Given a set of positive integers and a value sum,the task is to find if there is a subset with sum equal to the given value.

**Example-**

Given set of numbers- {3,34,4,12,5,2}

Sum=9.

The output will be ‘True’ because 4+5=9.

Algorithm-

The problem can be divided into two sub-problems-

1.By including the last element.

2.By excluding the last element.

**Recursive Solution-**

[cpp]

subsum(set, n, sum) = subsum(set, n-1, sum) //excluding the last element

|| subsum(arr, n-1, sum-set[n-1]) //including the last element

Base Cases:

subsum(set, n, sum) = false, if sum > 0 and n == 0

subsum(set, n, sum) = true, if sum == 0

[/cpp]

**Code-**

The above algorithm has exponential time complexity which can be converted to code having polynomial complexity using dynamic programming that is storing the intermediate results.

[cpp]

// 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[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];

}

[/cpp]