What is the time complexity of the optimal solution to Kth Largest Element in an Array?

The optimal solution runs in O(n log k). Sorting gives O(n log n) time and O(1) space (in-place), simple and predictable.

What pattern does Kth Largest Element in an Array use?

Kth Largest Element in an Array uses the heap pattern.

What companies ask Kth Largest Element in an Array?

Companies known to ask Kth Largest Element in an Array: Facebook, Amazon, Google, Microsoft, LinkedIn, Apple.

What are the edge cases in Kth Largest Element in an Array?

Watch for these edge cases: k equals array length; k equals 1; Array with all identical elements; Array with negative numbers; k is approximately n/2.

How would you explain Kth Largest Element in an Array in an interview?

Clarify whether duplicates count as separate elements—this affects the definition of 'kth largest'

What's the difference between the brute-force and optimal approach for Kth Largest Element in an Array?

Sorting gives O(n log n) time and O(1) space (in-place), simple and predictable. The min-heap approach achieves O(n log k) time with O(k) space—significantly faster when k << n. Quickselect offers O(n) average time with O(1) space but O(n²) worst case with adversarial inputs. For production systems, heap is often preferred for its predictable worst-case. Quickselect shines in average-case scenarios or when space is critical and worst-case is acceptable.

Problem Solution

Kth Largest Element in an Array

Medium ArrayDivide and ConquerSortingHeap (Priority Queue)Quickselect

Key Insight:You don't need to fully sort the array—maintaining just the k largest elements in a min-heap lets you efficiently track the kth largest as the heap's minimum, discarding smaller elements as you go.

Approaches

Insight Challenge

Why do we use a min-heap of size k instead of a max-heap?

Maintain only the top k candidates as we scan

Instead of sorting everything, we can maintain a collection of the k largest elements seen so far. The smallest element in this collection is our answer—if a new element is smaller than our current kth largest, it can't possibly be in the top k. A min-heap naturally surfaces the smallest element for efficient comparison and replacement.

Algorithm

Create an empty collection to hold candidates
Process each element in the array:
Add the element to our candidate collection
If we have more than k candidates, remove the smallest
The smallest remaining candidate is the kth largest overall

nums

Solution

Time O(n log k) Space O(k)

def find_kth_largest(nums, k):
    import heapq
    min_heap = []
    for num in nums:
        heapq.heappush(min_heap, num)
        if len(min_heap) > k:
            heapq.heappop(min_heap)
    return min_heap[0]