Find the minimum number of elements required so that their sum equals or exceeds S

Question

I know this can be done by sorting the array and taking the larger numbers until the required condition is met. That would take at least nlog(n) sorting time.

Is there a

Answer 1

One improvement (asymptotically) over Theta(nlogn) you can do is to get an O(n log K) time algorithm, where K is the required minimum number of elements.

Thus if K is constant, or say log n, this is better (asymptotically) than sorting. Of course if K is n^epsilon, then this is not better than Theta(n logn).

The way to do this is to use selection algorithms, which can tell you the i^th largest element in O(n) time.

Now do a binary search for K, starting with i=1 (the largest) and doubling i etc at each turn.

You find the i^th largest, and find the sum of the i largest elements and check if it is greater than S or not.

This way, you would run O(log K) runs of the selection algorithm (which is O(n)) for a total running time of O(n log K).