Merge pull request neetcode-gh#668 from KC10201/215-Kth-Largest-Eleme…

…nt-in-an-Array-refactor-Python-solution

215 kth largest element in an array: refactor python solution

python/215-Kth-Largest-Element-in-an-Array.py

@@ -1,26 +1,48 @@
-class Solution:
+# Solution: Sorting
+# Time Complexity:
+# - Best Case: O(n)
+# - Average Case: O(n*log(n))
+# - Worst Case:O(n*log(n))
+# Extra Space Complexity: O(n)
+class Solution1:
  def findKthLargest(self, nums: List[int], k: int) -> int:
  nums.sort()
  return nums[len(nums) - k]
 
  k = len(nums) - k
 
- def quickSelect(l, r):
- if l == r:
- return nums[l]
-
- pivot, p = nums[r], l
- for i in range(l, r):
- if nums[i] <= pivot:
- nums[p], nums[i] = nums[i], nums[p]
- p += 1
- nums[p], nums[r] = nums[r], nums[p]
-
- if p > k:
- return quickSelect(l, p - 1)
- elif p < k:
- return quickSelect(p + 1, r)
+# Solution: QuickSelect
+# Time Complexity: 
+# - Best Case: O(n)
+# - Average Case: O(n)
+# - Worst Case: O(n^2)
+# Extra Space Complexity: O(1)
+class Solution2:
+ def partition(self, nums: List[int], left: int, right: int) -> int:
+ pivot, fill = nums[right], left
+
+ for i in range(left, right):
+ if nums[i] <= pivot:
+ nums[fill], nums[i] = nums[i], nums[fill]
+ fill += 1
+
+ nums[fill], nums[right] = nums[right], nums[fill]
+
+ return fill
+
+ def findKthLargest(self, nums: List[int], k: int) -> int:
+ k = len(nums) - k
+ left, right = 0, len(nums) - 1
+
+ while left < right:
+ pivot = self.partition(nums, left, right)
+
+ if pivot < k:
+ left = pivot + 1
+ elif pivot > k:
+ right = pivot - 1
  else:
- return nums[p]
+ break
 
- return quickSelect(0, len(nums) - 1)
+ return nums[k]
+