Kth Largest Element

Sort everything

The direct route builds the entire leaderboard. Sort a copy in descending order, then read index k - 1.

def find_kth_largest(nums: list[int], k: int) -> int:
    ranked = sorted(nums, reverse=True)
    return ranked[k - 1]

Duplicates need no special handling. Sorting keeps both copies, so each one consumes a rank exactly as the statement requires.

Time O(n log n), space O(n) for the sorted copy.

Size-k min-heap — keep the live cutoff

You never need the full ranking. Keep only the best k values seen so far in a min-heap, a structure whose root is its smallest value.

Once the heap contains k values, that root is the cutoff for the current top k:

• A new value at or below the root can't improve the top k.
• A larger value belongs in the top k, so insert it and evict the old root.

import heapq
 
 
def find_kth_largest(nums: list[int], k: int) -> int:
    contenders: list[int] = []
    for score in nums:
        heapq.heappush(contenders, score)
        if len(contenders) > k:
            heapq.heappop(contenders)
    return contenders[0]

After every score, the heap holds the k largest values from the prefix you've read. At the end that prefix is the whole array, and the smallest of those contenders is the answer.

Time O(n log k), space O(k) for a heap that never grows beyond k + 1 entries.