Problem 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.
Key Insights
- The candies are distributed in a cyclic manner to the people.
- Each complete cycle distributes a total of a known number of candies, which can be calculated.
- The remaining candies after complete cycles can be distributed in a straightforward manner.
Space and Time Complexity
Time Complexity: O(num_people)
Space Complexity: O(num_people)
Solution
To solve this problem, we can use an array to keep track of the candies distributed to each person. We first calculate how many complete rounds of distribution can be done with the available candies. Each round distributes a total of n * (k + 1) candies to n people, where k is the number of complete rounds. After distributing candies for complete rounds, we handle the remaining candies by distributing them one by one in a cyclic manner until all candies are exhausted.