1103. Distribute Candies to People

Description

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 <= 1⁰⁹
• 1 <= num_people <= 1000

💡 Hint 1:
Give candy to everyone each "turn" first [until you can't], then give candy to one person per turn.

Submitted Code

class Solution:
    def distributeCandies(self, candies: int, num_people: int) -> List[int]:
        result = [0] * num_people
        count = 1

        # 1. 모든 사람에게 사탕을 나눠줄 수 있을 때까지 반복
        while True:
            gives = (count + (count + num_people - 1)) * num_people // 2    # 이번 턴에서 모두에게 줄 사탕
            if gives > candies:
                break
            for i in range(num_people):
                result[i] += count
                count += 1
            candies -= gives

        # 2. 사탕이 떨어질 때까지 한 사람씩 나눠주기 반복
        for i in range(num_people):
            give = count if candies - count > 0 else candies
            result[i] += give
            candies -= give
            if candies == 0:
                break
            count += 1

        return result

Runtime: 0 ms | Beats 100.00%
Memory: 17.17 MB | Beats 99.78%

힌트대로 두 단계로 나누면 한 사람씩 순서대로 나눠주는 것보다 더 효율적으로 풀 수 있다. 먼저 모든 사람에게 사탕을 줄 수 있을 때까지 최대한 반복한 이후, 그 다음부터 원래 규칙대로 한 사람씩 나눠주는 방법이다. 모든 사람에게 나눠줄 사탕 갯수를 구하는 공식은 (시작 숫자 + 끝 숫자) * 전체 인원 / 2이다.

Other Solutions

1st

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

time complexity: 𝑂(√candies)
space complexity: 𝑂(𝑛)

효율적인 방법은 아니지만 %를 이용하여 깔끔하게 작성된 코드이기 때문에 참고했다.

2nd

class Solution:
    def distributeCandies(self, candies, n):
        x = int(math.sqrt(candies * 2 + 0.25) - 0.5)    # 전체 지급 가능한 횟수
        res = [0] * n
        for i in xrange(n):
            m = x / n + (x % n > i)                     # 사람 i가 받는 횟수(기본횟수 + 마지막 사람 추가횟수)
            res[i] = m * (i + 1) + m * (m - 1) / 2 * n  # 사람 i가 받는 사탕(등차수열)
        res[x % n] += candies - x * (x + 1) / 2         # 배분 후 남은 사탕을 다음 사람에게 전부 지급
        return res

time complexity: 𝑂(𝑛) ← num_people
space complexity: 𝑂(𝑛)

수학 공식으로 더 간단히 풀 수도 있다. x값은 x(x+1)/2 <= candies 식을 변형해서 구한 것이다.