# Sum of cubes of all Subsets of given Array

• Last Updated : 30 Mar, 2020

Given an array arr[], the task is to calculate the sum of cubes of all possible non-empty subsets of the given array. Since, the answer can be large, print the value as mod 1000000007.

Examples:

Input: arr[] = {1, 2}
Output: 18
subset({1}) = 13 = 1
subsetval({2}) = 23 = 8
subset({1, 2}) = 13 + 23 = 1 + 8 = 9
Sum of cubes of all Subsets = 1 + 8 + 9 = 18

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 cube 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 each element of the original array appears in 2(N – 1) times in all subsets.
• Therefore contribution of any element arri in the final answer will be
`arri * 2(N – 1)`
• So, the Sum of cubes of all Subsets will be
```[arr03 + arr13 + arr23 + … + arr(N-1)3] * 2(N – 1)
```

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 cubes of subsets ` `long` `long` `subset_cube_sum(vector<``int``>& A) ` `{ ` ` `  `    ``int` `n = (``int``)A.size(); ` ` `  `    ``long` `long` `ans = 0; ` ` `  `    ``// cubing the elements ` `    ``// and adding it to ans ` `    ``for` `(``int` `i : A) { ` `        ``ans += (1LL * i * i * i) % mod; ` `        ``ans %= mod; ` `    ``} ` ` `  `    ``return` `(1LL * ans * power(n - 1)) ` `           ``% mod; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> A = { 1, 2 }; ` ` `  `    ``cout << subset_cube_sum(A); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of the approach  ` `mod ``=` `int``(``1e9``) ``+` `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 cubes of subsets  ` `def` `subset_cube_sum(A) :  ` ` `  `    ``n ``=` `len``(A);  ` ` `  `    ``ans ``=` `0``;  ` ` `  `    ``# cubing the elements  ` `    ``# and adding it to ans  ` `    ``for` `i ``in` `A : ` `        ``ans ``+``=` `(i ``*` `i ``*` `i) ``%` `mod;  ` `        ``ans ``%``=` `mod;  ` ` `  `    ``return` `(ans ``*` `power(n ``-` `1``)) ``%` `mod;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``A ``=` `[ ``1``, ``2` `];  ` ` `  `    ``print``(subset_cube_sum(A));  ` `     `  `# This code is contributed by Yash_R `

Output:

```18
```

Time Complexity: O(N)

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