<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 heapify(arr, n, i):
            smallest = i
            l = 2 * i + 1
            r = 2 * i + 2
            if l < n and arr[l] < arr[smallest]:
                smallest = l
            if r < n and arr[r] < arr[smallest]:
                smallest = r
            if i != smallest:
                arr[i], arr[smallest] = arr[smallest], arr[i]
                heapify(arr ,n ,smallest)
        def heapifier(arr):
            n = len(arr)
            for i in range(n//2-1, -1, -1):
                heapify(arr,n,i)
        num = nums[:k]
        heapifier(num)
        for i in range(k,len(nums)):
            if nums[i] > num[0]:
                num[0] = nums[i]
                heapify(num,k,0)
        return num[0]
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