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. Topics: Array, Hash Table, Binary Search, Sorting, Prefix Sum, Ordered Set. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 18 parametrized pytest cases, and a playground notebook:

lcpy gen -n 2251   # by problem number
lcpy gen -s number_of_flowers_in_full_bloom   # by problem name

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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.

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Examples

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.

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.

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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

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

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]

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Complexity

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

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Tags

NeetCode All.