/*Given a non-empty array of integers, return the k most frequent elements.

*For example,Given [1,1,1,2,2,3] and k = 2, return [1,2].

*Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements.

*Your algorithm's time complexity must be better than O(n log n), where n is the array's size.

*/

class Solution {

public:

vector<int> topKFrequent(vector<int>& nums, int k) {

unordered_map<int,int> map;

for(int num : nums){

map[num]++;

}

vector<int> res;

// pair<first, second>: first is frequency, second is number

priority_queue<pair<int,int>> pq;

for(auto it = map.begin(); it != map.end(); it++){

pq.push(make_pair(it->second, it->first));

if(pq.size() > (int)map.size() - k){

res.push_back(pq.top().second);

pq.pop();

}

}

return res;

}

};