Introduction

In this blog post, we will explore how to solve the LeetCode problem "2136" 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 have n flower seeds. Every seed must be planted first before it can begin to grow, then bloom. Planting a seed takes time and so does the growth of a seed. You are given two 0-indexed integer arrays plantTime and growTime, of length n each: plantTime[i] is the number of full days it takes you to plant the ith seed. Every day, you can work on planting exactly one seed. You do not have to work on planting the same seed on consecutive days, but the planting of a seed is not complete until you have worked plantTime[i] days on planting it in total. growTime[i] is the number of full days it takes the ith seed to grow after being completely planted. After the last day of its growth, the flower blooms and stays bloomed forever. From the beginning of day 0, you can plant the seeds in any order. Return the earliest possible day where all seeds are blooming. Example 1: Input: plantTime = [1,4,3], growTime = [2,3,1] Output: 9 Explanation: The grayed out pots represent planting days, colored pots represent growing days, and the flower represents the day it blooms. One optimal way is: On day 0, plant the 0th seed. The seed grows for 2 full days and blooms on day 3. On days 1, 2, 3, and 4, plant the 1st seed. The seed grows for 3 full days and blooms on day 8. On days 5, 6, and 7, plant the 2nd seed. The seed grows for 1 full day and blooms on day 9. Thus, on day 9, all the seeds are blooming. Example 2: Input: plantTime = [1,2,3,2], growTime = [2,1,2,1] Output: 9 Explanation: The grayed out pots represent planting days, colored pots represent growing days, and the flower represents the day it blooms. One optimal way is: On day 1, plant the 0th seed. The seed grows for 2 full days and blooms on day 4. On days 0 and 3, plant the 1st seed. The seed grows for 1 full day and blooms on day 5. On days 2, 4, and 5, plant the 2nd seed. The seed grows for 2 full days and blooms on day 8. On days 6 and 7, plant the 3rd seed. The seed grows for 1 full day and blooms on day 9. Thus, on day 9, all the seeds are blooming. Example 3: Input: plantTime = [1], growTime = [1] Output: 2 Explanation: On day 0, plant the 0th seed. The seed grows for 1 full day and blooms on day 2. Thus, on day 2, all the seeds are blooming. Constraints: n == plantTime.length == growTime.length 1 <= n <= 105 1 <= plantTime[i], growTime[i] <= 104

Explanation

Here's the solution:

• Key Idea: Sort the seeds based on their growTime in descending order. This greedy approach minimizes the overall blooming time. Planting seeds with longer grow times earlier allows them to grow in parallel with the planting of seeds that have shorter grow times.
• Calculation: Iterate through the sorted seeds, keeping track of the current planting end time. For each seed, calculate its bloom time (planting end time + grow time) and update the overall earliest bloom day to be the maximum of the current earliest bloom day and the seed's bloom time.
• Complexity:
  • Runtime: O(n log n) due to sorting.
  • Storage: O(n) to store the combined plant/grow time data for sorting (can be O(1) if in-place sorting is allowed/used).

Code

    def earliestFullBloom(plantTime, growTime):
    """
    Calculates the earliest day when all seeds bloom.

    Args:
        plantTime: A list of integers representing the planting time for each seed.
        growTime: A list of integers representing the grow time for each seed.

    Returns:
        The earliest day when all seeds bloom.
    """

    n = len(plantTime)
    seeds = []
    for i in range(n):
        seeds.append((plantTime[i], growTime[i]))

    # Sort seeds based on growTime in descending order
    seeds.sort(key=lambda x: x[1], reverse=True)

    current_plant_end_time = 0
    earliest_bloom_day = 0

    for plant_time, grow_time in seeds:
        current_plant_end_time += plant_time
        earliest_bloom_day = max(earliest_bloom_day, current_plant_end_time + grow_time)

    return earliest_bloom_day