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. 

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Example:

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

Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30
Output: False
There is no subset that add up to 30.

Method 1: Recursion.
Approach: For the recursive approach we will consider two cases. 

1. Consider the last element and now the required sum = target sum – value of ‘last’ element and number of elements = total elements – 1
2. Leave the ‘last’ element and now the required sum = target sum and number of elements = total elements – 1

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 

Let’s take a look at the simulation of above approach-: 

set[]={3, 4, 5, 2}
sum=9
(x, y)= 'x' is the left number of elements,
'y' is the required sum
  
              (4, 9)
             {True}
           /        \  
        (3, 6)       (3, 9)
               
        /    \        /   \ 
     (2, 2)  (2, 6)   (2, 5)  (2, 9)
     {True}  
     /   \ 
  (1, -3) (1, 2)  
{False}  {True} 
         /    \
       (0, 0)  (0, 2)
       {True} {False}      

C++

// A recursive solution for subset sum problem
#include <iostream>
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

C

// A recursive solution for subset sum problem
#include <stdio.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;
}

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)
            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 */

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):
        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
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)
            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

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)
        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
?>

Javascript

<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>

Output

Found a subset with given sum

Complexity Analysis: 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).

Method 2: To solve the problem in Pseudo-polynomial time use the Dynamic programming.
So we will create a 2D array of size (arr.size() + 1) * (target + 1) of type boolean. The state DP[i][j] will be true if there exists a subset of elements from A[0….i] with sum value = ‘j’. The approach for the problem is: 

if (A[i-1] > j)
DP[i][j] = DP[i-1][j]
else 
DP[i][j] = DP[i-1][j] OR DP[i-1][j-A[i-1]]

1. This means that if current element has value greater than ‘current sum value’ we will copy the answer for previous cases
2. And if the current sum value is greater than the ‘ith’ element we will see if any of previous states have already experienced the sum=’j’ OR any previous states experienced a value ‘j – A[i]’ which will solve our purpose.

The below simulation will clarify the above approach: 

set[]={3, 4, 5, 2}
target=6
 
    0    1    2    3    4    5    6

0   T    F    F    F    F    F    F

3   T    F    F    T    F    F    F
     
4   T    F    F    T    T    F    F   
      
5   T    F    F    T    T    T    F

2   T    F    T    T    T    T    T

Below is the implementation of the above approach: 

C

// A Dynamic Programming solution
// for subset sum problem
#include <stdio.h>

// Returns true if there is a subset of set[]
// with sum 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 bottom 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 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;
}
// This code is contributed by Arjun Tyagi.

C++

// A Dynamic Programming solution
// for subset sum problem
#include <iostream>
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)
{
    // 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 bottom 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]);
       cout <<"\n";
     }*/

    return subset[n][sum];
}

// 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

Java

// A Dynamic Programming 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)
    {
        // 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 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];
            }
        }

        /* // 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 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 */

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 sum 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 bottom 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 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
# 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 sum 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 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

PHP

<?php
// A Dynamic Programming 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)
{
    // 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 bottom
    // 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 code
$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.
?>

Javascript

<script>
    // A Dynamic Programming 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)
    {
        // The value of subset[i][j] will be
        // true if there is a subset of
        // set[0..j-1] with sum equal to i
        let subset = new Array(sum + 1);

        for(let i = 0; i < sum + 1; i++)
        {
            subset[i] = new Array(sum + 1);
            for(let j = 0; j < n + 1; j++)
            {
                subset[i][j] = 0;
            }
        }

        // If sum is 0, then answer is true
        for (let i = 0; i <= n; i++)
            subset[0][i] = true;

        // If sum is not 0 and set is empty,
        // then answer is false
        for (let i = 1; i <= sum; i++)
            subset[i][0] = false;

        // Fill the subset table in bottom
        // up manner
        for (let i = 1; i <= sum; i++) {
            for (let 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];
    }

    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 decode2207.
</script>

Output

Found a subset with given sum

Complexity Analysis:

Time Complexity: O(sum*n), where sum is the ‘target sum’ and ‘n’ is the size of array.
Auxiliary Space: O(sum*n), as the size of 2-D array is sum*n. + O(n) for recursive stack space

Memoization Technique for finding Subset Sum:

Method:

1. In this method, we also follow the recursive approach but In this method, we use another 2-D matrix in  we first initialize with -1 or any negative value.
2. In this method, we avoid the few of the recursive call which is repeated itself that’s why we use 2-D matrix. In this matrix we store the value of the previous call value.

