# 0kth-largest-element-in-an-array


Try it on <a href='https://leetcode.com/problems/0kth-largest-element-in-an-array'>leetcode</a>

## Description
<div class="description">
<div><p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>largest element in the array</em>.</p>

<p>Note that it is the <code>k<sup>th</sup></code> largest element in the sorted order, not the <code>k<sup>th</sup></code> distinct element.</p>

<p>Can you solve it without sorting?</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [3,2,1,5,6,4], k = 2
<strong>Output:</strong> 5
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [3,2,3,1,2,4,5,5,6], k = 4
<strong>Output:</strong> 4
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        return self.HeapMethod(nums, k)

    # Sorting method
    # Time Complexity: O(nlogn)
    # Space Complexity: O(n)

    def sorting(self, nums: List[int], k: int) -> int:
        nums.sort(reverse=true)
        return nums[k - 1]

    # Heap Method
    # Time Complexity : O(nlogk)
    # Space Complexity: O(k)
    def HeapMethod(self, nums: List[int], k: int) -> int:
        heap = []
        for num in nums:
            heapq.heappush(heap, num)
            if len(heap) > k:
                heapq.heappop(heap)
        return heap[0]

```