Given an array of strings words and an integer k, return the k most frequent strings.
Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order.
Example 1:
Input: words = ["i","love","leetcode","i","love","coding"], k = 2 Output: ["i","love"] Explanation: "i" and "love" are the two most frequent words. Note that "i" comes before "love" due to a lower alphabetical order.
Example 2:
Input: words = ["the","day","is","sunny","the","the","the","sunny","is","is"], k = 4 Output: ["the","is","sunny","day"] Explanation: "the", "is", "sunny" and "day" are the four most frequent words, with the number of occurrence being 4, 3, 2 and 1 respectively.
Constraints:
1 <= words.length <= 5001 <= words[i].length <= 10words[i] consists of lowercase English letters.k is in the range [1, The number of unique words[i]]
Follow-up: Could you solve it in O(n log(k)) time and O(n) extra space?
class Solution {
public:
struct comparator{
bool operator()(pair<string, int> a, pair<string, int> b){
return a.second!=b.second?a.second<b.second:a.first>b.first;
} // reverse sort by freq... if same ascending sort lexically
};
vector<string> topKFrequent(vector<string>& words, int k) {
unordered_map <string, int> m;
for(string s: words)
m[s]++;
priority_queue <pair<string, int>, vector<pair<string, int>>, comparator> pq;
for(auto [k, v]: m)
pq.push(make_pair(k, v));
vector <string> v(k);
int i;
while(!pq.empty() && i<k)
v[i++]=pq.top().first, pq.pop();
return v;
}
};