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2251. Number of Flowers in Full Bloom

Description

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 people of size n, where people[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]], people = [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]], people = [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 <= people.length <= 5 * 104
  • 1 <= people[i] <= 109

Solutions

We sort the flowers by their start and end times. Then, for each person, we can use binary search to find the number of flowers in bloom when they arrive. This means finding the number of flowers that have started blooming by the time each person arrives, minus the number of flowers that have wilted by that time, to get the answer.

The time complexity is $O((m + n) \times \log n)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the arrays $\textit{flowers}$ and $\textit{people}$, respectively.

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class Solution:
    def fullBloomFlowers(
        self, flowers: List[List[int]], people: List[int]
    ) -> List[int]:
        start, end = sorted(a for a, _ in flowers), sorted(b for _, b in flowers)
        return [bisect_right(start, p) - bisect_left(end, p) for p in people]

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class Solution {
    public int[] fullBloomFlowers(int[][] flowers, int[] people) {
        int n = flowers.length;
        int[] start = new int[n];
        int[] end = new int[n];
        for (int i = 0; i < n; ++i) {
            start[i] = flowers[i][0];
            end[i] = flowers[i][1];
        }
        Arrays.sort(start);
        Arrays.sort(end);
        int m = people.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = search(start, people[i] + 1) - search(end, people[i]);
        }
        return ans;
    }

    private int search(int[] nums, int x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

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class Solution {
public:
    vector<int> fullBloomFlowers(vector<vector<int>>& flowers, vector<int>& people) {
        int n = flowers.size();
        vector<int> start;
        vector<int> end;
        for (auto& f : flowers) {
            start.push_back(f[0]);
            end.push_back(f[1]);
        }
        sort(start.begin(), start.end());
        sort(end.begin(), end.end());
        vector<int> ans;
        for (auto& p : people) {
            auto r = upper_bound(start.begin(), sta