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This approach iteratively distributes candies to people in a sequence, considering one cycle at a time until all candies are allocated.
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.
1#include <iostream>
2#include <vector>
3using namespace std;
4
5vector<int> distributeCandies(int candies, int num_people) {
6 vector<int> result(num_people, 0);
7 int i = 0;
8 int amount = 1;
9 while (candies > 0) {
10 result[i % num_people] += min(candies, amount);
11 candies -= amount;
12 amount++;
13 i++;
14 }
15 return result;
16}
17
18int main() {
19 int candies = 10;
20 int num_people = 3;
21 vector<int> result = distributeCandies(candies, num_people);
22 for (int num : result) {
23 cout << num << " ";
24 }
25 return 0;
26}The C++ code distributes candies using a vector to store the allocation for each person. The use of the modulo operator allows for cycling through each person in sequence until candies are depleted. The min operation ensures that if fewer candies remain than needed, the remaining candies are given.
Using a mathematical approach, calculate full rounds of distribution first to optimize the solution as opposed to simulating step-by-step distribution.
Time Complexity: O(sqrt(2 * candies)) due to converging arithmetic series.
Space Complexity: O(num_people).
1def
This Python solution effectively reduces the operational complexity by calculating in larger blocks, then using an arithmetic method to fill any remaining candies. It allows for quicker transition between distribution cycles.