<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:
        def quickselect(nums: List[int], k: int) -> int:
            if len(nums) == 1:
                return nums[0]  # Return the single element
            
            pivot = random.choice(nums)
            lows = [x for x in nums if x > pivot]
            highs = [x for x in nums if x < pivot]
            pivots = [x for x in nums if x == pivot]

            if k <= len(lows):
                return quickselect(lows, k)
            elif k > len(lows) + len(pivots):
                return quickselect(highs, k - len(lows) - len(pivots))
            else:
                return pivots[0]  # Return the first pivot element
        
        return quickselect(nums, k)
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