We use a Min-Heap of size k. We iterate through the array and push elements onto the heap. If the heap size exceeds k, we pop the smallest element. After processing all elements, the top of the Min-Heap will be the kth largest element.

1. Initialize an empty min-heap.
2. For each number in nums:
   • Push it to the heap.
   • If len(heap) > k, pop the smallest element.
3. Return the top of the heap.

• Time Complexity: O(N log K).
• Space Complexity: O(K).

import heapq

def find_kth_largest(nums, k):
    # Min-heap to store the k largest elements
    heap = []
    for num in nums:
        heapq.heappush(heap, num)
        if len(heap) > k:
            heapq.heappop(heap)
            
    return heap[0]