# 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,

1. (arr[i], arr[i]) is also considered as a valid pair.
2. (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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the sum of the elements ` `// of all possible pairs from the array ` `int` `sumPairs(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To store the required sum ` `    ``int` `sum = 0; ` ` `  `    ``// Nested loop for all possible pairs ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = 0; j < n; j++) { ` ` `  `            ``// Add the sum of the elements ` `            ``// of the current pair ` `            ``sum += (arr[i] + arr[j]); ` `        ``} ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << sumPairs(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to return the sum of the elements ` `    ``// of all possible pairs from the array ` `    ``static` `int` `sumPairs(``int` `arr[], ``int` `n) ` `    ``{ ` ` `  `        ``// To store the required sum ` `        ``int` `sum = ``0``; ` ` `  `        ``// Nested loop for all possible pairs ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``for` `(``int` `j = ``0``; j < n; j++)  ` `            ``{ ` ` `  `                ``// Add the sum of the elements ` `                ``// of the current pair ` `                ``sum += (arr[i] + arr[j]); ` `            ``} ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``1``, ``2``, ``3``}; ` `        ``int` `n = arr.length; ` ` `  `        ``System.out.println(sumPairs(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the summ of the elements ` `# of all possible pairs from the array ` `def` `summPairs(arr, n): ` ` `  `    ``# To store the required summ ` `    ``summ ``=` `0` ` `  `    ``# Nested loop for all possible pairs ` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(n): ` ` `  `            ``# Add the summ of the elements ` `            ``# of the current pair ` `            ``summ ``+``=` `(arr[i] ``+` `arr[j]) ` ` `  `    ``return` `summ ` ` `  `# Driver code ` `arr ``=` `[``1``, ``2``, ``3``] ` `n ``=` `len``(arr) ` ` `  `print``(summPairs(arr, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to return the sum of the elements ` `    ``// of all possible pairs from the array ` `    ``static` `int` `sumPairs(``int` `[]arr, ``int` `n) ` `    ``{ ` ` `  `        ``// To store the required sum ` `        ``int` `sum = 0; ` ` `  `        ``// Nested loop for all possible pairs ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``for` `(``int` `j = 0; j < n; j++)  ` `            ``{ ` ` `  `                ``// Add the sum of the elements ` `                ``// of the current pair ` `                ``sum += (arr[i] + arr[j]); ` `            ``} ` `        ``} ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``int` `[]arr = {1, 2, 3}; ` `        ``int` `n = arr.Length; ` ` `  `        ``Console.WriteLine(sumPairs(arr, n)); ` `    ``} ` `} ` `     `  `// This code is contributed by PrinciRaj1992 `

Output:

```36
```

Time Complexity: O(N2)

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++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the sum of the elements ` `// of all possible pairs from the array ` `int` `sumPairs(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To store the required sum ` `    ``int` `sum = 0; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// It appears (2 * n) times ` `        ``sum = sum + (arr[i] * (2 * n)); ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << sumPairs(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` `     `  `class` `GFG ` `{ ` ` `  `// Function to return the sum of the elements ` `// of all possible pairs from the array ` `static` `int` `sumPairs(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To store the required sum ` `    ``int` `sum = ``0``; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// It appears (2 * n) times ` `        ``sum = sum + (arr[i] * (``2` `* n)); ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `static` `public` `void` `main(String []arg) ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``3` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(sumPairs(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the sum of the elements  ` `# of all possible pairs from the array  ` `def` `sumPairs(arr, n) :  ` ` `  `    ``# To store the required sum  ` `    ``sum` `=` `0``;  ` ` `  `    ``# For every element of the array  ` `    ``for` `i ``in` `range``(n) : ` ` `  `        ``# It appears (2 * n) times  ` `        ``sum` `=` `sum` `+` `(arr[i] ``*` `(``2` `*` `n)); ` ` `  `    ``return` `sum``;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``1``, ``2``, ``3` `];  ` `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(sumPairs(arr, n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach  ` `using` `System;          ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the sum of the elements ` `// of all possible pairs from the array ` `static` `int` `sumPairs(``int` `[]arr, ``int` `n) ` `{ ` ` `  `    ``// To store the required sum ` `    ``int` `sum = 0; ` ` `  `    ``// For every element of the array ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// It appears (2 * n) times ` `        ``sum = sum + (arr[i] * (2 * n)); ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main(String []arg) ` `{ ` `    ``int` `[]arr = { 1, 2, 3 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(sumPairs(arr, n)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

Output:

```36
```

Time Complexity: O(N)

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