# Sum of subsets of all the subsets of an array | O(N)

Given an array arr[] of length N, the task is to find the overall sum of subsets of all the subsets of the array.

Examples:

Input: arr[] = {1, 1}
Output: 6
All possible subsets:
a) {} : 0
All the possible subsets of this subset
will be {}, Sum = 0
b) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
c) {1} : 1
All the possible subsets of this subset
will be {} and {1}, Sum = 0 + 1 = 1
d) {1, 1} : 4
All the possible subsets of this subset
will be {}, {1}, {1} and {1, 1}, Sum = 0 + 1 + 1 + 2 = 4
Thus, ans = 0 + 1 + 1 + 4 = 6

Input: arr[] = {1, 4, 2, 12}
Output: 513

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

Approach: In this article, an approach with O(N) time complexity to solve the given problem will be discussed.
The key is observing the number of times an element will repeat in all the subsets.

Let’s magnify the view. It is known that every element will appear 2(N – 1) times in the sum of subsets. Now, let’s magnify the view even further and see how the count varies with the subset size.

There are N – 1CK – 1 subsets of size K for every index that include it.
Contribution of an element for a subset of size K will be equal to 2(K – 1) times its value. Thus, total contribution of an element for all the subsets of length K will be equal to N – 1CK – 1 * 2(K – 1)
Total contribution among all the subsets will be equal to:

N – 1CN – 1 * 2(N – 1) + N – 1CN – 2 * 2(N – 2 + N – 1CN – 3 * 2(N – 3) + … + N – 1C0 * 20.

Now, the contribution of each element in the final answer is known. So, multiply it to the sum of all the elements of the array which will give the required answer.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define maxN 10 ` ` `  `// To store factorial values ` `int` `fact[maxN]; ` ` `  `// Function to return ncr ` `int` `ncr(``int` `n, ``int` `r) ` `{ ` `    ``return` `(fact[n] / fact[r]) / fact[n - r]; ` `} ` ` `  `// Function to return the required sum ` `int` `findSum(``int``* arr, ``int` `n) ` `{ ` `    ``// Intialising factorial ` `    ``fact = 1; ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``fact[i] = i * fact[i - 1]; ` ` `  `    ``// Multiplier ` `    ``int` `mul = 0; ` ` `  `    ``// Finding the value of multipler ` `    ``// according to the formula ` `    ``for` `(``int` `i = 0; i <= n - 1; i++) ` `        ``mul += (``int``)``pow``(2, i) * ncr(n - 1, i); ` ` `  `    ``// To store the final answer ` `    ``int` `ans = 0; ` ` `  `    ``// Calculate the final answer ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``ans += mul * arr[i]; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << findSum(arr, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `static` `int` `maxN = ``10``; ` ` `  `// To store factorial values ` `static` `int` `[]fact = ``new` `int``[maxN]; ` ` `  `// Function to return ncr ` `static` `int` `ncr(``int` `n, ``int` `r) ` `{ ` `    ``return` `(fact[n] / fact[r]) / fact[n - r]; ` `} ` ` `  `// Function to return the required sum ` `static` `int` `findSum(``int``[] arr, ``int` `n) ` `{ ` `    ``// Intialising factorial ` `    ``fact[``0``] = ``1``; ` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `        ``fact[i] = i * fact[i - ``1``]; ` ` `  `    ``// Multiplier ` `    ``int` `mul = ``0``; ` ` `  `    ``// Finding the value of multipler ` `    ``// according to the formula ` `    ``for` `(``int` `i = ``0``; i <= n - ``1``; i++) ` `        ``mul += (``int``)Math.pow(``2``, i) * ncr(n - ``1``, i); ` ` `  `    ``// To store the final answer ` `    ``int` `ans = ``0``; ` ` `  `    ``// Calculate the final answer ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``ans += mul * arr[i]; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args) ` `{ ` `    ``int` `arr[] = { ``1``, ``1` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(findSum(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

 `# Python3 implementation of the approach  ` `maxN ``=` `10` ` `  `# To store factorial values  ` `fact ``=` `[``0``]``*``maxN;  ` ` `  `# Function to return ncr  ` `def` `ncr(n, r) :  ` ` `  `    ``return` `(fact[n] ``/``/` `fact[r]) ``/``/` `fact[n ``-` `r];  ` ` `  `# Function to return the required sum  ` `def` `findSum(arr, n) :  ` ` `  `    ``# Intialising factorial  ` `    ``fact[``0``] ``=` `1``;  ` `    ``for` `i ``in` `range``(``1``, n) :  ` `        ``fact[i] ``=` `i ``*` `fact[i ``-` `1``];  ` ` `  `    ``# Multiplier  ` `    ``mul ``=` `0``;  ` ` `  `    ``# Finding the value of multipler  ` `    ``# according to the formula  ` `    ``for` `i ``in` `range``(n) : ` `        ``mul ``+``=` `(``2` `*``*` `i) ``*` `ncr(n ``-` `1``, i);  ` ` `  `    ``# To store the final answer  ` `    ``ans ``=` `0``;  ` ` `  `    ``# Calculate the final answer  ` `    ``for` `i ``in` `range``(n) : ` `        ``ans ``+``=` `mul ``*` `arr[i];  ` ` `  `    ``return` `ans;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``1``, ``1` `];  ` `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(findSum(arr, n));  ` ` `  `# This code is contributed by AnkitRai01 `

 `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG ` `{ ` `static` `int` `maxN = 10; ` ` `  `// To store factorial values ` `static` `int` `[]fact = ``new` `int``[maxN]; ` ` `  `// Function to return ncr ` `static` `int` `ncr(``int` `n, ``int` `r) ` `{ ` `    ``return` `(fact[n] / fact[r]) / fact[n - r]; ` `} ` ` `  `// Function to return the required sum ` `static` `int` `findSum(``int``[] arr, ``int` `n) ` `{ ` `    ``// Intialising factorial ` `    ``fact = 1; ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``fact[i] = i * fact[i - 1]; ` ` `  `    ``// Multiplier ` `    ``int` `mul = 0; ` ` `  `    ``// Finding the value of multipler ` `    ``// according to the formula ` `    ``for` `(``int` `i = 0; i <= n - 1; i++) ` `        ``mul += (``int``)Math.Pow(2, i) * ncr(n - 1, i); ` ` `  `    ``// To store the final answer ` `    ``int` `ans = 0; ` ` `  `    ``// Calculate the final answer ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``ans += mul * arr[i]; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``int` `[]arr = { 1, 1 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(findSum(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:
```6
```

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Improved By : AnkitRai01, Rajput-Ji, 29AjayKumar

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