Algorithm on interview

Question

Recently I was asked the following interview question: You have two sets of numbers of the same length N, for example A = [3, 5, 9] and B = [7, 5, 1]. Next, for each position i in range 0..N-1, you can pick either number A[i] or B[i], so at the end you will have another array C of length N which consists in elements from A and B. If sum of all elements in C is less than or equal to K, then such array is good. Please write an algorithm to figure out the total number of good arrays by given arrays A, B and number K.

The only solution I've come up is Dynamic Programming approach, when we have a matrix of size NxK and M[i][j] represents how many combinations could we have for number X[i] if current sum is equal to j. But looks like they expected me to come up with a formula. Could you please help me with that? At least what direction should I look for? Will appreciate any help. Thanks.

Answer 1

After some consideration, I believe this is an NP-complete problem. Consider:

A = [0, 0, 0, ..., 0]
B = [b1, b2, b3, ..., bn]

Note that every construction of the third set C = ( A[i] or B[i] for i = 0..n ) is is just the union of some subset of A and some subset of B. In this case, since every subset of A sums to 0, the sum of C is the same as the sum of some subset of B.

Now your question "How many ways can we construct C with a sum less than K?" can be restated as "How many subsets of B sum to less than K?". Solving this problem for K = 1 and K = 0 yields the solution to the subset sum problem for B (the difference between the two solutions is the number of subsets that sum to 0).

By similar argument, even in the general case where A contains nonzero elements, we can construct an array S = [b1-a1, b2-a2, b3-a3, ..., bn-an], and the question becomes "How many subsets of S sum to less than K - sum(A)?"

Since the subset sum problem is NP-complete, this problem must be also. So with that in mind, I would venture that the dynamic programming solution you proposed is the best you can do, and certainly no magic formula exists.