MaxNonoverlappingSegments.md

Located on a line are N segments, numbered from 0 to N − 1, whose positions are given in zero-indexed arrays A and B. For each I (0 ≤ I < N) the position of segment I is from A[I] to B[I] (inclusive). The segments are sorted by their ends, which means that B[K] ≤ B[K + 1] for K such that 0 ≤ K < N − 1.

Two segments I and J, such that I ≠ J, are overlapping if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

We say that the set of segments is non-overlapping if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

    A[0] = 1    B[0] = 5
    A[1] = 3    B[1] = 6
    A[2] = 7    B[2] = 8
    A[3] = 9    B[3] = 9
    A[4] = 9    B[4] = 10

The segments are shown in the figure below.

The size of a non-overlapping set containing a maximal number of segments is 3. For example, possible sets are {0, 2, 3}, {0, 2, 4}, {1, 2, 3} or {1, 2, 4}. There is no non-overlapping set with four segments.

Write a function:

function solution(A, B);

that, given two zero-indexed arrays A and B consisting of N integers, returns the size of a non-overlapping set containing a maximal number of segments.

For example, given arrays A, B shown above, the function should return 3, as explained above.

Assume that:

• N is an integer within the range [0..30,000];
• each element of arrays A, B is an integer within the range [0..1,000,000,000];
• A[I] ≤ B[I], for each I (0 ≤ I < N);
• B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

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(A, B) {
		if (A.length ==0) {
            return A.length;
        }

        var count = 1;
        var ending = B[0];
        for (var i = 1; i < B.length; i++) {
            if (A[i] > ending) {
                ending = B[i];
                count++;
            }
        }
        
        return count;
}