K Empty Slots

lc 683

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.

Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.

For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.

Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.

If there isn't such day, output -1.

Example 1: Input: flowers: [1,3,2] k: 1 Output: 2 Explanation: In the second day, the first and the third flower have become blooming. Example 2: Input: flowers: [1,2,3] k: 1 Output: -1

class Solution {
    public int kEmptySlots(int[] flowers, int k) {
        TreeSet<Integer> bloom = new TreeSet<>();

        for (int i = 1; i <= flowers.length; i++ ) {
            bloom.add(flowers[i - 1]);
            Integer lower = bloom.lower(flowers[i - 1]);
            Integer higher = bloom.higher(flowers[i - 1]);
            if ((lower != null && flowers[i - 1] - lower - 1 == k) ||(higher != null && higher - flowers[i - 1] - 1 == k)) {
                return i;
            }
        }
        return -1;
    }
}