Introduction

In this blog post, we will explore how to solve the LeetCode problem "1488" 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

Your country has an infinite number of lakes. Initially, all the lakes are empty, but when it rains over the nth lake, the nth lake becomes full of water. If it rains over a lake that is full of water, there will be a flood. Your goal is to avoid floods in any lake. Given an integer array rains where: rains[i] > 0 means there will be rains over the rains[i] lake. rains[i] == 0 means there are no rains this day and you can choose one lake this day and dry it. Return an array ans where: ans.length == rains.length ans[i] == -1 if rains[i] > 0. ans[i] is the lake you choose to dry in the ith day if rains[i] == 0. If there are multiple valid answers return any of them. If it is impossible to avoid flood return an empty array. Notice that if you chose to dry a full lake, it becomes empty, but if you chose to dry an empty lake, nothing changes. Example 1: Input: rains = [1,2,3,4] Output: [-1,-1,-1,-1] Explanation: After the first day full lakes are [1] After the second day full lakes are [1,2] After the third day full lakes are [1,2,3] After the fourth day full lakes are [1,2,3,4] There's no day to dry any lake and there is no flood in any lake. Example 2: Input: rains = [1,2,0,0,2,1] Output: [-1,-1,2,1,-1,-1] Explanation: After the first day full lakes are [1] After the second day full lakes are [1,2] After the third day, we dry lake 2. Full lakes are [1] After the fourth day, we dry lake 1. There is no full lakes. After the fifth day, full lakes are [2]. After the sixth day, full lakes are [1,2]. It is easy that this scenario is flood-free. [-1,-1,1,2,-1,-1] is another acceptable scenario. Example 3: Input: rains = [1,2,0,1,2] Output: [] Explanation: After the second day, full lakes are [1,2]. We have to dry one lake in the third day. After that, it will rain over lakes [1,2]. It's easy to prove that no matter which lake you choose to dry in the 3rd day, the other one will flood. Constraints: 1 <= rains.length <= 105 0 <= rains[i] <= 109

Explanation

Here's a breakdown of the solution:

• Core Idea: The solution iterates through the rains array. When it encounters rain on a lake, it marks the lake as full. When it encounters a dry day (rains[i] == 0), it looks for the next opportunity to dry a lake that will flood in the future.
• Data Structures: A set is used to store the indices of dry days, and a dictionary is used to store the next rain day for each lake. The set allows for efficient searching of available dry days and removal when a dry day is utilized.

• Flood Detection & Avoidance: For each rainy day, the code checks if the lake is already full. If it is, a flood would occur, and the function returns an empty array. If not, it records the next rainy day for that lake. When a dry day is encountered, the algorithm finds the next lake that needs to be dried before it floods (using the sorted set of available dry days and the next rain days dictionary).

• Runtime & Space Complexity: O(N log N) time, O(N) space, where N is the length of the rains array. The log N factor comes from set operations (specifically, bisect_left, add, and remove).

Code

    def avoidFlood(rains):
    n = len(rains)
    ans = [-1] * n
    full_lakes = {}
    dry_days = []

    for i, rain in enumerate(rains):
        if rain > 0:
            if rain in full_lakes:
                return []  # Flood!
            full_lakes[rain] = i
        else:
            dry_days.append(i)

    for rain in full_lakes:
        next_rain = full_lakes[rain]

        if not dry_days:
            return []

        dry_day_index = -1
        for i, day in enumerate(dry_days):
          if day > next_rain:
            dry_day_index = i
            break

        if dry_day_index == -1:
          return []

        ans[dry_days[dry_day_index]] = rain
        dry_days.pop(dry_day_index)


    for i in range(n):
        if rains[i] == 0 and ans[i] == -1:
            ans[i] = 1  # Dry any arbitrary lake

    return ans