Maximum Candies You Can Get From Boxes - Leetcode Solution

Problem Statement:

You are given n boxes, each box is given in the form of a binary array containing 0s and 1s. You are also given an array of candies containing m elements, where candies[i] represents the number of candies in the ith box (0-indexed).

You can perform the following operation any number of times:

Choose two boxes with the same number of candies and the number of candies in one of the boxes is greater than zero. Remove one candy from each box.

Return the maximum number of candies you can get after performing the above operation any number of times.

Solution:

To maximize the number of candies, we can follow the following steps:

1. Initialize two arrays, candies and counts, where candies stores the number of candies in each box and counts stores the number of boxes having a specific number of candies.

2. Sort the counts array in descending order.

3. Initialize two variables, max_candies and curr_count, where max_candies stores the maximum number of candies we can get and curr_count stores the number of boxes having the same number of candies as the current count.

4. Iterate over the counts array in descending order and perform the following operations:

   a. If curr_count is zero, break the loop.

   b. If candies[curr_count-1] is greater than zero, add candies[curr_count-1] to max_candies and remove one candy from both boxes with curr_count-1 candies.

   c. Decrement curr_count.

5. Return max_candies.

Here is the Python code to implement the above algorithm:

class Solution: def maxCandies(self, status: List[int], candies: List[int], keys: List[List[int]], containedBoxes: List[List[int]], initialBoxes: List[int]) -> int: n = len(status) boxes = [0] * n for box in initialBoxes: boxes[box] = 1 counts = [0] * 1001 for box in boxes: if box: counts[candies[boxes.index(box)]] += 1 curr_count = max([i for i in range(len(counts)) if counts[i] > 0]) max_candies = 0 while curr_count >= 0: if curr_count == 0 or boxes.count(1) == 0: break if candies[boxes.index(1)] > 0: max_candies += candies[boxes.index(1)] boxes[boxes.index(1)] = 0 for key in keys[boxes.index(1)]: status[key] = 1 if boxes[key] == 0 and status[key] == 1: boxes[key] = 1 counts[candies[key]] += 1 for box in containedBoxes[boxes.index(1)]: if boxes[box] == 0 and status[box] == 1: boxes[box] = 1 counts[candies[box]] += 1 counts[curr_count] -= 1 if counts[curr_count] == 0: curr_count -= 1 return max_candies

Time Complexity Analysis:

The time complexity of the above algorithm is O(n), where n is the total number of boxes. We iterate over the counts array, which has a maximum length of 1001, and perform constant time operations in each iteration. Therefore, the time complexity is linear in the size of the input.

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