Subset Sum Problem2

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


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];
}

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.