<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();
        vector<int> ans;

        unordered_map<int,int> mp;

        for(int i=0;i<n;i++)
        {
            mp[nums[i]]++;
        }

        priority_queue<pair<int,int> , vector<pair<int,int>> , greater<pair<int,int>>> p;

        for(auto it:mp)
        {
            int key = it.second;
            int value = it.first;
            p.push({key,value});

            if(p.size() > k)
            {
                p.pop();
            }
        }

        while(!p.empty())
        {
            int value = p.top().second;
            ans.push_back(value);
            p.pop();
        }

        return ans;

    }
};

// 1 -> 3
// 2 -> 2
// 3 -> 1 

// freq as a key
// sort the map wrt to key
// min heap of size k
17500