> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Number of Flowers in Full Bloom

> Tested Python solution for LeetCode 2251 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 2251, [Hard](/catalog/hard). Topics: [Array](/catalog/topics/array), [Hash Table](/catalog/topics/hash-table), [Binary Search](/catalog/topics/binary-search), [Sorting](/catalog/topics/sorting), [Prefix Sum](/catalog/topics/prefix-sum), [Ordered Set](/catalog/topics/ordered-set). [View on LeetCode](https://leetcode.com/problems/number-of-flowers-in-full-bloom/description/).

Generate this problem as a practice environment: tested reference solution, 18 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 2251   # by problem number
lcpy gen -s number_of_flowers_in_full_bloom   # by problem name
```

## Problem

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.

### Examples

![Example 1](https://assets.leetcode.com/uploads/2022/03/02/ex1new.jpg)

```
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](https://assets.leetcode.com/uploads/2022/03/02/ex2new.jpg)

```
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 * 10^4`
* `flowers[i].length == 2`
* `1 <= starti <= endi <= 10^9`
* `1 <= people.length <= 5 * 10^4`
* `1 <= people[i] <= 10^9`

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/number_of_flowers_in_full_bloom/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/number_of_flowers_in_full_bloom/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
from bisect import bisect_left, bisect_right


class Solution:
    # Time: O(m log m + n log n + (m + n) log m), m = len(flowers), n = len(people)
    # Space: O(m)
    def full_bloom_flowers(self, flowers: list[list[int]], people: list[int]) -> list[int]:
        starts = sorted(s for s, _ in flowers)
        ends = sorted(e for _, e in flowers)
        return [bisect_right(starts, t) - bisect_left(ends, t) for t in people]
```

## Complexity

| Time | Space |
| - | - |
| O(m log m + n log n + (m + n) log m), m = len(flowers), n = len(people) | O(m) |

## Tags

[NeetCode All](/catalog/neetcode).


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