Question

I would like to know if there exist algorithms that solves this issue. It is little bit similar to knapsack 0-1 problem, or power set problem however it is different.

Given a finite set of sorted real numbers we need to generate all possible subsets whose sum <= k. Here k is real, the sorted real numbers are all positive. For example an array = {1.48, 2.21 3.07, 4.35, 4.46} and k = 5.94 Output is : {4.46}, {4.46, 1.48}, {4.35}, {4.35, 1.48}, {3.07}, {3.07, 2.21}, {2.21}, {2,21, 1.48} and {1.48} .

One way to solve is to simply traverse from highest number {4.46} to see how many you can inlude in the basket then continue going next lowest number {4.35} and so on. Is there an efficient way to do this? let me know

Answer 1

Greedy algorithm can definitely work. To take advantage of the fact that the input is sorted, binary search can be used.

The basic idea is: first search for the biggest number in the array that is smaller than K by binary search, push the result element to a stack, then recursively search the subarray that ends at that element for a sum K – the value of that element. After this is done, search in the subarray for a sum K to cover the situations in which that element is not chosen.

Sample code:

void boundedSumSubarray(float * arr, int size, float K, stack S) {
    int pos=binarySearch(arr,size,K);
    if (pos>=0) {
        pushStack(S,arr[pos]);
        boundedSumSubarray(arr,pos-1,K-arr[pos],S);
        popStack(S);
        boundedSumSubarray(arr,pos-1,K,S);
    } else
        printStack(S);
}