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Solution 6: Perpetual Ranker

Approach Explanation

We maintain a Min-Heap of size k. The top of the heap is always the kth largest element in the stream.

Step-by-Step Logic

  1. __init__: Heapify the given nums and pop elements until the size is k.
  2. add(val):
    • Push val to heap.
    • If size > k, pop.
    • Return heap[0].

Complexity

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

Code

import heapq

class KthLargest:
    def __init__(self, k: int, nums: list[int]):
        self.k = k
        self.heap = nums
        heapq.heapify(self.heap)
        
        while len(self.heap) > k:
            heapq.heappop(self.heap)

    def add(self, val: int) -> int:
        heapq.heappush(self.heap, val)
        if len(self.heap) > self.k:
            heapq.heappop(self.heap)
        return self.heap[0]