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The goal is to find the maximum number of unique types of candies Alice can eat. We can take the following steps:
maxCandies = n / 2.maxCandies.Time Complexity: O(n) because we iterate over the array to create the set.
Space Complexity: O(n) for storing the unique types in a set.
1def distributeCandies(candyType):
2 return min(len(set(candyType)), len(candyType) // 2)This solution creates a set from the list of candies to determine how many unique types there are. The minimum of the number of unique types and n / 2 candies Alice can eat is the answer.
For another perspective:
n / 2.Time Complexity: O(n) for creating the counter.
Space Complexity: O(n) for storing unique types in the counter.
1#include <vector>
using namespace std;
class Solution {
public:
int distributeCandies(vector<int>& candyType) {
unordered_map<int, int> candyMap;
for (int candy : candyType) {
candyMap[candy]++;
}
return min(candyMap.size(), candyType.size() / 2);
}
};The C++ implementation uses an unordered_map for easy frequency counting, comparing entry size to n / 2 for final decision.