algorithm - Given a set of n integers, list all possible subsets with sum>=k

Question

Given an unsorted set of integers in the form of array, find all possible subsets whose sum is greater than or equal to a const integer k, eg:- Our set is {1,2,3} and k=2

Possible subsets:-

 {2},
 {3},
 {1,2},
 {1,3},
 {2,3}, 
 {1,2,3}

I can only think of a naive algorithm which lists all the subsets of set and checks if sum of subset is >=k or not, but its an exponential algorithm and listing all subsets requires O(2^N). Can I use dynamic programming to solve it in polynomial time?

Answer 1

Listing all the subsets is going to be still O(2^N) because in the worst case you may still have to list all subsets apart from the empty one.

Dynamic programming can help you count the number of sets that have sum >= K

You go bottom-up keeping track of how many subsets summed to some value from range [1..K]. An approach like this will be O(N*K) which is going to be only feasible for small K.

The idea with the dynamic programming solution is best illustrated with an example. Consider this situation. Assume you know that out of all the sets composed of the first i elements you know that t1 sum to 2 and t2 sum to 3. Let's say that the next i+1 element is 4. Given all the existing sets we can build all the new sets by either appending the element i+1 or leaving it out. If we leave it out we get t1 subsets that sum to 2 and t2 subsets that sum to 3. If we append it then we obtain t1 subsets that sum to 6 (2 + 4) and t2 that sum to 7 (3 + 4) and one subset which contains just i+1 which sums to 4. That gives us the numbers of subsets that sum to (2,3,4,6,7) consisting of the first i+1 elements. We continue until N.

In pseudo-code this could look something like this:

int DP[N][K];
int set[N];

//go through all elements in the set by index
for i in range[0..N-1]
   //count the one element subset consisting only of set[i]
   DP[i][set[i]] = 1

   if (i == 0) continue;

   //case 1. build and count all subsets that don't contain element set[i]
   for k in range[1..K-1]
       DP[i][k] += DP[i-1][k]

    //case 2. build and count subsets that contain element set[i]
    for k in range[0..K-1] 
       if k + set[i] >= K then break inner loop                     
       DP[i][k+set[i]] += DP[i-1][k]

//result is the number of all subsets - number of subsets with sum < K
//the -1 is for the empty subset
return 2^N - sum(DP[N-1][1..K-1]) - 1