Maximum Candies Allocated To K Children - Leetcode Solution

Problem statement: Given an array of integers candies where candies[i] represents the number of candies that the ith kid has and an integer k where k is the number of kids you want to allocate the candies to. You need to allocate the candies to the k kids such that:

Each child gets exactly one candy. There is at least one candy to every child. All the children have the maximum number of candies.

Return an integer array res of length k where res[i] is the maximum number of candies the ith kid has.

Solution: To solve this problem, we need to distribute candies to k children in such a way that each child gets exactly one candy, and the number of candies distributed is maximum possible. We can follow the following steps to solve the problem:

Step 1: Determine the minimum number of candies that can be distributed to each child. To do this, we need to divide the total number of candies by k and take the floor value of the result.

Step 2: Allocate the minimum number of candies to each child.

Step 3: Distribute the remaining candies to the children who have not yet received the maximum number of candies.

Step 4: Return the maximum number of candies allocated to each child.

Let's now implement the solution in Python:

def maxCandies(candies: List[int], k: int) -> List[int]: total_candies = sum(candies) min_candies = total_candies // k res = [min_candies] * k remaining_candies = total_candies % k i = 0 while remaining_candies: res[i] += 1 i += 1 remaining_candies -= 1 if i == k: i = 0 return res

In the above implementation, we first calculate the total number of candies and the minimum number of candies that can be distributed to each child. We then initialize the answer array with the minimum number of candies for each child.

We then calculate the remaining candies after allocating the minimum number of candies to each child and distribute the remaining candies to the children who have not yet received the maximum number of candies in a round-robin fashion. We keep track of the current child using the variable i and increment it by 1 after allocating a candy to each child.

After distributing all the candies, we return the result array containing the maximum number of candies allocated to each child.

Time Complexity: The time complexity of the above implementation is O(n), where n is the length of the candies array, as we need to iterate through the candies array to calculate the total number of candies and the remaining candies after the minimum allocation.

Space Complexity: The space complexity of the above implementation is O(k), as we only need to store the answer array of length k, where k is the number of children.

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