Distribute Candies - Leetcode Solution

Problem:

You have n candies, and you want to distribute them equally between a brother and a sister. You can only give one candy to one person at a time, and you must alternate between the brother and the sister. That is, first, the brother gets a candy; then, the sister gets a candy; and so on until all the candies are gone.

Return an array ans of length 2, where ans[0] is the number of candies the brother gets, and ans[1] is the number of candies the sister gets.

Constraints:

1. 2 <= n <= 10^9

Solution:

To distribute candies equally among a brother and a sister, we need to divide the total number of candies by two. We will then give one candy to the brother and one candy to the sister alternatively until we run out of candies.

Let's calculate the maximum number of candies each person can get. If the total number of candies is even, then we can divide them equally between the brother and the sister. For example, if the total number of candies is 8, then each person can get 4 candies.

If the total number of candies is odd, then one person will get an extra candy. We can give this extra candy to the brother or the sister, depending on which person got the last candy. For example, if the total number of candies is 7, and we give the last candy to the sister, then the brother will get 3 candies, and the sister will get 4 candies.

To implement this algorithm in code, we can use the following steps:

1. Calculate the maximum number of candies each person can get by dividing the total number of candies by two. If the total number of candies is odd, then add 1 to this number.

2. Determine which person got the last candy. If the total number of candies is odd, and the last candy was given to the sister, then add 1 to the number of candies the brother gets.

3. Return an array containing the number of candies each person gets, in the order [brother, sister].

Here is the Python code for the solution:

def distributeCandies(candies: int) -> List[int]: # Calculate the maximum number of candies each person can get max_candies = candies // 2 if candies % 2 != 0: max_candies += 1

# Determine which person got the last candy
last_person = candies % 2

# Determine the number of candies each person gets
brother_candies = max_candies if last_person == 0 or last_person == 1 else max_candies - 1
sister_candies = max_candies if last_person == 0 or last_person == -1 else max_candies - 1

# Return the result as an array
return [brother_candies, sister_candies]

Time Complexity:

The time complexity of this solution is O(1), as we are performing only constant-time operations to calculate the number of candies each person gets.

Space Complexity:

The space complexity of this solution is O(1), as we are using only a few variables to store the result.

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