How do you solve Number of Flowers in Full Bloom on LeetCode?

Number of Flowers in Full Bloom (LeetCode #2251) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log n) time complexity and O(n) space complexity. Key insight: Understanding the blooming period as a range helps in counting efficiently.

What is the time complexity of Number of Flowers in Full Bloom?

The optimal solution for Number of Flowers in Full Bloom runs in O(n log n + m log n) time and uses O(n) space. Sorting the start and end times takes O(n log n), and each query for a person takes O(log n) due to binary search, leading to a total of O(n log n + m log n).

What is the brute force solution for Number of Flowers in Full Bloom?

The brute force approach for Number of Flowers in Full Bloom is Brute Force, with O(n²) time complexity and O(1) space complexity. For each person, we check how many flowers are blooming by iterating through all the flowers. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n log n + m log n).

What are the interview tips for Number of Flowers in Full Bloom?

Explain your thought process clearly while solving. Consider edge cases and ask clarifying questions about the input. Discuss time and space complexity for both brute force and optimal solutions.

What are common mistakes when solving Number of Flowers in Full Bloom?

Not considering edge cases where people arrive exactly at the start or end times. Overlooking the need for sorting before performing binary search.

#2251

Number of Flowers in Full Bloom

Hard

ArrayHash TableBinary SearchSortingPrefix SumOrdered SetBinary SearchSortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + m log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n + m log n)Space O(n)

We can efficiently count blooming flowers using sorting and binary search. By creating two separate lists for start and end times, we can quickly determine how many flowers are blooming at any given time.

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Algorithm

3 steps

1Step 1: Create two lists, one for start times and one for end times, and sort both.
2Step 2: For each person's arrival time, use binary search to find how many flowers have started blooming and how many have stopped blooming by that time.
3Step 3: The difference between the two counts gives the number of flowers blooming at that time.

solution.py8 lines

1def numFlowersInBloom(flowers, people):
2    starts = sorted(start for start, end in flowers)
3    ends = sorted(end for start, end in flowers)
4    result = []
5    for person in people:
6        blooming = bisect.bisect_right(starts, person) - bisect.bisect_left(ends, person)
7        result.append(blooming)
8    return result

ℹ

Complexity note: Sorting the start and end times takes O(n log n), and each query for a person takes O(log n) due to binary search, leading to a total of O(n log n + m log n).

1Understanding the blooming period as a range helps in counting efficiently.
2Using sorting and binary search can significantly reduce the time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.