Below is the implementation of the above approach:

C++

// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Taking the matrix as globally
int tab[2000][2000];

// Check if possible subset with
// given sum is possible or not
int subsetSum(int a[], int n, int sum)
{

    // If the sum is zero it means
    // we got our expected sum
    if (sum == 0)
        return 1;

    if (n <= 0)
        return 0;

    // If the value is not -1 it means it
    // already call the function
    // with the same value.
    // it will save our from the repetition.
    if (tab[n - 1][sum] != -1)
        return tab[n - 1][sum];

    // if the value of a[n-1] is
    // greater than the sum.
    // we call for the next value
    if (a[n - 1] > sum)
        return tab[n - 1][sum] = subsetSum(a, n - 1, sum);
    else
    {

        // Here we do two calls because we
        // don't know which value is
        // full-fill our criteria
        // that's why we doing two calls
        return tab[n - 1][sum] = subsetSum(a, n - 1, sum) ||
                       subsetSum(a, n - 1, sum - a[n - 1]);
    }
}

// Driver Code
int main()
{
    // Storing the value -1 to the matrix
    memset(tab, -1, sizeof(tab));
    int n = 5;
    int a[] = {1, 5, 3, 7, 4};
    int sum = 12;

    if (subsetSum(a, n, sum))
    {
        cout << "YES" << endl;
    }
    else
        cout << "NO" << endl;

    /* This Code is Contributed by Ankit Kumar*/
}

Java

// Java program for the above approach
class GFG {

    // Check if possible subset with
    // given sum is possible or not
    static int subsetSum(int a[], int n, int sum)
    {
        // Storing the value -1 to the matrix
        int tab[][] = new int[n + 1][sum + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= sum; j++) {
                tab[i][j] = -1;
            }
        }

        // If the sum is zero it means
        // we got our expected sum
        if (sum == 0)
            return 1;

        if (n <= 0)
            return 0;

        // If the value is not -1 it means it
        // already call the function
        // with the same value.
        // it will save our from the repetition.
        if (tab[n - 1][sum] != -1)
            return tab[n - 1][sum];

        // if the value of a[n-1] is
        // greater than the sum.
        // we call for the next value
        if (a[n - 1] > sum)
            return tab[n - 1][sum]
                = subsetSum(a, n - 1, sum);
        else {

            // Here we do two calls because we
            // don't know which value is
            // full-fill our criteria
            // that's why we doing two calls
            if (subsetSum(a, n - 1, sum) != 0
                || subsetSum(a, n - 1, sum - a[n - 1])
                       != 0) {
                return tab[n - 1][sum] = 1;
            }
            else
                return tab[n - 1][sum] = 0;
        }
    }

      // Driver Code
    public static void main(String[] args)
    {

        int n = 5;
        int a[] = { 1, 5, 3, 7, 4 };
        int sum = 12;

        if (subsetSum(a, n, sum) != 0) {
            System.out.println("YES\n");
        }
        else
            System.out.println("NO\n");
    }
}

// This code is contributed by rajsanghavi9.

