<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:
        # nums.sort()
        # return nums[-k]
        def heapify(idx, heap_size):
            n = len(nums)
            left = 2*idx+1
            right = 2*idx+2
            largest = idx
            
            if left < heap_size and nums[left] > nums[largest]:
                largest = left
            if right < heap_size and nums[right] > nums[largest]:
                largest = right

            if largest != idx:
                nums[idx], nums[largest] = nums[largest], nums[idx]
                heapify(largest, heap_size)

        def build_heap():
            n = len(nums)
            idx = (n-2)//2
            while idx>=0:
                heapify(idx, n)
                idx-=1
            
        def extract_max(heap_size):
            maxi = nums[0]
            nums[0], nums[heap_size-1] = nums[heap_size-1], None
            heapify(0, heap_size-1)
            return maxi
    

        build_heap()
        heap_size = len(nums)
        for i in range(k):
            p = extract_max(heap_size)
            heap_size -= 1

        return p    
            


32937