Problem Description

In this problem, we have a certain number of candies that need to be distributed to num_people people arranged in a row. The distribution starts with the first person receiving one candy, the second person receiving two candies, and so on, increasing the count of candies by one for each subsequent person until the nth person receives n candies. After reaching the last person, the distribution continues from the first person again, but this time each person gets one more candy than the previous cycle (so the first person now gets n+1 candies, the second gets n+2, and so on). This process repeats until we run out of candies. If there are not enough candies to give the next person in the sequence their "full" amount, they receive the remaining candies, and the distribution ends.

The goal is to return an array of length num_people, with each element representing the total number of candies that each person receives at the end of the distribution process.

Intuition

To solve this problem, we want to simulate the described candy distribution process. We keep handing out candies until we have none left. Each person gets a certain number of candies based on the round of distribution we are in. In the first round, person 1 gets 1 candy, person 2 gets 2 candies, and so on. Once we reach num_people, we wrap around and start from person 1 again, increasing the amount of candy given out by num_people each round.

The solution involves iterating over the people in a loop and incrementing the number of candies each person gets by the distribution rule given. We maintain a counter i to keep track of how many candies have been given out so far, and a list ans to store the total candies for each person.

With each person's turn, we give out the number of candies equal to the counter i + 1, but if we have fewer candies left than i + 1, we give out all the remaining candies. After that, we update the total number of candies left by subtracting the number given out. If the candies finish during someone's turn, we stop the distribution and return our ans list to show the final distribution of candies.

The intuition is to replicate the physical process of handing out the candies in a loop, ensuring that conditions such as running out of candies are properly handled.

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Solution Approach

The solution to this problem uses a simple iterative approach as our algorithm. Here are the steps and the reasoning in detail:

1. Initialize the Answer List: We start by initializing an array ans of length num_people with all elements set to 0. This array will be used to keep track of the number of candies each person receives.

2. Starting the Distribution: We need to keep track of two things - the index of the person to whom we're currently giving candies, and the number of candies we're currently handing out. We start the distribution by setting a counter i to 0, which will increase with each iteration to represent the amount of candy to give.

3. Iterative Distribution: We use a while loop to continue distributing candies until we run out (candies > 0). During each iteration of the loop:

   • Compute i % num_people to find the index of the current person. This ensures that after the last person, we start again from the first person.
   • Determine the number of candies to give to the current person. We use min(candies, i + 1) to decide this amount because we either give i + 1 candies or the remaining candies if we have less than i + 1.
   • Subtract the number of distributed candies from candies.
   • Increment i by 1 to update the count for the next iteration.

4. Updating Answer List: In each iteration, update ans[i % num_people] with the number of candies distributed in that iteration.

5. Handling Remaining Candies: If we deplete our supply of candies, we give out the remaining candies to the last person. This is built into the allocation step with min(candies, i + 1).

6. Returning the Final Distribution: Once the loop ends (no more candies are left), we exit the loop. The array ans now contains the total number of candies received by each person, which we return as the final answer.

This problem does not require any complex data structures or patterns. The concept is straightforward and only uses basic array manipulation to achieve the goal. It focuses on handling the loop correctly and ensuring that the distribution of candies is done according to the specified rules.

class Solution:
    def distributeCandies(self, candies: int, num_people: int) -> List[int]:
        ans = [0] * num_people  # Initialize the answer list
        i = 0                   # Counter for the distribution process
        # Distribute candies until we run out
        while candies:          
            # Give out min(candies, i + 1) candies to the (i % num_people)th person
            ans[i % num_people] += min(candies, i + 1) 
            candies -= min(candies, i + 1) # Subtract the candies given from the total
            i += 1  # Move to the next person

        return ans  # Return the final distribution

The solution makes effective use of modulo operation to cycle through the indices repeatedly while the while loop condition ensures that the distribution halts at the right time.

Example Walkthrough

Let's use a small example to illustrate the solution approach.

Suppose we have candies = 7 and num_people = 4.

We want to distribute these candies across 4 people as described. Let's walk through the process using the provided algorithm.

1. Initialize the Answer List:
   • ans = [0, 0, 0, 0]
2. Starting the Distribution:
   • candies = 7
   • Distribution counter i = 0
3. Iterative Distribution:
   1. During the first iteration (i = 0):
      • Current person index: 0 % 4 = 0 (first person)
      • Candies to give out: min(7, 0 + 1) = 1
      • Remaining candies: 7 - 1 = 6
      • Updated ans list: [1, 0, 0, 0]
      • Increment i to 1
   2. In the second iteration (i = 1):
      • Current person index: 1 % 4 = 1 (second person)
      • Candies to give out: min(6, 1 + 1) = 2
      • Remaining candies: 6 - 2 = 4
      • Updated ans list: [1, 2, 0, 0]
      • Increment i to 2
   3. In the third iteration (i = 2):
      • Current person index: 2 % 4 = 2 (third person)
      • Candies to give out: min(4, 2 + 1) = 3
      • Remaining candies: 4 - 3 = 1
      • Updated ans list: [1, 2, 3, 0]
      • Increment i to 3
   4. In the fourth iteration (i = 3):
      • Current person index: 3 % 4 = 3 (fourth person)
      • Candies to give out: min(1, 3 + 1) = 1
      • Remaining candies: 1 - 1 = 0 (no more candies)
      • Updated ans list: [1, 2, 3, 1]
      • Candies are now depleted, we stop the distribution.
4. Return the Final Distribution:
   • The final ans list is [1, 2, 3, 1].

Each element in the ans list represents the total number of candies each person receives after the distribution is done. The algorithm successfully mimics the handing out of candies until there are no more left, while following the rules set out in the problem description.

Solution Implementation

Time and Space Complexity

The time complexity of the given code can be determined by the while loop, which continues until all candies are distributed. In each iteration of the loop, i is incremented by 1, and the amount of candies distributed is also incremented by 1 until all candies are exhausted. This forms an arithmetic sequence from 1 to n where n is the turn where the candies run out. The total number of candies distributed by this sequence can be represented by the sum of the first n natural numbers formula n*(n+1)/2. So the time complexity is governed by the smallest n such that n*(n+1)/2 >= candies. Therefore, the time complexity is O(sqrt(candies)) because we need to find an n such that n^2 is asymptotically equal to the total number of candies.

The space complexity of the code is determined by the list ans that has a size equal to num_people. Since the size of this list does not change and does not depend on the number of candies, the space complexity is O(num_people), which is the space required to store the final distribution of the candies among the people.

Learn more about how to find time and space complexity quickly using problem constraints.