LeetCode 1103. Distribute Candie

We distribute some number of candies, to a row of n = num_people people in the following way:

We then give 1 candy to the first person, 2 candies to the second person, and so on until we give n candies to the last person.

Then, we go back to the start of the row, giving n + 1 candies to the first person, n + 2 candies to the second person, and so on until we give 2 * n candies to the last person.

This process repeats (with us giving one more candy each time, and moving to the start of the row after we reach the end) until we run out of candies. The last person will receive all of our remaining candies (not necessarily one more than the previous gift).

Return an array (of length num_people and sum candies) that represents the final distribution of candies.

Example 1:
Input: candies = 7, num_people = 4
Output: [1,2,3,1]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0,0].
On the third turn, ans[2] += 3, and the array is [1,2,3,0].
On the fourth turn, ans[3] += 1 (because there is only one candy left), and the final array is [1,2,3,1].

Example 2:
Input: candies = 10, num_people = 3
Output: [5,2,3]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0].
On the third turn, ans[2] += 3, and the array is [1,2,3].
On the fourth turn, ans[0] += 4, and the final array is [5,2,3].

Constraints:
1 <= candies <= 10^9
1 <= num_people <= 1000

Solution:

class Solution:
    def distributeCandies(self, candies: int, num_people: int) -> List[int]:
        res = [0] * num_people
        c = 0
        while candies > 0:
            res[c % num_people] += min(c+1, candies)
            c += 1
            candies -= c
        return res
        

Explanation
First, we initialize an result array, then we use mod to target the index of the i-th person. As long as we still have candies, we give next person the candies he should have, and update the candies remaining for next loop. We always return min(c+1, candies), meaning we choose the candies left, if the candies are not enough.