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:
This approach iteratively distributes candies to people in a sequence, considering one cycle at a time until all candies are allocated.
The C solution utilizes a simple while loop for the distribution of candies. It tracks how many candies to give based on a running total. The modulo operation (%num_people) ensures allocation is cycled through the people repeatedly until candies run out.
C++
Java
Python
C#
JavaScript
Time Complexity: O(sqrt(2 * candies)). This is because the linear sum of the sequence continues until all candies are exhausted, which resembles the behavior of an arithmetic series.
Space Complexity: O(num_people), since we maintain an array to store the number of candies for each person.
Using a mathematical approach, calculate full rounds of distribution first to optimize the solution as opposed to simulating step-by-step distribution.
Although this solution follows a similar process to the iterative one, it stops calculating after wrapping through all the necessary candies in full cycles using arithmetic logic for each complete set of distributions.
C++
Java
Python
C#
JavaScript
Time Complexity: O(sqrt(2 * candies)) due to converging arithmetic series.
Space Complexity: O(num_people).
| Approach | Complexity |
|---|---|
| Iterative Distribution Approach | Time Complexity: O(sqrt(2 * candies)). This is because the linear sum of the sequence continues until all candies are exhausted, which resembles the behavior of an arithmetic series. Space Complexity: O(num_people), since we maintain an array to store the number of candies for each person. |
| Mathematical Distribution Approach | Time Complexity: O(sqrt(2 * candies)) due to converging arithmetic series. Space Complexity: O(num_people). |
LeetCode 1103. Distribute Candies to People (Algorithm Explained) • Nick White • 12,873 views views
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