diff --git a/python/215-Kth-Largest-Element-in-an-Array.py b/python/215-Kth-Largest-Element-in-an-Array.py
index 910a1db38..d5cb796c1 100644
--- a/python/215-Kth-Largest-Element-in-an-Array.py
+++ b/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]
+