Sponsored
Sponsored
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.
1function distributeCandies(candyType) {
2 const uniqueCandies = new Set(candyType);
3 return Math.min(uniqueCandies.size, candyType.length / 2);
4}This JavaScript solution uses a Set to derive the unique candy types, then returns the min between the size of the set and half of the candy array length, which is the maximum Alice can eat.
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.