TieRopes.md

There are N ropes numbered from 0 to N − 1, whose lengths are given in a zero-indexed array A, lying on the floor in a line. For each I (0 ≤ I < N), the length of rope I on the line is A[I].

We say that two ropes I and I + 1 are adjacent. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For a given integer K, the goal is to tie the ropes in such a way that the number of ropes whose length is greater than or equal to K is maximal.

For example, consider K = 4 and array A such that:

    A[0] = 1
    A[1] = 2
    A[2] = 3
    A[3] = 4
    A[4] = 1
    A[5] = 1
    A[6] = 3

The ropes are shown in the figure below.

We can tie:

• rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
• rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

After that, there will be three ropes whose lengths are greater than or equal to K = 4. It is not possible to produce four such ropes.

Write a function:

function solution(K, A);

that, given an integer K and a non-empty zero-indexed array A of N integers, returns the maximum number of ropes of length greater than or equal to K that can be created.

For example, given K = 4 and array A such that:

    A[0] = 1
    A[1] = 2
    A[2] = 3
    A[3] = 4
    A[4] = 1
    A[5] = 1
    A[6] = 3

the function should return 3, as explained above.

Assume that:

• N is an integer within the range [1..100,000];
• K is an integer within the range [1..1,000,000,000];
• each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

function solution(K, A) {
    if (K === 1)
        return A.length;

    var count = 0;
    var sum = 0;

    for (var i = 0; i < A.length; i++) {
        sum += A[i];

        if (sum >= K) {
            sum = 0;
            count++;
        }
    }

    return count;
}