<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:
        # Step 1: Create a min-heap with the first 'k' elements
        min_heap = nums[:k]
        heapq.heapify(min_heap)
        
        # Step 2: Process the remaining elements
        for num in nums[k:]:
            if num > min_heap[0]:
                heapq.heappushpop(min_heap, num)  # Push the current number and pop the smallest
            
        # Step 3: The root of the min-heap is the kth largest element
        return min_heap[0]
        
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