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Subset Sum Problem | DP-25
• Difficulty Level : Medium
• Last Updated : 15 Feb, 2021

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.

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 ` `// 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);``    ``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

 ` ``\$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: 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] > j)
DP[i][j] = DP[i-1][j]
else
DP[i][j] = DP[i-1][j] OR DP[i-1][j-A[i]]```
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 ` `// 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] = ``true``;` `    ``// If sum is not 0 and set is empty,``    ``// then answer is false``    ``for` `(``int` `i = 1; i <= sum; i++)``        ``subset[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 code``int` `main()``{``    ``int` `set[] = { 3, 34, 4, 12, 5, 2 };``    ``int` `sum = 9;``    ``int` `n = ``sizeof``(set) / ``sizeof``(set);``    ``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 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 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 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 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 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

 `= ``\$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.``?>`
Output
`Found a subset with given sum`

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 ``using` `namespace` `std;` `// Taking the matrix as globally``int` `tab;` `// 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 repetation.``    ``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 critaria``        ``// 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*/``}`
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), as the size of 2-D array is 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.

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