<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:
   static bool cmp(pair<int,int>p1, pair<int,int>p2){
        return p2.second < p1.second;
    }
    vector<int> topKFrequent(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int>ans;
        if(n==0 || n<k)
            return ans;
        sort(nums.begin(), nums.end());  
        vector<pair<int, int>>vpr;
        int i=0;            
        while(i<n){
            int j = upper_bound(nums.begin(), nums.end(), nums[i])-nums.begin();
            // cout<<"e- "<<j<<endl;
            vpr.push_back({nums[i], j-i});
            i=j;
        }
        sort(vpr.begin(), vpr.end(), cmp);
        for(int i=0; i<k; i++){
            ans.push_back(vpr[i].first);
        }
        return ans;
    }
};
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