Sum over Subsets | Dynamic Programming

Prerequisite: Basic Dynamic Programming, Bitmasks 
Consider the following problem where we will use Sum over subset Dynamic Programming to solve it. 
Given an array of 2ⁿ integers, we need to calculate function F(x) = ?A_i such that x&i==i for all x. i.e, i is a bitwise subset of x. i will be a bitwise subset of mask x, if x&i==i.
Examples: 
 

Input: A[] = {7, 12, 14, 16} , n = 2
Output: 7, 19, 21, 49
Explanation: There will be 4 values of x: 0,1,2,3
So, we need to calculate F(0),F(1),F(2),F(3).
Now, F(0) = A0 = 7 
F(1) = A0 + A1 = 19
F(2) = A0 + A2 = 21
F(3) = A0 + A1 + A2 + A3 = 49

Input: A[] = {7, 11, 13, 16} , n = 2
Output: 7, 18, 20, 47 
Explanation: There will be 4 values of x: 0,1,2,3
So, we need to calculate F(0),F(1),F(2),F(3).
Now, F(0) = A0 = 7 
F(1) = A0 + A1 = 18
F(2) = A0 + A2 = 20
F(3) = A0 + A1 + A2 + A3 = 47

 

Brute-Force Approach: 
Iterate for all the x from 0 to (2ⁿ-1) . Calculate the bitwise subsets of all the x and sum it up for every x.
Time-Complexity: O(4^n)
Below is the implementation of above idea:
 

Output: 
 

7 19 21 49 

. 
Sub-Optimal Approach: 
The brute-force algorithm can be easily improved by just iterating over bitwise subsets. Instead of iterating for every i, we can simply iterate for the bitwise subsets only. Iterating backward for i=(i-1)&x gives us every bitwise subset, where i starts from x and ends at 1. If the mask x has k set bits, we do 2^k iterations. A number of k set bits will have 2^k bitwise subsets. Therefore total number of mask x with k set bits is. Therefore the total number of iterations is ?2^k = 3ⁿ 
Time Complexity: O(3ⁿ)
Below is the implementation of above idea: 
 

Output: 
 

7 19 21 49 

Time Complexity: O(n*2ⁿ) 

Auxiliary Space: O(2ⁿ)

Reference: