ]
n = len (arr)
print (summPairs(arr, n))
|
C#
using System;
class GFG
{
static int sumPairs( int []arr, int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
sum += (arr[i] + arr[j]);
}
}
return sum;
}
public static void Main(String[] args)
{
int []arr = {1, 2, 3};
int n = arr.Length;
Console.WriteLine(sumPairs(arr, n));
}
}
|
Javascript
<script>
function sumPairs(arr, n)
{
var sum = 0;
var i, j;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
sum += (arr[i] + arr[j]);
}
}
return sum;
}
var arr = [ 1, 2, 3 ];
var n = arr.length;
document.write(sumPairs(arr, n));
</script>
|
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient approach: It can be observed that each element appears exactly (2 * N) times as one of the elements of the pair (x, y). Exactly N times as x and exactly N times as y.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int sumPairs( int arr[], int n)
{
int sum = 0;
for ( int i = 0; i < n; i++) {
sum = sum + (arr[i] * (2 * n));
}
return sum;
}
int main()
{
int arr[] = { 1, 2, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << sumPairs(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int sumPairs( int arr[], int n)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
{
sum = sum + (arr[i] * ( 2 * n));
}
return sum;
}
static public void main(String []arg)
{
int arr[] = { 1 , 2 , 3 };
int n = arr.length;
System.out.println(sumPairs(arr, n));
}
}
|
Python3
def sumPairs(arr, n) :
sum = 0 ;
for i in range (n) :
sum = sum + (arr[i] * ( 2 * n));
return sum ;
if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 ];
n = len (arr);
print (sumPairs(arr, n));
|
C#
using System;
class GFG
{
static int sumPairs( int []arr, int n)
{
int sum = 0;
for ( int i = 0; i < n; i++)
{
sum = sum + (arr[i] * (2 * n));
}
return sum;
}
static public void Main(String []arg)
{
int []arr = { 1, 2, 3 };
int n = arr.Length;
Console.WriteLine(sumPairs(arr, n));
}
}
|
Javascript
<script>
function sumPairs(arr, n)
{
let sum = 0;
for (let i = 0; i < n; i++) {
sum = sum + (arr[i] * (2 * n));
}
return sum;
}
let arr = [ 1, 2, 3 ];
let n = arr.length;
document.write(sumPairs(arr, n));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
07 Jan, 2024
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