<p>Given an array of strings <code>words</code> and an integer <code>k</code>, return <em>the </em><code>k</code><em> most frequent strings</em>.</p>

<p>Return the answer <strong>sorted</strong> by <strong>the frequency</strong> from highest to lowest. Sort the words with the same frequency by their <strong>lexicographical order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;i&quot;,&quot;love&quot;,&quot;leetcode&quot;,&quot;i&quot;,&quot;love&quot;,&quot;coding&quot;], k = 2
<strong>Output:</strong> [&quot;i&quot;,&quot;love&quot;]
<strong>Explanation:</strong> &quot;i&quot; and &quot;love&quot; are the two most frequent words.
Note that &quot;i&quot; comes before &quot;love&quot; due to a lower alphabetical order.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;the&quot;,&quot;day&quot;,&quot;is&quot;,&quot;sunny&quot;,&quot;the&quot;,&quot;the&quot;,&quot;the&quot;,&quot;sunny&quot;,&quot;is&quot;,&quot;is&quot;], k = 4
<strong>Output:</strong> [&quot;the&quot;,&quot;is&quot;,&quot;sunny&quot;,&quot;day&quot;]
<strong>Explanation:</strong> &quot;the&quot;, &quot;is&quot;, &quot;sunny&quot; and &quot;day&quot; are the four most frequent words, with the number of occurrence being 4, 3, 2 and 1 respectively.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 500</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li><code>k</code> is in the range <code>[1, The number of <strong>unique</strong> words[i]]</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Could you solve it in <code>O(n log(k))</code> time and <code>O(n)</code> extra space?</p>

class Solution {
public:
    struct cmp {
        bool operator()(pair<int, string> x, pair<int, string> y) {
            if (x.first == y.first) {
                return x.second > y.second;
            } else {
                return x.first < y.first;
            }
        }
    };
    vector<string> topKFrequent(vector<string>& words, int k) {
        map<string, int> M;
        for (auto word : words) {
            M[word] += 1;
        }
        priority_queue<pair<int, string>, vector<pair<int, string>>, cmp> pq;
        for (auto it = M.begin(); it != M.end(); it++) {
            pq.push({(*it).second, (*it).first});
        }

        vector<string> ans;
        for (int i = 0; i < k; i++) {
            ans.push_back(pq.top().second);
            pq.pop();
        }

        return ans;
    }
};
17700