Count of subsets with sum equal to X using Recursion

Given an array arr[] of length N and an integer X, the task is to find the number of subsets with sum equal to X using recursion.

Examples:

Input: arr[] = {2, 3, 5, 6, 8, 10}, X = 10
Output: 3
Explanation:
All possible subsets with sum 10 are {2, 3, 5}, {2, 8}, {10}



Input: arr[] = {1, 2, 3, 4, 5}, X = 7
Output: 3
Explanation:
All possible subsets with sum 7 are {2, 5}, {3, 4}, {1, 2, 4}

Approach: The idea is to recursively check all the subsets. If any subset has the sum equal to N, then increment the count by 1. Else, continue.

In order to form a subset, there are two cases for every element:

  • Include the element in the set.
  • Exclude the element in the set.

Therefore, the following steps can be followed to compute the answer:

  1. Get the array for which the subsets with the sum equal to K is to be found.
  2. Recursively count the subsets with the sum equal to K in the following way:
    • Base Case: The base case will be when the end of the array has been reached. If here the sum has been found as X, then increase the count of the subset by 1. Return the count evaluated in the base condition.
      if (i == n) {
          if (sum == 0)
              count++;
          return count;
      }
      
    • Recursive Call: If the base case is not satisfied, then call the function twice. Once by including the element at index ‘i’ and once by not including the element. Find the count for both these cases and then return the final count.
      count = subsetSum(arr, n, i + 1, sum - arr[i], count);
      count = subsetSum(arr, n, i + 1, sum, count);
      
    • Return Statement: At every step, the count of subsets by either including a particular element or not including a particular element is returned. Finally, when the entire recursion stack is executed, the total count is returned.

From the above approach, it can be clearly analyzed that if there are N elements in the array, then a total of 2N cases arise. Every element is checked for the above cases using recursion.

Below is the implementation of the above approach:

C++

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// C++ program to print the count of
// subsets with sum equal to the given value X
  
#include <iostream>
using namespace std;
  
// Recursive function to return the count
// of subsets with sum equal to the given value
int subsetSum(int arr[], int n, int i,
              int sum, int count)
{
    // The recursion is stopped at N-th level
    // where all the subsets of the given array
    // have been checked
    if (i == n) {
  
        // Incrementing the count if sum is
        // equal to 0 and returning the count
        if (sum == 0) {
            count++;
        }
        return count;
    }
  
    // Recursively calling the function for two cases
    // Either the element can be counted in the subset
    // If the element is counted, then the remaining sum
    // to be checked is sum - the selected element
    // If the element is not included, then the remaining sum
    // to be checked is the total sum
    count = subsetSum(arr, n, i + 1, sum - arr[i], count);
    count = subsetSum(arr, n, i + 1, sum, count);
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int sum = 10;
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << subsetSum(arr, n, 0, sum, 0);
}

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Java

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// Java program to print the count of
// subsets with sum equal to the given value X
import java.util.*;
  
class GFG
{
  
// Recursive function to return the count
// of subsets with sum equal to the given value
static int subsetSum(int arr[], int n, int i,
                    int sum, int count)
{
    // The recursion is stopped at N-th level
    // where all the subsets of the given array
    // have been checked
    if (i == n) 
    {
  
        // Incrementing the count if sum is
        // equal to 0 and returning the count
        if (sum == 0
        {
            count++;
        }
        return count;
    }
  
    // Recursively calling the function for two cases
    // Either the element can be counted in the subset
    // If the element is counted, then the remaining sum
    // to be checked is sum - the selected element
    // If the element is not included, then the remaining sum
    // to be checked is the total sum
    count = subsetSum(arr, n, i + 1, sum - arr[i], count);
    count = subsetSum(arr, n, i + 1, sum, count);
    return count;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int sum = 10;
    int n = arr.length;
  
    System.out.print(subsetSum(arr, n, 0, sum, 0));
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program to prthe count of
# subsets with sum equal to the given value X
  
# Recursive function to return the count
# of subsets with sum equal to the given value
def subsetSum(arr, n, i,sum, count):
      
    # The recursion is stopped at N-th level
    # where all the subsets of the given array
    # have been checked
    if (i == n):
  
        # Incrementing the count if sum is
        # equal to 0 and returning the count
        if (sum == 0):
            count += 1
        return count
  
    # Recursively calling the function for two cases
    # Either the element can be counted in the subset
    # If the element is counted, then the remaining sum
    # to be checked is sum - the selected element
    # If the element is not included, then the remaining sum
    # to be checked is the total sum
    count = subsetSum(arr, n, i + 1, sum - arr[i], count)
    count = subsetSum(arr, n, i + 1, sum, count)
    return count
  
# Driver code
arr = [1, 2, 3, 4, 5]
sum = 10
n = len(arr)
  
print(subsetSum(arr, n, 0, sum, 0))
  
# This code is contributed by mohit kumar 29

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C#

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// C# program to print the count of
// subsets with sum equal to the given value X
using System;
  
class GFG
{
  
// Recursive function to return the count
// of subsets with sum equal to the given value
static int subsetSum(int []arr, int n, int i,
                    int sum, int count)
{
    // The recursion is stopped at N-th level
    // where all the subsets of the given array
    // have been checked
    if (i == n) 
    {
  
        // Incrementing the count if sum is
        // equal to 0 and returning the count
        if (sum == 0) 
        {
            count++;
        }
        return count;
    }
  
    // Recursively calling the function for two cases
    // Either the element can be counted in the subset
    // If the element is counted, then the remaining sum
    // to be checked is sum - the selected element
    // If the element is not included, then the remaining sum
    // to be checked is the total sum
    count = subsetSum(arr, n, i + 1, sum - arr[i], count);
    count = subsetSum(arr, n, i + 1, sum, count);
    return count;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int sum = 10;
    int n = arr.Length;
  
    Console.Write(subsetSum(arr, n, 0, sum, 0));
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

3

Efficient Approach:
An efficient method to solve the problem using Dynamic Programming has been discussed in this article.

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