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You need to distribute n candies to k people in a specific way. The i-th person gets i candies (1-indexed) in the first round, the i-th person gets k + i candies in the second round, and so on until we run out of candies. The distribution process continues in rounds until all candies are distributed. Return an array representing the distribution of candies to each person.

Example:

Input: candies = 7, num_people = 4
Output: [1, 2, 3, 1]

Explanation:
On the first turn, the first person gets 1 candy, the second person gets 2 candies, the third person gets 3 candies, and the fourth person gets 1 candy (because there is only one candy left).

Note:

Clarifying Questions

  1. Can num_people be zero?
    • No, according to the constraints, 1 <= k, so num_people will always be at least 1.
  2. Are we distributing the candies in a cyclic manner?
    • Yes, once we reach the last person, we start again with the first person.
  3. Should the remaining candies be distributed according to the sequence?
    • Yes, any leftover candies go to the next person in the sequence as per the required distribution rule.

Strategy

  1. Initialize an array of size num_people with all zeros. This array will store the candies distributed to each person.
  2. Use a variable i to keep track of the current number of candies to distribute and another variable index to track the current person.
  3. Use a loop to distribute candies:
    • Each iteration will distribute i candies to the index-th person until candies < i.
    • Reduce the remaining candies by i.
    • Increment i by 1 to distribute in the next round.
    • Move to the next person (index), using modulo operation to cycle through the people.
  4. If the remaining candies are less than i, distribute all remaining candies to the current person and end.

Let’s implement this step-by-step in Python.

Code

def distributeCandies(candies, num_people):
    distribution = [0] * num_people
    i = 0
    while candies > 0:
        distribution[i % num_people] += min(candies, i + 1)
        candies -= i + 1
        i += 1
    return distribution

Explanation

  1. Initialization:
    • distribution = [0] * num_people to create a list of zeros of size num_people.
  2. Loop through to distribute candies:
    • i starts from 0 and increases with each iteration.
    • In each iteration, distribute min(candies, i + 1) candies to the (i % num_people)-th person. This ensures if candies is less than i + 1, we only distribute the remaining candies.
    • Subtract i + 1 from candies.
    • Increase i by 1 to move to the next turn.
  3. Exit condition:
    • The loop exits when candies becomes 0 or less.

Time Complexity

The time complexity of this solution is O(sqrt(2n)). The reason being the number of iterations i grows until the sum of the first i natural numbers is greater than or equal to n, which grows roughly with the order of the square root of n.

Thus, the time complexity is efficient and suitable for the given constraints.

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