789 Kth Largest Element In A Stream

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.

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.h = []
        self.limit = k
        for l in nums:
            heapq.heappush(self.h, l)
        
        

    def add(self, val: int) -> int:
        heapq.heappush(self.h, val)
        while len(self.h) > self.limit:
            heapq.heappop(self.h)
        return self.h[0]
        

        


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