Kth Largest Element In A Stream

Design a class to find the k^th largest element in a stream. Note that it is the k^th largest element in the sorted order, not the k^th 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 k^th 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 k^th element.

 class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.heap = nums
        heapq.heapify(self.heap)
        self.k = k

        while len(self.heap)>k:
            heapq.heappop(self.heap)

    def add(self, val: int) -> int:

        if len(self.heap) < self.k:
            heapq.heappush(self.heap, val)
            return self.heap[0]

        if val < self.heap[0]:
            return self.heap[0]

        heapq.heapreplace(self.heap, val)        
        return self.heap[0]


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

Kth Largest Element In A Stream

Random Note