Given an array arr[] consisting of N integers and a positive integer K, the task is to find the sum of all the subsets of size K.
Examples:
Input: arr[] = {1, 2, 4, 5}, K = 2
Output: 36
Explanation:
The subsets of size K(= 2) are = {1, 2}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {4, 5}. Now, the sum of all subsets sum = 3 + 5 + 6 + 6 + 7 + 9 = 36.Input: arr[] = {2, 4, 5, 6, 8}, K=3
Output: 150
Naive Approach: The simplest approach to solve the given problem is to generate all possible subsets of the given array and find the sum of elements of those subsets whose size is K. After finding the sum of all K sized subsets, print the sum of all the sums obtained as the result.
Time Complexity: O(K*2N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by observing the fact that the number of occurrences of each element arr[i] in the summation series depends on value of N and K.
Let’s find the occurrence of a general element x in summation series:
- Occurrence of x in summation series of all size=k subsets = n-1Ck-1
Therefore, the frequency of each element of arr[] is same in the summation equation = n-1Ck-1. Hence, the sum of all the subsets = (Sum of all elements of the array) * n-1Ck-1.
Follow the steps below to solve the given problem:
- Initialize the variable, say freq as 0 and calculate n-1Ck-1
- Initialize the variable, say sum as 0 to store the sum of all the array elements.
- After performing the above steps, print the value of sum*freq as the resultant sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of all// sub-sets of size Kvoid findSumOfAllSubsets(int arr[], int n, int k){ // Frequency of each array element // in summation equation. int factorial_N=1, factorial_d=1, factorial_D=1; //calculate factorial of n-1 for(int i=1; i<=n-1; i++) factorial_N*=i; //calculate factorial of k-1 for(int i=1; i<=k-1; i++) factorial_d*=i; //calculate factorial of n-k for(int i=1; i<=n-k; i++) factorial_D*=i; int freq = factorial_N/(factorial_d * factorial_D); // Calculate sum of array. int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // Sum of all subsets of size k. sum = sum * freq; cout <<"Sum of all subsets of size = "<<k<<" is => "<< sum << endl;}// Driver Codeint main(){ int arr[] = { 1, 2, 4, 5 }; int n = 4, k = 2; findSumOfAllSubsets(arr, n, k); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG{ // Function to find the sum of all // sub-sets of size K static void findSumOfAllSubsets(int[] arr, int n, int k) { // Frequency of each array element // in summation equation. int factorial_N = 1, factorial_d = 1, factorial_D = 1; // calculate factorial of n-1 for (int i = 1; i <= n - 1; i++) factorial_N *= i; // calculate factorial of k-1 for (int i = 1; i <= k - 1; i++) factorial_d *= i; // calculate factorial of n-k for (int i = 1; i <= n - k; i++) factorial_D *= i; int freq = factorial_N / (factorial_d * factorial_D); // Calculate sum of array. int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // Sum of all subsets of size k. sum = sum * freq; System.out.println("Sum of all subsets of size = " + k + " is => " + sum); } // Driver Code public static void main(String[] args) { int[] arr = { 1, 2, 4, 5 }; int n = 4, k = 2; findSumOfAllSubsets(arr, n, k); }}// This code is contributed by maddler. |
Python3
# Python 3 program for the above approach# Function to find the sum of all# sub-sets of size Kdef findSumOfAllSubsets(arr, n, k): # Frequency of each array element # in summation equation. factorial_N = 1 factorial_d = 1 factorial_D = 1 # calculate factorial of n-1 for i in range(1, n, 1): factorial_N *= i # calculate factorial of k-1 for i in range(1, k, 1): factorial_d *= i # calculate factorial of n-k for i in range(1, n - k + 1, 1): factorial_D *= i freq = factorial_N//(factorial_d * factorial_D) # Calculate sum of array. sum = 0 for i in range(n): sum += arr[i] # Sum of all subsets of size k. sum = sum * freq print("Sum of all subsets of size = ",k," is =>",sum)# Driver Codeif __name__ == '__main__': arr = [1, 2, 4, 5] n = 4 k = 2 findSumOfAllSubsets(arr, n, k) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the sum of all // sub-sets of size K static void findSumOfAllSubsets(int[] arr, int n, int k) { // Frequency of each array element // in summation equation. int factorial_N = 1, factorial_d = 1, factorial_D = 1; // calculate factorial of n-1 for (int i = 1; i <= n - 1; i++) factorial_N *= i; // calculate factorial of k-1 for (int i = 1; i <= k - 1; i++) factorial_d *= i; // calculate factorial of n-k for (int i = 1; i <= n - k; i++) factorial_D *= i; int freq = factorial_N / (factorial_d * factorial_D); // Calculate sum of array. int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // Sum of all subsets of size k. sum = sum * freq; Console.WriteLine("Sum of all subsets of size = " + k + " is => " + sum); } // Driver Code public static void Main() { int[] arr = { 1, 2, 4, 5 }; int n = 4, k = 2; findSumOfAllSubsets(arr, n, k); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find the sum of all // sub-sets of size K function findSumOfAllSubsets(arr, n, k) { // Frequency of each array element // in summation equation. let factorial_N = 1, factorial_d = 1, factorial_D = 1; // calculate factorial of n-1 for (let i = 1; i <= n - 1; i++) factorial_N *= i; // calculate factorial of k-1 for (let i = 1; i <= k - 1; i++) factorial_d *= i; // calculate factorial of n-k for (let i = 1; i <= n - k; i++) factorial_D *= i; let freq = factorial_N / (factorial_d * factorial_D); // Calculate sum of array. let sum = 0; for (let i = 0; i < n; i++) sum += arr[i]; // Sum of all subsets of size k. sum = sum * freq; document.write("Sum of all subsets of size = " + k + " is => " + sum + "<br>"); } // Driver Code let arr = [1, 2, 4, 5]; let n = 4, k = 2; findSumOfAllSubsets(arr, n, k);// This code is contributed by Potta Lokesh </script> |
Output
36
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
