<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>


bool customSort(pair<string, int>& a, pair<string, int>& b) {
    // Compare based on string length
    if(a.second>b.second) return true;
    else if(a.second==b.second) return a.first<b.first;
    else return false;
}
class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        map<string, int> mp;
        for (auto i : words)
            mp[i] = -1;
        for (auto i : words) {
            if (mp[i] != -1)
                mp[i]++;
            else
                mp[i] = 1;
        }
        vector<pair<string, int>> vec(mp.begin(), mp.end());
        // Sort the vector using the custom comparator
        sort(vec.begin(), vec.end(), customSort);
        vector<string>ans;
        ans.push_back(vec[0].first);
        for (int i = 1; i < k; i++) {
            ans.push_back(vec[i].first);
        }
        
        return ans;
    }
};
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