# Sum of squares of all Subsets of given Array

Given an array arr[]. The value of a subset of array A is defined as the sum of squares of all the numbers in that subset. The task is to calculate the sum of values of all possible non-empty subsets of the given array.
Since, the answer can be large print the val mod 1000000007.

Examples:

Input: arr[] = {3, 7}
Output: 116
val({3}) = 32 = 9
val({7}) = 72 = 49
val({3, 7}) = 32 + 72 = 9 + 49 = 58
9 + 49 + 58 = 116

Input: arr[] = {1, 1, 1}
Output: 12

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

Naive approach: A simple approach is to find all the subset and then square each element in that subset and add it to the result. The time complexity of this approach will be O(2N)

Efficient approach: It can be observed that in all the possible subsets of the given array, every element will occur 2N – 1 times where N is the size of the array.
So the contribution of any element X in the sum will be 2N – 1 * X2.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `mod = 1e9 + 7; ` ` `  `// Function to return (2^P % mod) ` `long` `long` `power(``int` `p) ` `{ ` `    ``long` `long` `res = 1; ` `    ``for` `(``int` `i = 1; i <= p; ++i) { ` `        ``res *= 2; ` `        ``res %= mod; ` `    ``} ` `    ``return` `res % mod; ` `} ` ` `  `// Function to return the sum of squares of subsets ` `long` `long` `subset_square_sum(vector<``int``>& A) ` `{ ` ` `  `    ``int` `n = (``int``)A.size(); ` ` `  `    ``long` `long` `ans = 0; ` ` `  `    ``// Sqauaring the elements ` `    ``// and adding it to ans ` `    ``for` `(``int` `i : A) { ` `        ``ans += (1LL * i * i) % mod; ` `        ``ans %= mod; ` `    ``} ` ` `  `    ``return` `(1LL * ans * power(n - 1)) % mod; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> A = { 3, 7 }; ` ` `  `    ``cout << subset_square_sum(A); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `    ``static` `final` `int` `mod = (``int``)(1e9 + ``7``);  ` `     `  `    ``// Function to return (2^P % mod)  ` `    ``static` `long` `power(``int` `p)  ` `    ``{  ` `        ``long` `res = ``1``;  ` `        ``for` `(``int` `i = ``1``; i <= p; ++i) ` `        ``{  ` `            ``res *= ``2``;  ` `            ``res %= mod;  ` `        ``}  ` `        ``return` `res % mod;  ` `    ``}  ` `     `  `    ``// Function to return the sum of squares of subsets  ` `    ``static` `long` `subset_square_sum(``int` `A[])  ` `    ``{  ` `        ``int` `n = A.length;  ` `     `  `        ``long` `ans = ``0``;  ` `     `  `        ``// Sqauaring the elements  ` `        ``// and adding it to ans  ` `        ``for` `(``int` `i : A)  ` `        ``{  ` `            ``ans += (``1` `* i * i) % mod;  ` `            ``ans %= mod;  ` `        ``}  ` `        ``return` `(``1` `* ans * power(n - ``1``)) % mod;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `A[] = { ``3``, ``7` `};  ` `     `  `        ``System.out.println(subset_square_sum(A));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` `mod ``=` `10``*``*``9` `+` `7` ` `  `# Function to return (2^P % mod) ` `def` `power(p): ` ` `  `    ``res ``=` `1` `    ``for` `i ``in` `range``(``1``, p ``+` `1``): ` `        ``res ``*``=` `2` `        ``res ``%``=` `mod ` ` `  `    ``return` `res ``%` `mod ` ` `  `# Function to return the sum of ` `# squares of subsets ` `def` `subset_square_sum(A): ` ` `  `    ``n ``=` `len``(A) ` ` `  `    ``ans ``=` `0` ` `  `    ``# Squaring the elements ` `    ``# and adding it to ans ` `    ``for` `i ``in` `A: ` `        ``ans ``+``=` `i ``*` `i ``%` `mod ` `        ``ans ``%``=` `mod ` ` `  `    ``return` `ans ``*` `power(n ``-` `1``) ``%` `mod ` ` `  `# Driver code ` `A ``=` `[``3``, ``7``] ` ` `  `print``(subset_square_sum(A)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `    ``static` `readonly` `int` `mod = (``int``)(1e9 + 7);  ` `     `  `    ``// Function to return (2^P % mod)  ` `    ``static` `long` `power(``int` `p)  ` `    ``{  ` `        ``long` `res = 1;  ` `        ``for` `(``int` `i = 1; i <= p; ++i) ` `        ``{  ` `            ``res *= 2;  ` `            ``res %= mod;  ` `        ``}  ` `        ``return` `res % mod;  ` `    ``}  ` `     `  `    ``// Function to return the sum of squares of subsets  ` `    ``static` `long` `subset_square_sum(``int` `[]A)  ` `    ``{  ` `        ``int` `n = A.Length;  ` `     `  `        ``long` `ans = 0;  ` `     `  `        ``// Sqauaring the elements  ` `        ``// and adding it to ans  ` `        ``foreach` `(``int` `i ``in` `A)  ` `        ``{  ` `            ``ans += (1 * i * i) % mod;  ` `            ``ans %= mod;  ` `        ``}  ` `        ``return` `(1 * ans * power(n - 1)) % mod;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args) ` `    ``{  ` `        ``int` `[]A = { 3, 7 };  ` `     `  `        ``Console.WriteLine(subset_square_sum(A));  ` `    ``}  ` `} ` `     `  `// This code is contributed by 29AjayKumar `

Output:

```116
```

Time Complexity: O(N)

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