<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:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        int n=nums.size();
        map<int,int>mp;
        for(int i=0;i<nums.size();i++){
            mp[nums[i]]++;
        }
        vector<vector<int>>bucket(n+1);
        for(auto it:mp){
            int val=it.first;
            int freq=it.second;
            bucket[freq].push_back(val);
        }
        vector<int>ans;
        for(int i=n;i>=0&&ans.size()<k;i--){
            if(!bucket[i].empty()){
                for(int j=0;j<bucket[i].size();j++){
                      ans.push_back(bucket[i][j]);
                      if(ans.size()==k){
                        return ans;
                      }
                }
            }
        }
    return ans;
    }
};
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