Algorithm on interview - algorithm

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.

Answer 2

" Please write an algorithm to figure out the total number of good
arrays by given arrays A, B and number K."
Is it not the goal?
int A[];
int B[];
int N;
int K;
int Solutions = 0;
void FindSolutons(int Depth, int theSumSoFar) {
if (theSumSoFar > K) return;
if (Depth >= N) {
Solutions++;
return;
}
FindSolutions(Depth+1,theSumSoFar+A[Depth]);
FindSolutions(Depth+1,theSumSoFar+B[Depth]);
}
Invoke FindSolutions with both arguments set to zero. On return the Solutions will be equal to the number of good arrays;

Answer 3

this is how i would try to solve the problem
(Sorry if its stupid)
think of arrays
A=[3,5,9,8,2]
B=[7,5,1,8,2]
if
elements
0..N-1
number of choices
2^N
C1=0,C2=0
for all A[i]=B[i]
{
C1++
C2+=A[i]+B[i]
}
then create new two arrays like
A1=[3,5,9]
B1=[7,5,1]
also now C2 is 10
now number of all choices are reduced to 2^(N-C1)
now calculate all good numbers
using 'K' as K=K-C2
unfortunately
no matter what method you use, you have
to calculate sum 2^(N-C1) times

Answer 4

So there's 2^N choices, since at each point you either pick from A or from B. In the specific example you give where N happens to be 3 there are 8. For discussion you can characterise each set of decisions as a bit pattern.
So as a brute-force approach would try every single bit pattern.
But what should be obvious is that if the first few bits produce a number too large then every subsequent possible group of tail bits will also produce a number that is too large. So probably a better way to model it is a tree where you don't bother walking down the limbs that have already grown beyond your limit.
You can also compute the maximum totals that can be reached from each bit to the end of the table. If at any point your running total plus the maximum that you can obtain from here on down is less than K then every subtree from where you are is acceptable without any need for traversal. The case, as discussed in the comments, where every single combination is acceptable is a special case of this observation.
As pointed out by Serge below, a related observation is to us minimums and use the converse logic to cancel whole subtrees without traversal.
A potential further optimisation rests behind the observation that, as long as we shuffle each in the same way, changing the order of A and B has no effect because addition is commutative. You can therefore make an effort to ensure either that the maximums grow as quickly as possible or the minimums grow as slowly as possible, to try to get the earliest possible exit from traversal. In practice you'd probably want to apply a heuristic comparing the absolute maximum and minimum (both of which you've computed anyway) to K.
That being the case, a recursive implementation is easiest, e.g. (in C)
/* assume A, B and N are known globals */
unsigned int numberOfGoodArraysFromBit(
unsigned int bit,
unsigned int runningTotal,
unsigned int limit)
{
// have we ended up in an unacceptable subtree?
if(runningTotal > limit) return 0;
// have we reached the leaf node without at any
// point finding this subtree to be unacceptable?
if(bit >= N) return 1;
// maybe every subtree is acceptable?
if(runningTotal + MAXV[bit] <= limit)
{
return 1 << (N - bit);
}
// maybe no subtrees are acceptable?
if(runningTotal + MINV[bit] > limit)
{
return 0;
}
// if we can't prima facie judge the subtreees,
// we'll need specifically to evaluate them
return
numberOfGoodArraysFromBit(bit+1, runningTotal+A[bit], limit) +
numberOfGoodArraysFromBit(bit+1, runningTotal+B[bit], limit);
}
// work out the minimum and maximum values at each position
for(int i = 0; i < N; i++)
{
MAXV[i] = MAX(A[i], B[i]);
MINV[i] = MIN(A[i], B[i]);
}
// hence work out the cumulative totals from right to left
for(int i = N-2; i >= 0; i--)
{
MAXV[i] += MAXV[i+1];
MINV[i] += MINV[i+1];
}
// to kick it off
printf("Total valid combinations is %u", numberOfGoodArraysFromBit(0, 0, K));
I'm just thinking extemporaneously; it's likely better solutions exist.