Number of Flowers in Full Bloom LeetCode Solution

Problem – Number of Flowers in Full Bloom LeetCode Solution

You are given a 0-indexed 2D integer array flowers, where flowers[i] = [starti, endi] means the ith flower will be in full bloom from starti to endi (inclusive). You are also given a 0-indexed integer array persons of size n, where persons[i] is the time that the ith person will arrive to see the flowers.

Return an integer array answer of size n, where answer[i] is the number of flowers that are in full bloom when the ith person arrives.

Example 1:

Input: flowers = [[1,6],[3,7],[9,12],[4,13]], persons = [2,3,7,11]
Output: [1,2,2,2]
Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.

Example 2:

Input: flowers = [[1,10],[3,3]], persons = [3,3,2]
Output: [2,2,1]
Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.

Constraints:

• 1 <= flowers.length <= 5 * 104
• flowers[i].length == 2
• 1 <= starti <= endi <= 109
• 1 <= persons.length <= 5 * 104
• 1 <= persons[i] <= 109

Number of Flowers in Full Bloom LeetCode Solution in C++

    vector<int> fullBloomFlowers(vector<vector<int>>& flowers, vector<int>& persons) {
        vector<int> start, end;
        for (auto& t : flowers)
            start.push_back(t[0]), end.push_back(t[1]);
        sort(start.begin(), start.end());
        sort(end.begin(), end.end());
        vector<int> res;
        for (int t : persons) {
            int started = upper_bound(start.begin(), start.end(), t) - start.begin();
            int ended = lower_bound(end.begin(), end.end(), t) - end.begin();
            res.push_back(started - ended);
        }
        return res;
    }

Number of Flowers in Full Bloom LeetCode Solution in Python

    def fullBloomFlowers(self, A: List[List[int]], persons: List[int]) -> List[int]:
        start, end = sorted(a for a,b in A), sorted(b for a,b in A)
        return [bisect_right(start, t) - bisect_left(end, t) for t in persons]

Number of Flowers in Full Bloom LeetCode Solution in Java

public int[] fullBloomFlowers(int[][] flowers, int[] persons) {
    /*3 events:
        0 - Bloom
        1 - Person
        2 - Die
	*/
    //{time position, event type, if person -> index}
	//first sort by time, then by event type
    Queue<int[]> pq = new PriorityQueue<>((a,b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
	
	//add the person events
    for(int i = 0; i < persons.length; i++){
        //for person also add what index they were - {time position, human, index}
        pq.offer(new int[]{persons[i], 1, i});
    }
    //add the blossom events
    for(int[] flower : flowers){
        pq.offer(new int[]{flower[0], 0});
        pq.offer(new int[]{flower[1], 2});
    }

    //traverse them all
    int[] ret = new int[persons.length];
    int numEvents = pq.size();
    int blooms = 0;
    for(int i = 0; i < numEvents; i++){
        int[] cur = pq.poll();
		//if bloom, increment blooms
        if(cur[1] == 0){
            blooms++;
		//if die, decrement blooms
        }else if(cur[1] == 2){
            blooms--;
		//if person, set their index to # blooms
        }else{
			int personIndex = cur[2];
            ret[personIndex] = blooms;
        }
    }
    return ret;
}

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