## Kth Largest Element in an Array

###### DESCRIPTION (credit Leetcode.com)

Write a function that takes an array of unsorted integers nums and an integer k, and returns the kth largest element in the array. This function should run in O(n log k) time, where n is the length of the array.

###### EXAMPLES

**Example 1:**

Inputs:

`nums = [5, 3, 2, 1, 4] k = 2`

Output:

`4`

Run your code to see results here

## Explanation

### Approach 1: Sorting

The simplest approach is to sort the array in descending order and return the kth element. This approach has a time complexity of O(n log n) where n is the number of elements in the array, and a space complexity of O(1).

### Approach 2: Min Heap

By using a **min-heap**, we can reduce the time complexity to O(n log k), where n is the number of elements in the array and k is the value of k.

The idea behind this solution is to iterate over the elements in the array while storing the k largest elements we've seen so far in a min-heap. At each element, we check if it is greater than the smallest element (the root) of the heap. If it is, we pop the smallest element from the heap and push the current element into the heap. This way, the heap will always contain the k largest elements we've seen so far.

After iterating over all the elements, the root of the heap will be the kth largest element in the array.

## Solution

import heapqdef kth_largest(nums, k):if not nums:returnheap = []for num in nums:if len(heap) < k:heapq.heappush(heap, num)elif num > heap[0]:heapq.heappushpop(heap, num)return heap[0]

## Complexity Analysis

**Time Complexity:** O(n log k). We iterate over all the elements in the array. Comparing the current element with the smallest element in the heap takes O(1) time. In the worst case, we both push and pop each element from the heap, which takes O(log k) time.

**Space Complexity:** O(k). The space used by the heap to store the k largest elements.

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