Kth Largest Element in a Stream
Created: March 9, 2019 by [lek-tin]
Last updated: March 9, 2019
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.
Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.
Example:
int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3); // returns 4
kthLargest.add(5); // returns 5
kthLargest.add(10); // returns 5
kthLargest.add(9); // returns 8
kthLargest.add(4); // returns 8
Note
You may assume that nums’ length ≥ k-1 and k ≥ 1.
Solution:
Java
class KthLargest {
private PriorityQueue<Integer> pQueue = null;
private int k = 0;
public KthLargest(int k, int[] nums) {
pQueue = new PriorityQueue<Integer>(k+1);
this.k = k;
for (int num: nums) {
add(num);
}
}
public int add(int val) {
pQueue.offer(val);
if (pQueue.size() > k) {
pQueue.poll();
}
return pQueue.peek();
}
}
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest obj = new KthLargest(k, nums);
* int param_1 = obj.add(val);
*/
Python
import queue as Q
class KthLargest:
def __init__(self, k: int, nums: List[int]):
self.heap = []
self.k = k
self.q = Q.PriorityQueue()
for num in nums:
self.add(num)
def add(self, val: int) -> int:
self.q.put(val)
if self.q.qsize() > self.k:
self.q.get()
ans = self.q.get()
self.q.put(ans)
return ans
# Your KthLargest object will be instantiated and called as such:
# obj = KthLargest(k, nums)
# param_1 = obj.add(val)