Kth Largest Element in a Stream

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Implement KthLargest class:

• KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
• int add(int val) Appends the integer val to the stream and returns the element representing the kth largest element in the stream.

Example 1:

Input
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3); // return 4
kthLargest.add(5); // return 5
kthLargest.add(10); // return 5
kthLargest.add(9); // return 8
kthLargest.add(4); // return 8

Constraints:

• 1 <= k <= 104
• 0 <= nums.length <= 104
• -104 <= nums[i] <= 104
• -104 <= val <= 104
• At most 104 calls will be made to add.
• It is guaranteed that there will be at least k elements in the array when you search for the kth element.

SOLUTION:

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] # Your KthLargest object will be instantiated and called as such: # obj = KthLargest(k, nums) # param_1 = obj.add(val)