<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k</code> <em>most frequent elements</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,1,2,2,3], k = 2
<strong>Output:</strong> [1,2]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> [1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>k</code> is in the range <code>[1, the number of unique elements in the array]</code>.</li>
	<li>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Your algorithm&#39;s time complexity must be better than <code>O(n log n)</code>, where n is the array&#39;s size.</p>

class Solution {
public:
    bool comparator(pair<int, int> a, pair<int, int> b) {
        return a.first > b.first;
    }
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        for(auto num: nums) {
            mp[num]++;
        }
        priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int>>> pq;
        auto it=mp.begin();
        for(auto i=0 ; i<k && it!=mp.end() ; i++) {
            pq.push({it->second, it->first});
            it++;
        }
        for(int i=k;i<nums.size() && it!=mp.end();i++) {
            if(pq.top().first < it->second) {
                pq.pop();
                pq.push({it->second, it->first});
            }
            it++;
        }

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