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

Given an array of numbers from 1 to N. Each number in the array is the slot where the flower will open in (i+1)-th day. For example, the array [5, 3, 1, 4, 2] means flow 5 will bloom in day 1, flower 3 will bloom in day 2, flower 1 will bloom in day 3, etc.

Given an integer k, return the earliest day in which there exists two flowers blooming and also the flowers between them is k and these flowers are not blooming.

**Java Solution 1**

public int kEmptySlots(int[] flowers, int k) { int[] days = new int[flowers.length + 1]; for (int i = 0; i < flowers.length; i++) { days[flowers[i]] = i + 1; } int result = Integer.MAX_VALUE; for (int i = 1; i < days.length - k - 1; i++) { int l = days[i]; int r = days[i + k + 1]; int max = Math.max(l, r); int min = Math.min(l, r); boolean flag = true; for (int j = 1; j <= k; j++) { if (days[i + j] < max) { flag = false; break; } } if (flag && max < result) { result = max; } } return result == Integer.MAX_VALUE ? -1 : result; } |

Time complexity is O(N*k) and space complexity is O(N).

**Java Solution 2**

public int kEmptySlots2(int[] flowers, int k) { TreeSet<Integer> set = new TreeSet<>(); for (int i = 0; i < flowers.length; i++) { Integer higher = set.higher(flowers[i]); Integer lower = set.lower(flowers[i]); if (lower != null && flowers[i] - k - 1 == lower) { return i + 1; } if (higher != null && flowers[i] + k + 1 == higher) { return i + 1; } set.add(flowers[i]); } return -1; } |

Time complexity is O(N*log(N)) and space complexity is O(N).