<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>

import heapq
class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:

        return self.quick_select(nums, k)


        # quick sort / quick select
        
        # we could sort it and return the k-th postion
        # we could use queue to maintain the k q, if k 
        # quick select and find the k the with binary search 
        # h = []
        # for num in nums:
        #     if len(h) < k:
        #         heapq.heappush(h, num)
        #     else:
        #         heapq.heappush(h, max(heapq.heappop(h), num))
        
        # return h[0]

        # binary search 
        
    def quick_select(self, nums, k):
        pivot = random.choice(nums)
        left, mid, right = [], [], []
        for num in nums:
            if num > pivot:
                left.append(num)
            elif num < pivot:
                right.append(num)
            else:
                mid.append(num)
        
        if len(left) >= k:
            return self.quick_select(left, k)
        elif len(left) + len(mid) >= k:
            return pivot
        else:
            return self.quick_select(right, k - len(left) - len(mid))

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