leetcode_gym/_problems/kthlargestElementInAStream.md
2026-08-25 19:14:07 -06:00

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layout title last_solved category leetcode_url status
page Kth Largest Element in a Stream heap https://leetcode.com/problems/kth-largest-element-in-a-stream Attempted

Problem

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

Solution

Sorting Solution = nlogn

{% highlight python %}

class KthLargest:

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

def add(self, val: int) -> int:
    
    self.nums.append(val)
    
    # sort
    self.nums = sorted(self.nums, reverse=True)
    
    return self.nums[self.k-1]

{% endhighlight %}

__________-

use heap -> log(n) for insertion and removal, o(1) for peeking at root

{% highlight python %}

class KthLargest:

def __init__(self, k: int, nums: List[int]):
    
    self.k = k
    # self.nums = nums
    self.nums = heapq.nlargest(self.k, nums)[::-1]
    

def add(self, val: int) -> int:
    
    # push new element onto heap
    heapq.heappush(self.nums, val)

    # print(self.nums)
    
    # if length exceeds k, pop off the smallest (make room for a larger one)
    if len(self.nums)>self.k:
        heapq.heappop(self.nums)
           

    # print(self.nums)
    # print()
            
    return self.nums[0]     # kth largest element is sitting at root (min of non evicted elements)

{% endhighlight %}

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