Python3

# Python program for the above approach

# Taking the matrix as globally
tab = [[-1 for i in range(2000)] for j in range(2000)]

# Check if possible subset with
# given sum is possible or not
def subsetSum(a, n, sum):

    # If the sum is zero it means
    # we got our expected sum
    if (sum == 0):
        return 1

    if (n <= 0):
        return 0

    # If the value is not -1 it means it
    # already call the function
    # with the same value.
    # it will save our from the repetition.
    if (tab[n - 1][sum] != -1):
        return tab[n - 1][sum]

    # if the value of a[n-1] is
    # greater than the sum.
    # we call for the next value
    if (a[n - 1] > sum):
        tab[n - 1][sum] = subsetSum(a, n - 1, sum)
        return tab[n - 1][sum]
    else:

        # Here we do two calls because we
        # don't know which value is
        # full-fill our criteria
        # that's why we doing two calls
        tab[n - 1][sum] = subsetSum(a, n - 1, sum)
        return tab[n - 1][sum] or subsetSum(a, n - 1, sum - a[n - 1])

# Driver Code

n = 5
a = [1, 5, 3, 7, 4]
sum = 12

if (subsetSum(a, n, sum)):
    print("YES")
else:
    print("NO")

# This code is contributed by shivani.

C#

// C# program for the above approach
using System;
class GFG
{

    // Check if possible subset with
    // given sum is possible or not
    static int subsetSum(int []a, int n, int sum)
    {

        // Storing the value -1 to the matrix
        int [,]tab = new int[n + 1,sum + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= sum; j++) {
                tab[i,j] = -1;
            }
        }

        // If the sum is zero it means
        // we got our expected sum
        if (sum == 0)
            return 1;

        if (n <= 0)
            return 0;

        // If the value is not -1 it means it
        // already call the function
        // with the same value.
        // it will save our from the repetition.
        if (tab[n - 1,sum] != -1)
            return tab[n - 1,sum];

        // if the value of a[n-1] is
        // greater than the sum.
        // we call for the next value
        if (a[n - 1] > sum)
            return tab[n - 1,sum]
                = subsetSum(a, n - 1, sum);
        else {

            // Here we do two calls because we
            // don't know which value is
            // full-fill our criteria
            // that's why we doing two calls
            if (subsetSum(a, n - 1, sum) != 0
                || subsetSum(a, n - 1, sum - a[n - 1])
                       != 0) {
                return tab[n - 1,sum] = 1;
            }
            else
                return tab[n - 1,sum] = 0;
        }
    }

      // Driver Code
    public static void Main(String[] args)
    {

        int n = 5;
        int []a = { 1, 5, 3, 7, 4 };
        int sum = 12;

        if (subsetSum(a, n, sum) != 0) {
            Console.Write("YES\n");
        }
        else
            Console.Write("NO\n");
    }
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

        // JavaScript Program for the above approach


        // Check if possible subset with
        // given sum is possible or not
        function subsetSum(a, n, sum) {

            // If the sum is zero it means
            // we got our expected sum
            if (sum == 0)
                return 1;

            if (n <= 0)
                return 0;

            // If the value is not -1 it means it
            // already call the function
            // with the same value.
            // it will save our from the repetition.
            if (tab[n - 1][sum] != -1)
                return tab[n - 1][sum];

            // if the value of a[n-1] is
            // greater than the sum.
            // we call for the next value
            if (a[n - 1] > sum)
                return tab[n - 1][sum] = subsetSum(a, n - 1, sum);
            else {

                // Here we do two calls because we
                // don't know which value is
                // full-fill our criteria
                // that's why we doing two calls
                return tab[n - 1][sum] = subsetSum(a, n - 1, sum) ||
                    subsetSum(a, n - 1, sum - a[n - 1]);
            }
        }

        // Driver Code

        // Storing the value -1 to the matrix
        let tab = Array(2000).fill().map(() => Array(2000).fill(-1));

        let n = 5;
        let a = [1, 5, 3, 7, 4];
        let sum = 12;

        if (subsetSum(a, n, sum)) {
            document.write("YES" + "<br>");
        }
        else {
            document.write("NO" + "<br>");
        }

    // This code is contributed by Potta Lokesh

</script>

Output

YES

Complexity Analysis:

• Time Complexity: O(sum*n), where sum is the ‘target sum’ and ‘n’ is the size of array.
• Auxiliary Space: O(sum*n) + O(n) -> O(sum*n) = the size of 2-D array is sum*n and O(n)=auxiliary stack space.

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