<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>largest element in the array</em>.</p>

<p>Note that it is the <code>k<sup>th</sup></code> largest element in the sorted order, not the <code>k<sup>th</sup></code> distinct element.</p>

<p>Can you solve it without sorting?</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [3,2,1,5,6,4], k = 2
<strong>Output:</strong> 5
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [3,2,3,1,2,4,5,5,6], k = 4
<strong>Output:</strong> 4
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &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>
</ul>

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        vector<int>ans;
        priority_queue<int, vector<int>, greater<int>>pq;
        for(int i=0; i<k; i++)
        {
            pq.push(nums[i]);
        }

        for(int i=k; i<nums.size(); i++)
        {
            if(pq.top()<nums[i])
            {
                pq.pop();
                pq.push(nums[i]);
            }
        }
        while(!pq.empty())
        {
            ans.push_back(pq.top());
            pq.pop();
        }
        reverse(ans.begin(), ans.end());

        return ans[k-1];

    }
};
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