# O(nlogn) time using priority queue and O(n) time using bucket sort (C++)

• // (1) use a priority queue to find top K frequent words.
//     O(n log k) time - use priority queue with O(n) space
struct cmp{
bool operator()(const pair<int,string> a, const pair<int,string> b){
if( a.first != b.first) return a.first > b.first; // number sorted from small to large
else{
return a.second < b.second; // string sorted with the higher alphabetical order comes first.
}
}
};
vector<string> topKFrequent(vector<string>& words, int k) {
if( words.size() == 0) return vector<string>();

unordered_map<string, int> freq;
for(auto s : words) freq[s]++;

priority_queue<pair<int,string>, vector<pair<int,string>>, cmp> pq;
for(auto it = freq.begin(); it != freq.end(); it++){
pq.push(make_pair(it->second, it->first));
if(pq.size() > k ) pq.pop();
}

vector<string> res;
for(int i = 0; i < k; i++){
res.insert(res.begin(), pq.top().second);
pq.pop();
}
return res;
}

// solution 2: using bucket sort with O (n ) time
vector<string> topKFrequent(vector<string>& words, int k) {
if( words.size() == 0) return vector<string>();

unordered_map<string, int> freq;
for(auto s : words) freq[s]++;

vector<set<string>> bucket(words.size(), set<string>());
for(auto it = freq.begin(); it != freq.end(); it++){
bucket[it->second].insert(it->first);
}

vector<string> res;
bool findK = false;
for(int i = bucket.size() - 1; i >= 0; i--){
if(bucket[i].size() > 0 ){
for(auto e : bucket[i]){
res.push_back(e);
if(res.size() == k ){
findK = true;
break;
}
}
if( findK ) break;
}
}
return res;
}

• @fancy1984

Your bucket solution is not O(n)

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