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KEmptySlots.java
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KEmptySlots.java
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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<Integer> 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;
}
}