<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:
    def findKthLargest(self, nums: List[int], k: int) -> int:

        maxx = max(nums)
        minn = min(nums)

        pivot = (maxx + minn) // 2

        less_than = list()
        middle = list()
        greater_than = list()

        for num in nums:
            if num < pivot:
                less_than.append(num)
            elif num == pivot:
                middle.append(num)
            else:
                greater_than.append(num)
        # print(pivot)
        # print("l", less_than)
        # print("m", middle)
        # print("g", greater_than)
        if k <= len(greater_than):
            return self.findKthLargest(greater_than, k)
            
        if k > len(greater_than) and k <= len(greater_than) + len(middle):
            return pivot
        else:
            return self.findKthLargest(less_than, k - len(middle) - len(greater_than))
        
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