# Dynamic Programming | Set 25 (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,
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,
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:

```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)