Sum of products of all possible K size subsets of the given array

Given an array arr[] of N non-negative integers and an integer 1 ≤ K ≤ N. The task is to find the sum of the products of all possible subsets of arr[] of size K.

Examples:

Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 35
(1 * 2) + (1 * 3) + (1 * 4) + (2 * 3) + (2 * 4)
+ (3 * 4) = 2 + 3 + 4 + 6 + 8 + 12 = 35

Input: arr[] = {1, 2, 3, 4}, K = 3
Output: 50

Naive approach: Generate all possible subsets of size K and find the resultant product of each subset. Then sum the product obtained for each subset. The time complexity of this solution would be exponential.



Efficient approach: Take the example of an array a[] = {1, 2, 3} and K = 3. Then,

k = 1, answer = 1 + 2 + 3 = 6
k = 2, answer = 1 * (2 + 3) + 2 * 3 + 0 = 11
k = 3, answer = 1 * (2 * 3 + 0) + 0 + 0 = 6

In the example, if the contribution of 1 is needed to be obtained in the answer for K = 2 then the sum of all elements after the index of element 1 is required in the previously computed values for K = 1. It can be seen that the sum of elements 2 and 3 is required. Thus, for any K, the answer obtained for K – 1 is required.
So, bottom up dynamic programming approach can be used to solve this problem. Create a table dp[][] and fill it in bottom up manner where dp[i][j] will store the contribution of an element arr[j – 1] to the answer for K = i. Hence, the recurrence relation will be,

dp[i][j] = arr[j-1] * \sum_{k=j+1}^{n} dp[i-1][k]

answer[k] = \sum_{i=1}^{n} dp[k][i]

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the sum of products of
// all the possible k size subsets
int sumOfProduct(int arr[], int n, int k)
{
    // Initialising all the values to 0
    int dp[n + 1][n + 1] = { 0 };
  
    // To store the answer for
    // current value of k
    int cur_sum = 0;
  
    // For k = 1, the answer will simply
    // be the sum of all the elements
    for (int i = 1; i <= n; i++) {
        dp[1][i] = arr[i - 1];
        cur_sum += arr[i - 1];
    }
  
    // Filling the table in bottom up manner
    for (int i = 2; i <= k; i++) {
  
        // To store the elements of the current
        // row so that we will be able to use this sum
        // for subsequent values of k
        int temp_sum = 0;
  
        for (int j = 1; j <= n; j++) {
  
            // We will subtract previously computed value
            // so as to get the sum of elements from j + 1
            // to n in the (i - 1)th row
            cur_sum -= dp[i - 1][j];
  
            dp[i][j] = arr[j - 1] * cur_sum;
            temp_sum += dp[i][j];
        }
        cur_sum = temp_sum;
    }
    return cur_sum;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(int);
    int k = 2;
  
    cout << sumOfProduct(arr, n, k);
  
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 implementation of the approach
   
# Function to return the sum of products of
# all the possible k size subsets
def sumOfProduct(arr, n, k):
      
    # Initialising all the values to 0
    dp = [ [ 0 for x in range(n + 1)] for y in range(n + 1)]
      
    # To store the answer for
    # current value of k
    cur_sum = 0
      
    # For k = 1, the answer will simply
    # be the sum of all the elements
    for i in range(1, n + 1):
        dp[1][i] = arr[i - 1]
        cur_sum += arr[i - 1]
   
    # Filling the table in bottom up manner
    for i in range(2 , k + 1):
   
        # To store the elements of the current
        # row so that we will be able to use this sum
        # for subsequent values of k
        temp_sum = 0
   
        for j in range( 1,  n + 1):
   
            # We will subtract previously computed value
            # so as to get the sum of elements from j + 1
            # to n in the (i - 1)th row
            cur_sum -= dp[i - 1][j]
   
            dp[i][j] = arr[j - 1] * cur_sum
            temp_sum += dp[i][j]
        cur_sum = temp_sum
    return cur_sum
   
# Driver code
if __name__ == "__main__":
      
    arr = [ 1, 2, 3, 4 ]
    n = len(arr)
    k = 2
    print(sumOfProduct(arr, n, k))
  
# This code is contributed by chitranayal

chevron_right


Time Complexity: O(N2)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : nihalmittal47, chitranayal