Introduction

In this blog post, we will explore how to solve the LeetCode problem "1298" 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 boxes labeled from 0 to n - 1. You are given four arrays: status, candies, keys, and containedBoxes where: status[i] is 1 if the ith box is open and 0 if the ith box is closed, candies[i] is the number of candies in the ith box, keys[i] is a list of the labels of the boxes you can open after opening the ith box. containedBoxes[i] is a list of the boxes you found inside the ith box. You are given an integer array initialBoxes that contains the labels of the boxes you initially have. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it. Return the maximum number of candies you can get following the rules above. Example 1: Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0] Output: 16 Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you do not have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2. In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed. Total number of candies collected = 7 + 4 + 5 = 16 candy. Example 2: Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0] Output: 6 Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6. Constraints: n == status.length == candies.length == keys.length == containedBoxes.length 1 <= n <= 1000 status[i] is either 0 or 1. 1 <= candies[i] <= 1000 0 <= keys[i].length <= n 0 <= keys[i][j] < n All values of keys[i] are unique. 0 <= containedBoxes[i].length <= n 0 <= containedBoxes[i][j] < n All values of containedBoxes[i] are unique. Each box is contained in one box at most. 0 <= initialBoxes.length <= n 0 <= initialBoxes[i] < n

Explanation

• Initialization and Box Processing: Start with the initial boxes and iteratively process them. Keep track of available keys and contained boxes.
  • Candy Collection and Key Acquisition: When a box is opened (either initially open or opened with a key), collect the candies and acquire new keys and contained boxes.
  • Iterative Opening: Continue opening boxes using acquired keys, and adding contained boxes to the list of boxes to process. Stop when no new boxes can be opened or processed.

• Runtime Complexity: O(N), where N is the number of boxes. Storage Complexity: O(N)

Code

    def maxCandies(status, candies, keys, containedBoxes, initialBoxes):
    """
    Calculates the maximum number of candies that can be collected.

    Args:
        status (list): A list of integers representing the status of each box (1 for open, 0 for closed).
        candies (list): A list of integers representing the number of candies in each box.
        keys (list): A list of lists, where keys[i] is a list of keys found in box i.
        containedBoxes (list): A list of lists, where containedBoxes[i] is a list of boxes found in box i.
        initialBoxes (list): A list of integers representing the initial boxes.

    Returns:
        int: The maximum number of candies that can be collected.
    """

    n = len(status)
    total_candies = 0
    available_keys = set()
    boxes_to_process = set(initialBoxes)
    can_open = True

    while can_open:
        can_open = False
        boxes_to_process_copy = set(boxes_to_process)  # Iterate over a copy to avoid modification during iteration
        for box_index in boxes_to_process_copy:
            if status[box_index] == 1 or box_index in available_keys:
                can_open = True
                boxes_to_process.remove(box_index)
                total_candies += candies[box_index]

                # Add keys
                for key in keys[box_index]:
                    available_keys.add(key)

                # Add contained boxes
                for contained_box in containedBoxes[box_index]:
                    boxes_to_process.add(contained_box)

    return total_candies