Subset Sum Problem using Recursion
For the recursive approach, there will be two cases.
- Consider the ‘last’ element to be a part of the subset. Now the new required sum = required sum – value of ‘last’ element.
- Don’t include the ‘last’ element in the subset. Then the new required sum = old required sum.
In both cases, the number of available elements decreases by 1.
Mathematically the recurrence relation will look like the following:
isSubsetSum(set, n, sum) = isSubsetSum(set, n-1, sum) | isSubsetSum(set, n-1, sum-set[n-1])
Base Cases:
isSubsetSum(set, n, sum) = false, if sum > 0 and n = 0
isSubsetSum(set, n, sum) = true, if sum = 0
The structure of the recursion tree will be like the following:
Structure of the recursion tree of the above recursion formula
Follow the below steps to implement the recursion:
- Build a recursive function and pass the index to be considered (here gradually moving from the last end) and the remaining sum amount.
- For each index check the base cases and utilize the above recursive call.
- If the answer is true for any recursion call, then there exists such a subset. Otherwise, no such subset exists.
Below is the implementation of the above approach.
// A recursive solution for subset sum problem
#include <bits/stdc++.h>
using namespace std;
// Returns true if there is a subset
// of set[] with sum equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than sum,
// then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
// Else, check if sum can be obtained by any
// of the following:
// (a) including the last element
// (b) excluding the last element
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
// Driver code
int main()
{
int set[] = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = sizeof(set) / sizeof(set[0]);
if (isSubsetSum(set, n, sum) == true)
cout << "Found a subset with given sum";
else
cout << "No subset with given sum";
return 0;
}
// This code is contributed by shivanisinghss2110
// A recursive solution for subset sum problem
#include <stdio.h>
#include <stdbool.h>
// Returns true if there is a subset
// of set[] with sum equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than sum,
// then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
// Else, check if sum can be obtained by any
// of the following:
// (a) including the last element
// (b) excluding the last element
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
// Driver code
int main()
{
int set[] = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = sizeof(set) / sizeof(set[0]);
if (isSubsetSum(set, n, sum) == true)
printf("Found a subset with given sum");
else
printf("No subset with given sum");
return 0;
}
// A recursive solution for subset sum
import java.io.*;
class GFG {
// Returns true if there is a subset
// of set[] with sum equal to given sum
static boolean isSubsetSum(int set[], int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than
// sum, then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
// Else, check if sum can be obtained
// by any of the following
// (a) including the last element
// (b) excluding the last element
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
// Driver code
public static void main(String args[])
{
int set[] = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = set.length;
if (isSubsetSum(set, n, sum) == true)
System.out.println("Found a subset"
+ " with given sum");
else
System.out.println("No subset with"
+ " given sum");
}
}
/* This code is contributed by Rajat Mishra */
# A recursive solution for subset sum
# problem
# Returns true if there is a subset
# of set[] with sun equal to given sum
def isSubsetSum(set, n, sum):
# Base Cases
if (sum == 0):
return True
if (n == 0):
return False
# If last element is greater than
# sum, then ignore it
if (set[n - 1] > sum):
return isSubsetSum(set, n - 1, sum)
# Else, check if sum can be obtained
# by any of the following
# (a) including the last element
# (b) excluding the last element
return isSubsetSum(
set, n-1, sum) or isSubsetSum(
set, n-1, sum-set[n-1])
# Driver code
if __name__ == '__main__':
set = [3, 34, 4, 12, 5, 2]
sum = 9
n = len(set)
if (isSubsetSum(set, n, sum) == True):
print("Found a subset with given sum")
else:
print("No subset with given sum")
# This code is contributed by Nikita Tiwari.
// A recursive solution for subset sum problem
using System;
class GFG {
// Returns true if there is a subset of set[]
// with sum equal to given sum
static bool isSubsetSum(int[] set, int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than sum,
// then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
// Else, check if sum can be obtained
// by any of the following
// (a) including the last element
// (b) excluding the last element
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
// Driver code
public static void Main()
{
int[] set = { 3, 34, 4, 12, 5, 2 };
int sum = 9;
int n = set.Length;
if (isSubsetSum(set, n, sum) == true)
Console.WriteLine(
"Found a subset with given sum");
else
Console.WriteLine("No subset with given sum");
}
}
// This code is contributed by Sam007
<script>
// A recursive solution for subset sum problem
// Returns true if there is a subset of set[] with sum
// equal to given sum
function isSubsetSum(set, n, sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0)
return false;
// If last element is greater than sum,
// then ignore it
if (set[n - 1] > sum)
return isSubsetSum(set, n - 1, sum);
// Else, check if sum can be obtained
// by any of the following
// (a) including the last element
// (b) excluding the last element
return isSubsetSum(set, n - 1, sum)
|| isSubsetSum(set, n - 1, sum - set[n - 1]);
}
let set = [ 3, 34, 4, 12, 5, 2 ];
let sum = 9;
let n = set.length;
if (isSubsetSum(set, n, sum) == true)
document.write("Found a subset with given sum");
else
document.write("No subset with given sum");
// This code is contributed by mukesh07.
</script>
<?php
// A recursive solution for subset sum problem
// Returns true if there is a subset of set
// with sun equal to given sum
function isSubsetSum($set, $n, $sum)
{
// Base Cases
if ($sum == 0)
return true;
if ($n == 0)
return false;
// If last element is greater
// than sum, then ignore it
if ($set[$n - 1] > $sum)
return isSubsetSum($set, $n - 1, $sum);
// Else, check if sum can be
// obtained by any of the following
// (a) including the last element
// (b) excluding the last element
return isSubsetSum($set, $n - 1, $sum) ||
isSubsetSum($set, $n - 1,
$sum - $set[$n - 1]);
}
// Driver Code
$set = array(3, 34, 4, 12, 5, 2);
$sum = 9;
$n = 6;
if (isSubsetSum($set, $n, $sum) == true)
echo"Found a subset with given sum";
else
echo "No subset with given sum";
// This code is contributed by Anuj_67
?>
Output
Found a subset with given sum
Complexity Analysis:
- Time Complexity: O(2n) The above solution may try all subsets of the given set in worst case. Therefore time complexity of the above solution is exponential. The problem is in-fact NP-Complete (There is no known polynomial time solution for this problem).
- Auxiliary Space: O(n) where n is recursion stack space.
Subset Sum Problem
Given a set of non-negative integers and a value sum, the task is to check if there is a subset of the given set whose sum is equal to the given sum.
Examples:
Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True
Explanation: There is a subset (4, 5) with sum 9.Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30
Output: False
Explanation: There is no subset that add up to 30.
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