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.
Example:

Input:  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 */

Python3

# 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.  

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 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 

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. 
?> 