Minor formatting fixes

Author: darpan-jain

sliding_window/minimum_window_substring.py

@@ -1,11 +1,11 @@
-'''
+"""
 Question: https://leetcode.com/problems/minimum-window-substring/
-'''
+"""
 
 class Solution:
     def minWindow(self, s: str, t: str) -> str:
-        '''
-        Sliding window with two pointers - 
+        """
+        Approach: Sliding window with two pointers - 
         
         1. Count the frequency of characters in `t`
         2. Maintain two counters - `have` & `need`
@@ -18,8 +18,8 @@ def minWindow(self, s: str, t: str) -> str:
 
         So, we move `right` pointer in `have != need` (the for loop at line 43) and `left` pointer if `have == need` (line 56)
         
-        Time complexity: O(n)
-        '''
+        Time complexity: O(N)
+        """
 
         # Edge case
         if not t: 

stack/valid_parentheses.py

@@ -1,6 +1,7 @@
-'''
+"""
 Question: https://leetcode.com/problems/valid-parentheses/
-'''
+"""
+
 
 class Solution:
     def isValid(self, s: str) -> bool:
@@ -14,9 +15,10 @@ def isValid(self, s: str) -> bool:
         # Iterate through the input string
         for b in s:
             """
-            If stack isn't empty AND we have a valid bracket AND if last brack 
-            from stack is equal to closing brack's key from dict i.e. opening, 
-            then remove that pair of brackets from the stack
+            If `stack` isn't empty AND we have a valid bracket 
+            AND if last brack from stack is equal to closing brack's key
+            from dict i.e. opening, 
+            then remove that pair of brackets from the stack.
             """
 
             # We check if it's a valid bracket and if the stack is non-empty
@@ -35,17 +37,17 @@ def isValid(self, s: str) -> bool:
             else:
                 stack.append(b)
 
-        ''' Returns True if stack is empty (checked using 'not stack') and vice-versa'''
+        """ Returns True if stack is empty (checked using 'not stack') and vice-versa """
         return True if not stack else False
 
-        ''' Alternative code for same approach '''
+        """ Alternative code for same approach """
 
         # closeToOpen = {")": "(", "]": "[", "}": "{"}
         # stack = []
 
         # for c in s:
 
-        #     # If c not in closeToOpen dict, that means it's an opening bracket
+        #     # If c not in `closeToOpen` dict, that means it's an opening bracket
         #     if c not in closeToOpen:
         #         # Add to stack and skip to next iteration
         #         stack.append(c)
@@ -64,5 +66,3 @@ def isValid(self, s: str) -> bool:
 
         # # `not stack` means stack is empty i.e. valid else False if stack is not empty
         # return True if not stack else False
-
-

trees/invert_binary_tree.py

@@ -1,31 +1,36 @@
-'''
+"""
 Question: https://leetcode.com/problems/invert-binary-tree/
-'''
+"""
+
+from typing import Optional
+
+class TreeNode:
+    """
+    Definition for a binary tree node.
+    """
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
 
-# Definition for a binary tree node.
-# class TreeNode:
-#     def __init__(self, val=0, left=None, right=None):
-#         self.val = val
-#         self.left = left
-#         self.right = right
 
 class Solution:
     def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
-        '''
-        Use Recursion to swap the children of the root. Done using DFS.
-        '''
+        """
+        Approach: Use Recursive DFS to swap the children of the root.
+        """
 
         # Base case for recursion
         if not root:
             return None
 
-        # Swap the Children
+        # Swap the Children of the current `root`
         root.left, root.right = root.right, root.left
 
         # Run invertion on left and right subtrees
         self.invertTree(root.left)
         self.invertTree(root.right)
 
-        # Return the inverted root
+        # At the end of all recursions, the final `root` will be the inverted tree.
         return root