Sum over Subsets | Dynamic Programming

Last Updated : 29 Mar, 2024

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:

C++

// CPP program for brute force // approach of SumOverSubsets DP #include <bits/stdc++.h> using namespace std;  // function to print the sum over subsets value void SumOverSubsets(int a[], int n) {    // array to store the SumOverSubsets   int sos[1 << n] = {0};    // iterate for all possible x   for (int x = 0; x < (1 << n); x++) {      // iterate for all possible bitwise subsets     for (int i = 0; i < (1 << n); i++) {        // if i is a bitwise subset of x       if ((x & i) == i)         sos[x] += a[i];     }   }    // printa all the subsets   for (int i = 0; i < (1 << n); i++)     cout << sos[i] << " "; }  // Driver Code int main() {   int a[] = {7, 12, 14, 16};   int n = 2;   SumOverSubsets(a, n);   return 0; }

Java

// Java program for brute force // approach of SumOverSubsets DP  class GFG{  // function to print the // sum over subsets value static void SumOverSubsets(int a[], int n) {  // array to store the SumOverSubsets int sos[] = new int [1 << n];   // iterate for all possible x for (int x = 0; x < (1 << n); x++) {      // iterate for all possible         // bitwise subsets     for (int i = 0; i < (1 << n); i++) {      // if i is a bitwise subset of x     if ((x & i) == i)         sos[x] += a[i];     } }  // printa all the subsets for (int i = 0; i < (1 << n); i++)     System.out.printf("%d ", sos[i]); }  // Driver Code public static void main(String[] args) { int a[] = {7, 12, 14, 16}; int n = 2; SumOverSubsets(a, n); } }  // This code is contributed by  // Smitha Dinesh Semwal

Python3

# Python 3 program # for brute force # approach of SumOverSubsets DP  # function to print the # sum over subsets value def SumOverSubsets(a, n):      # array to store     # the SumOverSubsets     sos = [0] * (1 << n)          # iterate for all possible x     for x in range(0,(1 << n)):               # iterate for all         # possible bitwise subsets         for i in range(0,(1 << n)):                    # if i is a bitwise subset of x             if ((x & i) == i):                 sos[x] += a[i]                            # printa all the subsets     for i in range(0,(1 << n)):          print(sos[i],end = " ")   # Driver Code a = [7, 12, 14, 16] n = 2 SumOverSubsets(a, n)  # This code is contributed by # Smitha Dinesh Semwal

// C# program for brute force // approach of SumOverSubsets DP using System;  class GFG {          // function to print the     // sum over subsets value     static void SumOverSubsets(int []a, int n)     {              // array to store the SumOverSubsets         int []sos = new int [1 << n];                           // iterate for all possible x         for (int x = 0; x < (1 << n); x++)          {                      // iterate for all possible             // bitwise subsets             for (int i = 0; i < (1 << n); i++)              {                          // if i is a bitwise subset of x                 if ((x & i) == i)                     sos[x] += a[i];             }         }                  // printa all the subsets         for (int i = 0; i < (1 << n); i++)             Console.Write(sos[i] + " ");     }          // Driver function     public static void Main()     {         int []a = {7, 12, 14, 16};         int n = 2;         SumOverSubsets(a, n);     } }  // This code is contributed by Sam007

PHP

<?php // PHP program for brute force // approach of SumOverSubsets DP  // function to print the sum  // over subsets value function SumOverSubsets($a, $n) {      // array to store the SumOverSubsets     $sos = array(1 << $n);          for($i = 0 ;$i < (1 << $n); $i++)         $sos[$i] = 0;           // iterate for all possible x     for ($x = 0; $x < (1 << $n); $x++)     {              // iterate for all possible          // bitwise subsets         for($i = 0; $i < (1 << $n); $i++)         {                  // if i is a bitwise             // subset of x             if (($x & $i) == $i)                 $sos[$x] += $a[$i];         }     }          // printa all the subsets     for ($i = 0; $i < (1 << $n); $i++)         echo $sos[$i] . " "; }  // Driver Code $a = array(7, 12, 14, 16); $n = 2; SumOverSubsets($a, $n);  // This code is contributed by Sam007 ?>

JavaScript

<script>     // Javascript program for brute force     // approach of SumOverSubsets DP          // function to print the     // sum over subsets value     function SumOverSubsets(a, n)     {               // array to store the SumOverSubsets         let sos = new Array(1 << n);         sos.fill(0);                             // iterate for all possible x         for (let x = 0; x < (1 << n); x++)         {                       // iterate for all possible             // bitwise subsets             for (let i = 0; i < (1 << n); i++)             {                           // if i is a bitwise subset of x                 if ((x & i) == i)                     sos[x] += a[i];             }         }                   // printa all the subsets         for (let i = 0; i < (1 << n); i++)             document.write(sos[i] + " ");     }          let a = [7, 12, 14, 16];     let n = 2;     SumOverSubsets(a, n);          </script>

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{n \choose k} . Therefore the total number of iterations is ?{n \choose k} 2^k = 3ⁿ
Time Complexity: O(3ⁿ)
Below is the implementation of above idea:

C++

// CPP program for sub-optimal // approach of SumOverSubsets DP #include <bits/stdc++.h> using namespace std;  // function to print the sum over subsets value void SumOverSubsets(int a[], int n) {    // array to store the SumOverSubsets   int sos[1 << n] = {0};    // iterate for all possible x   for (int x = 0; x < (1 << n); x++) {     sos[x] = a[0];      // iterate for the bitwise subsets only     for (int i = x; i > 0; i = (i - 1) & x)       sos[x] += a[i];   }    // print all the subsets   for (int i = 0; i < (1 << n); i++)     cout << sos[i] << " "; }  // Driver Code int main() {   int a[] = {7, 12, 14, 16};   int n = 2;   SumOverSubsets(a, n);   return 0; }

Java

// java program for sub-optimal // approach of SumOverSubsets DP public class GFG {          // function to print the sum over     // subsets value     static void SumOverSubsets(int a[], int n)     {              // array to store the SumOverSubsets         int sos[] = new int[(1 << n)];                  // iterate for all possible x         for (int x = 0; x < (1 << n); x++) {             sos[x] = a[0];                      // iterate for the bitwise subsets only             for (int i = x; i > 0; i = (i - 1) & x)                 sos[x] += a[i];         }                  // print all the subsets         for (int i = 0; i < (1 << n); i++)             System.out.print(sos[i] + " ");     }          // Driver code     public static void main(String args[])     {         int a[] = {7, 12, 14, 16};         int n = 2;                  SumOverSubsets(a, n);     } }  // This code is contributed by Sam007

Python3

# Python program for sub-optimal # approach of SumOverSubsets DP  # function to print sum over subsets value def SumOverSubsets(a, n):     sos = [0]*(1 << n)      # iterate for all possible x     for x in range((1 << n)):         sos[x] = a[0]                  # iterate for the bitwise subsets only         i = x          while i > 0:           sos[x] += a[i]           i = ((i - 1) & x)    # print all the subsets     for i in range(1<<n):         print(sos[i], end = " ")  # Driver Code if __name__ == '__main__':     a = [7, 12, 14, 16]     n = 2     SumOverSubsets(a, n)  # This code is contributed by mohit kumar 29.

// C# program for sub-optimal // approach of SumOverSubsets DP using System;  class GFG {          // function to print the sum over     // subsets value     static void SumOverSubsets(int []a, int n)     {              // array to store the SumOverSubsets         int []sos = new int[(1 << n)];                  // iterate for all possible x         for (int x = 0; x < (1 << n); x++) {             sos[x] = a[0];                      // iterate for the bitwise subsets only             for (int i = x; i > 0; i = (i - 1) & x)                 sos[x] += a[i];         }                  // print all the subsets         for (int i = 0; i < (1 << n); i++)         Console.Write(sos[i] + " ");     }          // Driver code     static void Main()     {         int []a = {7, 12, 14, 16};         int n = 2;                  SumOverSubsets(a, n);     } }  // This code is contributed by Sam007.

PHP

<?php // PHP program for sub-optimal // approach of SumOverSubsets DP  // function to print the  // sum over subsets value function SumOverSubsets($a,$n)  {      // array to store the SumOverSubsets     $sos=array(1 << $n);          // iterate for all possible x     for ($x = 0; $x < (1 << $n); $x++)      {         $sos[$x] = $a[0];              // iterate for the bitwise         // subsets only         for ($i = $x; $i > 0; $i = ($i - 1) & $x)         $sos[$x] += $a[$i];     }          // print all the subsets     for ($i = 0; $i < (1 << $n); $i++)         echo $sos[$i] . " "; }  // Driver Code $a = array(7, 12, 14, 16); $n = 2; SumOverSubsets($a, $n);  // This code is contributed by Sam007. ?>

JavaScript

<script>         // Javascript program for sub-optimal     // approach of SumOverSubsets DP          // function to print the sum over     // subsets value     function SumOverSubsets(a, n)     {               // array to store the SumOverSubsets         let sos = new Array((1 << n));         sos.fill(0);                   // iterate for all possible x         for (let x = 0; x < (1 << n); x++) {             sos[x] = a[0];                       // iterate for the bitwise subsets only             for (let i = x; i > 0; i = (i - 1) & x)                 sos[x] += a[i];         }                   // print all the subsets         for (let i = 0; i < (1 << n); i++)             document.write(sos[i] + " ");     }          let a = [7, 12, 14, 16];     let n = 2;      SumOverSubsets(a, n);  </script>

Output:

7 19 21 49

Time Complexity: O(n*2ⁿ)

Auxiliary Space: O(2ⁿ)

Reference:

Coin Change - Count Ways to Make Sum

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