---
title: '215. Kth Largest Element in an Array'
description: Given an integer array nums and an integer k, return the k^th largest element in the array
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
  - { name: "Sorting", slug: "sorting" }
  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
  - { name: "Quickselect", slug: "quickselect" }
video: "XEmy13g1Qxc"
sidebar:
  label: 'Kth Largest Element in an Array'
  badge: 'Medium'
---

### Example 1:
- Input: `nums = [3,2,1,5,6,4], k = 2`
- Output: `5`

### Example 2:
- Input: `nums = [3,2,3,1,2,4,5,5,6], k = 4`
- Output: `4`

### Constraints:

- `1 <= k <= nums.length <= 10^5`
- `-10^4 <= nums[i] <= 10^4`

## Approach

```mermaid
flowchart TD
  S(["findKthLargest(nums, k)"]) --> F{"more i in range(len(nums))?"}
  F -- yes --> N["nums[i] = -nums[i] — negate so the min-heap acts as a max-heap"]
  N --> F
  F -- no --> H["heapq.heapify(nums)"]
  H --> W{"more _ in range(k - 1)?"}
  W -- yes --> P["heapq.heappop(nums) — discard the k-1 largest"]
  P --> W
  W -- no --> E(["return -heapq.heappop(nums)"])
```

## Solution

```py
import heapq


class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        for i in range(len(nums)):
            nums[i] = -nums[i]

        heapq.heapify(nums)

        for _ in range(k - 1):
            heapq.heappop(nums)

        return -heapq.heappop(nums)
```