Subset Sum Problem | DP-25

Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.
Example:

Input:  set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output:  True  //There is a subset (4, 5) with sum 9.

Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with sum equal to sum. n is the number of elements in set[].

The isSubsetSum problem can be divided into two subproblems
…a) Include the last element, recur for n = n-1, sum = sum – set[n-1]
…b) Exclude the last element, recur for n = n-1.
If any of the above the above subproblems return true, then return true.

Following is the recursive formula for isSubsetSum() problem.



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 



Following is naive recursive implementation that simply follows the recursive structure mentioned above.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// A recursive solution for subset sum problem
#include <stdio.h>
  
// Returns true if there is a subset of set[] with sun equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
   // Base Cases
   if (sum == 0)
     return true;
   if (n == 0 && sum != 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 program to test above function
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// A recursive solution for subset sum
// problem
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 && sum != 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 program to test above function */
    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 */

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 and sum != 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 program to test above function
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.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 && sum != 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 program 
    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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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 && $sum != 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 
?>

chevron_right


Output:

Found a subset with given sum

The above solution may try all subsets of 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).

We can solve the problem in Pseudo-polynomial time using Dynamic programming. We create a boolean 2D table subset[][] and fill it in bottom up manner. The value of subset[i][j] will be true if there is a subset of set[0..j-1] with sum equal to i., otherwise false. Finally, we return subset[sum][n]

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Dynamic Programming solution for subset sum problem
#include <stdio.h>
   
// Returns true if there is a subset of set[] with sun equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
    // The value of subset[i][j] will be true if there is a 
    // subset of set[0..j-1] with sum equal to i
    bool subset[n+1][sum+1];
   
    // If sum is 0, then answer is true
    for (int i = 0; i <= n; i++)
      subset[i][0] = true;
   
    // If sum is not 0 and set is empty, then answer is false
    for (int i = 1; i <= sum; i++)
      subset[0][i] = false;
   
     // Fill the subset table in botton up manner
     for (int i = 1; i <= n; i++)
     {
       for (int j = 1; j <= sum; j++)
       {
         if(j<set[i-1])
         subset[i][j] = subset[i-1][j];
         if (j >= set[i-1])
           subset[i][j] = subset[i-1][j] || 
                                 subset[i - 1][j-set[i-1]];
       }
     }
   
     /*   // uncomment this code to print table
     for (int i = 0; i <= n; i++)
     {
       for (int j = 0; j <= sum; j++)
          printf ("%4d", subset[i][j]);
       printf("\n");
     }*/
   
     return subset[n][sum];
}
   
// Driver program to test above function
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;
}
// This code is contributed by Arjun Tyagi.

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Dynamic Programming solution for subset
// sum problem
class GFG {
      
    // Returns true if there is a subset of 
    // set[] with sun equal to given sum
    static boolean isSubsetSum(int set[], 
                             int n, int sum)
    {
        // The value of subset[i][j] will be
        // true if there is a subset of 
        // set[0..j-1] with sum equal to i
        boolean subset[][] = 
                     new boolean[sum+1][n+1];
      
        // If sum is 0, then answer is true
        for (int i = 0; i <= n; i++)
            subset[0][i] = true;
      
        // If sum is not 0 and set is empty,
        // then answer is false
        for (int i = 1; i <= sum; i++)
            subset[i][0] = false;
      
        // Fill the subset table in botton
        // up manner
        for (int i = 1; i <= sum; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                subset[i][j] = subset[i][j-1];
                if (i >= set[j-1])
                subset[i][j] = subset[i][j] || 
                     subset[i - set[j-1]][j-1];
            }
        }
      
        /* // uncomment this code to print table
        for (int i = 0; i <= sum; i++)
        {
        for (int j = 0; j <= n; j++)
            System.out.println (subset[i][j]);
        } */
      
        return subset[sum][n];
    }
  
    /* Driver program to test above function */
    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 */

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# A Dynamic Programming solution for subset sum problem
# Returns true if there is a subset of 
# set[] with sun equal to given sum 
  
# Returns true if there is a subset of set[] 
# with sun equal to given sum
def isSubsetSum(set,n,sum):
      
    # The value of subset[i][j] will be 
    # true if there is a
    # subset of set[0..j-1] with sum equal to i
    subset=([[False for i in range(sum+1)] 
            for i in range(n+1)])
      
    # If sum is 0, then answer is true 
    for i in range(n+1):
        subset[i][0] = True
          
        # If sum is not 0 and set is empty, 
        # then answer is false 
        for i in range(1,sum+1):
            subset[0][i]=False
              
        # Fill the subset table in botton up manner
        for i in range(1,n+1):
            for j in range(1,sum+1):
                if j<set[i-1]:
                    subset[i][j] = subset[i-1][j]
                if j>=set[i-1]:
                    subset[i][j] = (subset[i-1][j] or 
                                   subset[i - 1][j-set[i-1]])
      
        # uncomment this code to print table 
        # for i in range(n+1):
        # for j in range(sum+1):
        #     print (subset[i][j],end=" ")
        # print()
    return subset[n][sum]
          
# Driver program to test above function
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 
# sahil shelangia. 

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// A Dynamic Programming solution for subset sum problem
using System;
  
class GFG
{
    // Returns true if there is a subset 
    // of set[] with sun equal to given sum
    static bool isSubsetSum(int []set, int n, int sum)
    {
        // The value of subset[i][j] will be true if there 
        // is a subset of set[0..j-1] with sum equal to i
        bool [,]subset = new bool[sum+1,n+1];
      
        // If sum is 0, then answer is true
        for (int i = 0; i <= n; i++)
        subset[0, i] = true;
      
        // If sum is not 0 and set is empty, then answer is false
        for (int i = 1; i <= sum; i++)
        subset[i, 0] = false;
      
        // Fill the subset table in bottom up manner
        for (int i = 1; i <= sum; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                subset[i, j] = subset[i, j - 1];
                if (i >= set[j - 1])
                subset[i, j] = subset[i, j] || 
                               subset[i - set[j - 1], j - 1];
                                              
            }
        }
      
        return subset[sum,n];
    }
      
    // Driver program 
    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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// A Dynamic Programming 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)
{
    // The value of subset[i][j] will
    // be true if there is a subset of
    // set[0..j-1] with sum equal to i
    $subset = array(array());
  
    // If sum is 0, then answer is true
    for ( $i = 0; $i <= $n; $i++)
        $subset[$i][0] = true;
  
    // If sum is not 0 and set is empty,
    // then answer is false
    for ( $i = 1; $i <= $sum; $i++)
        $subset[0][$i] = false;
  
    // Fill the subset table in botton
    // up manner
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1; $j <= $sum; $j++)
        {
            if($j < $set[$i-1])
                $subset[$i][$j] = 
                      $subset[$i-1][$j];
            if ($j >= $set[$i-1])
                $subset[$i][$j] = 
                       $subset[$i-1][$j] || 
                       $subset[$i - 1][$j
                               $set[$i-1]];
        }
    }
  
    /* // uncomment this code to print table
    for (int i = 0; i <= n; i++)
    {
    for (int j = 0; j <= sum; j++)
        printf ("%4d", subset[i][j]);
    printf("n");
    }*/
  
    return $subset[$n][$sum];
}
  
// Driver program to test above function
$set = array(3, 34, 4, 12, 5, 2);
$sum = 9;
$n = count($set);
  
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.
?>

chevron_right


Output:

Found a subset with given sum

Time complexity of the above solution is O(sum*n).

Subset Sum Problem in O(sum) space
Perfect Sum Problem (Print all subsets with given sum)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up