How do you solve Kth Largest Element in a Stream on LeetCode?

Kth Largest Element in a Stream (LeetCode #703) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log k) time complexity and O(k) space complexity. Key insight: Using a min-heap allows for efficient tracking of the kth largest element as new scores are added.

What is the time complexity of Kth Largest Element in a Stream?

The optimal solution for Kth Largest Element in a Stream runs in O(log k) time and uses O(k) space. Maintaining a min-heap of size k takes O(log k) time for each addition and O(k) space for the heap.

What is the brute force solution for Kth Largest Element in a Stream?

The brute force approach for Kth Largest Element in a Stream is Brute Force, with O(n log n) time complexity and O(n) space complexity. The simplest way to find the kth largest element is to sort the entire list of scores every time a new score is added. This approach is straightforward but inefficient for large datasets. The optimal Optimal Solution approach improves this to O(log k).

What algorithm or data structure is used to solve Kth Largest Element in a Stream?

Kth Largest Element in a Stream uses the following concepts: Tree, Design, Binary Search Tree, Heap (Priority Queue), Binary Tree, Data Stream. The recommended patterns to study are: Heap (Min-Heap for this problem), Sorting (Brute-force approach).

What are the interview tips for Kth Largest Element in a Stream?

Always consider the data structure that best fits the problem requirements. Explain your thought process clearly, especially when transitioning from a brute-force to an optimal solution. Practice writing code for both approaches to strengthen your understanding.

What are common mistakes when solving Kth Largest Element in a Stream?

Forgetting to handle the case where the heap size is less than k when adding new elements. Not using a min-heap, which leads to inefficient solutions.

#703

Kth Largest Element in a Stream

Easy

TreeDesignBinary Search TreeHeap (Priority Queue)Binary TreeData StreamHeap (Min-Heap for this problem)Sorting (Brute-force approach)

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n log n)	O(log k)
Space	O(n)	O(k)

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Intuition

Time O(log k)Space O(k)

Using a min-heap allows us to efficiently keep track of the kth largest element without needing to sort the entire list every time. We maintain a heap of size k, which gives us O(log k) time complexity for adding new scores.

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Algorithm

3 steps

1Step 1: Initialize a min-heap of size k with the first k elements from the stream.
2Step 2: For each new score, if the score is larger than the smallest in the heap, replace the smallest.
3Step 3: The root of the min-heap will always be the kth largest element.

solution.py16 lines

1import heapq
2
3class KthLargest:
4    def __init__(self, k: int, nums: List[int]):
5        self.k = k
6        self.min_heap = nums[:k]
7        heapq.heapify(self.min_heap)
8        for num in nums[k:]:
9            self.add(num)
10
11    def add(self, val: int) -> int:
12        if len(self.min_heap) < self.k:
13            heapq.heappush(self.min_heap, val)
14        elif val > self.min_heap[0]:
15            heapq.heappushpop(self.min_heap, val)
16        return self.min_heap[0]

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Complexity note: Maintaining a min-heap of size k takes O(log k) time for each addition and O(k) space for the heap.

1Using a min-heap allows for efficient tracking of the kth largest element as new scores are added.
2Maintaining a fixed-size heap ensures that we only keep the necessary elements for our kth largest calculation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.