package array; import java.util.TreeSet; /** * Created by gouthamvidyapradhan on 01/01/2018. * 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 Note: The given array will be in the range [1, 20000]. Solution: O(n log n). Maintain a tree-set of bloomed flowers and for every element in the array find the upper and lower bound bloomed flowers and calculate their difference with the current. If the difference is k return the current day, if none found then return -1 */ public class KEmptySlots { /** * Main method * @param args * @throws Exception */ public static void main(String[] args) throws Exception{ int[] A = {1, 3, 2}; System.out.println(new KEmptySlots().kEmptySlots(A, 2)); } public int kEmptySlots(int[] flowers, int k) { TreeSet set = new TreeSet<>(); for(int i = 0; i < flowers.length; i ++){ Integer lowerBound = set.floor(flowers[i]); Integer upperBound = set.ceiling(flowers[i]); if(lowerBound != null){ if((Math.abs(flowers[i] - lowerBound) + 1) == k){ return i + 1; } } if(upperBound != null){ if((Math.abs(flowers[i] - upperBound) + 1) == k){ return i + 1; } } set.add(flowers[i]); } return -1; } }