Introduction

In this blog post, we will explore how to solve the LeetCode problem "2251" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the solution:

• Separate Starts and Ends: Extract the start and end times of the flowers into separate sorted arrays. This allows us to use binary search efficiently.
• Binary Search for Active Flowers: For each person, use binary search on the start times to find how many flowers have started blooming by their arrival time. Also, use binary search on the end times to find how many flowers have already stopped blooming. The difference gives the number of blooming flowers.

• Efficient Counting: The binary searches return the number of flowers that satisfy the condition (started or ended), which directly provides the count needed.

• Runtime Complexity: O(n log m + m log m), where n is the number of people and m is the number of flowers. O(m log m) for sorting the start and end times, and O(n log m) for the binary searches for each person. Storage Complexity: O(m), to store the sorted start and end times.

Code

    def fullBloomFlowers(flowers, people):
    """
    Calculates the number of flowers in full bloom when each person arrives.

    Args:
        flowers: A list of lists, where each inner list represents a flower's bloom period [start, end].
        people: A list of integers representing the arrival times of people.

    Returns:
        A list of integers representing the number of flowers in bloom for each person.
    """
    starts = sorted([flower[0] for flower in flowers])
    ends = sorted([flower[1] for flower in flowers])
    result = []

    def binary_search(arr, target, find_start=True):
        """
        Performs binary search to find the number of elements <= target.

        Args:
            arr: The sorted array to search.
            target: The target value.
            find_start: True to find the position to insert target to the left,
                         False to find the position to insert target to the right.
        Returns: The index of the rightmost element less than or equal to the target (if find_start is true)
                 or the index of the rightmost element strictly less than the target (if find_start is false).
        """
        left, right = 0, len(arr)
        while left < right:
            mid = (left + right) // 2
            if arr[mid] <= target:
                left = mid + 1
            else:
                right = mid
        return left

    for person in people:
        started = binary_search(starts, person)
        ended = binary_search(ends, person)
        result.append(started - ended)

    return result