<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) {
       unordered_map<int,int> mp;
       vector<vector<int>> freq(nums.size()+1); 
       for(int num : nums){
            mp[num]++;
       }
       //after the first loop we got the pair key - value is num-freq
       //  1 - 3
       //  2 -2
       //3 - 1
       //we need to sort the value
       //to do this we need to put the value into the vector freq
        for(auto& pair : mp){
            freq[pair.second].push_back(pair.first);
        }
        //afrer this we got bunch of vector :
        // 3=[1,1,1],2=[2,2],1=[3];
        vector<int> result;//this vector store the top k freq
        for(int i=freq.size()-1 ; i>=0;i--){
            for(int j : freq[i]){
                result.push_back(j);
                if(result.size()==k){
                    return result;
                }
            }
        }
        return {};

    }

};
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