Dynamic Programming - Subset Sum Problem

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.

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

// 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;
}

Java

// 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 */

Python3

# 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.

C#

// 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

PHP



<?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 
?>

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

// 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.

Java

// 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 */

C#

// 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 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];
                                            
            }
        }
    
        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

PHP

<?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.
?>

Output: