Improved swift readmes

Author: javadev

src/main/swift/g0001_0100/s0019_remove_nth_node_from_end_of_list/readme.md

@@ -49,46 +49,32 @@ Here's the implementation:
 
 ```swift
 class Solution {
-    func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
-        
-          guard ((head?.next) != nil), n > 0 else {return nil}
-        
+    func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {        
+        guard ((head?.next) != nil), n > 0 else {return nil}        
         var count = 0
-        var current = head
-        
+        var current = head        
         while let currentNode = current{
             count += 1
             current = currentNode.next
-        }
-        
+        }        
         current = head
-        count = count - n + 1
-      
-        var prev: ListNode?
-        
-        while let currentNode = current{
-            
-            count -= 1
-            
-            if count == 0{
-            
-                if prev == nil{
+        count = count - n + 1      
+        var prev: ListNode?        
+        while let currentNode = current {            
+            count -= 1            
+            if count == 0 {            
+                if prev == nil {
                     current = current?.next
                     return current
-                }else{
+                } else {
                     prev?.next = current?.next
                 }
                 break
-            }
-            
+            }            
             prev = current
-            current = current?.next
-            
-        }
-        
-        return head
-
-        
+            current = current?.next            
+        }        
+        return head        
     }
 }
 ```

src/main/swift/g0101_0200/s0102_binary_tree_level_order_traversal/readme.md

@@ -27,76 +27,4 @@ Given the `root` of a binary tree, return _the level order traversal of its node
 **Constraints:**
 
 *   The number of nodes in the tree is in the range `[0, 2000]`.
-*   `-1000 <= Node.val <= 1000`
-
-To solve the "Binary Tree Level Order Traversal" problem in Java with a `Solution` class, we'll perform a breadth-first search (BFS) traversal of the binary tree. Below are the steps:
-
-1. **Create a `Solution` class**: Define a class named `Solution` to encapsulate our solution methods.
-
-2. **Create a `levelOrder` method**: This method takes the root node of the binary tree as input and returns the level order traversal of its nodes' values.
-
-3. **Initialize a queue**: Create a queue to store the nodes during BFS traversal.
-
-4. **Check for null root**: Check if the root is null. If it is, return an empty list.
-
-5. **Perform BFS traversal**: Enqueue the root node into the queue. While the queue is not empty:
-   - Dequeue the front node from the queue.
-   - Add the value of the dequeued node to the current level list.
-   - Enqueue the left and right children of the dequeued node if they exist.
-   - Move to the next level when all nodes in the current level are processed.
-
-6. **Return the result**: After the BFS traversal is complete, return the list containing the level order traversal of the binary tree.
-
-Here's the Java implementation:
-
-```java
-import java.util.ArrayList;
-import java.util.LinkedList;
-import java.util.List;
-import java.util.Queue;
-
-class Solution {
-    public List<List<Integer>> levelOrder(TreeNode root) {
-        List<List<Integer>> result = new ArrayList<>(); // Initialize list to store level order traversal
-        if (root == null) return result; // Check for empty tree
-        
-        Queue<TreeNode> queue = new LinkedList<>(); // Initialize queue for BFS traversal
-        queue.offer(root); // Enqueue the root node
-        
-        while (!queue.isEmpty()) {
-            int levelSize = queue.size(); // Get the number of nodes in the current level
-            List<Integer> level = new ArrayList<>(); // Initialize list for the current level
-            
-            for (int i = 0; i < levelSize; i++) {
-                TreeNode node = queue.poll(); // Dequeue the front node
-                level.add(node.val); // Add node value to the current level list
-                
-                // Enqueue the left and right children if they exist
-                if (node.left != null) queue.offer(node.left);
-                if (node.right != null) queue.offer(node.right);
-            }
-            
-            result.add(level); // Add the current level list to the result list
-        }
-        
-        return result; // Return the level order traversal
-    }
-    
-    // Definition for a TreeNode
-    public class TreeNode {
-        int val;
-        TreeNode left;
-        TreeNode right;
-        
-        TreeNode() {}
-        TreeNode(int val) { this.val = val; }
-        TreeNode(int val, TreeNode left, TreeNode right) {
-            this.val = val;
-            this.left = left;
-            this.right = right;
-        }
-    }
-}
-```
-
-This implementation follows the steps outlined above and efficiently computes the level order traversal of the binary tree in Java using BFS.
+*   `-1000 <= Node.val <= 1000`

src/main/swift/g0101_0200/s0104_maximum_depth_of_binary_tree/readme.md

@@ -35,46 +35,4 @@ A binary tree's **maximum depth** is the number of nodes along the longest path
 **Constraints:**
 
 *   The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.
-*   `-100 <= Node.val <= 100`
-
-To solve the "Maximum Depth of Binary Tree" problem in Java with a `Solution` class, we'll perform a depth-first search (DFS) traversal of the binary tree. Below are the steps:
-
-1. **Create a `Solution` class**: Define a class named `Solution` to encapsulate our solution methods.
-
-2. **Create a `maxDepth` method**: This method takes the root node of the binary tree as input and returns its maximum depth.
-
-3. **Check for null root**: Check if the root is null. If it is, return 0 as the depth.
-
-4. **Perform DFS traversal**: Recursively compute the depth of the left and right subtrees. The maximum depth of the binary tree is the maximum depth of its left and right subtrees, plus 1 for the current node.
-
-5. **Return the result**: After the DFS traversal is complete, return the maximum depth of the binary tree.
-
-Here's the Java implementation:
-
-```java
-class Solution {
-    public int maxDepth(TreeNode root) {
-        if (root == null) return 0; // Check for empty tree
-        int leftDepth = maxDepth(root.left); // Compute depth of left subtree
-        int rightDepth = maxDepth(root.right); // Compute depth of right subtree
-        return Math.max(leftDepth, rightDepth) + 1; // Return maximum depth of left and right subtrees, plus 1 for the current node
-    }
-    
-    // Definition for a TreeNode
-    public class TreeNode {
-        int val;
-        TreeNode left;
-        TreeNode right;
-        
-        TreeNode() {}
-        TreeNode(int val) { this.val = val; }
-        TreeNode(int val, TreeNode left, TreeNode right) {
-            this.val = val;
-            this.left = left;
-            this.right = right;
-        }
-    }
-}
-```
-
-This implementation follows the steps outlined above and efficiently computes the maximum depth of the binary tree in Java using DFS traversal.
+*   `-100 <= Node.val <= 100`

src/main/swift/g0101_0200/s0105_construct_binary_tree_from_preorder_and_inorder_traversal/readme.md

@@ -26,80 +26,4 @@ Given two integer arrays `preorder` and `inorder` where `preorder` is the preord
 *   `preorder` and `inorder` consist of **unique** values.
 *   Each value of `inorder` also appears in `preorder`.
 *   `preorder` is **guaranteed** to be the preorder traversal of the tree.
-*   `inorder` is **guaranteed** to be the inorder traversal of the tree.
-
-To solve the "Construct Binary Tree from Preorder and Inorder Traversal" problem in Java with a `Solution` class, we'll use a recursive approach. Below are the steps:
-
-1. **Create a `Solution` class**: Define a class named `Solution` to encapsulate our solution methods.
-
-2. **Create a `buildTree` method**: This method takes two integer arrays, `preorder` and `inorder`, as input and returns the constructed binary tree.
-
-3. **Check for empty arrays**: Check if either of the arrays `preorder` or `inorder` is empty. If so, return null, as there's no tree to construct.
-
-4. **Define a helper method**: Define a recursive helper method `build` to construct the binary tree.
-   - The method should take the indices representing the current subtree in both `preorder` and `inorder`.
-   - The start and end indices in `preorder` represent the current subtree's preorder traversal.
-   - The start and end indices in `inorder` represent the current subtree's inorder traversal.
-   
-5. **Base case**: If the start index of `preorder` is greater than the end index or if the start index of `inorder` is greater than the end index, return null.
-
-6. **Find the root node**: The root node is the first element in the `preorder` array.
-
-7. **Find the root's position in `inorder`**: Iterate through the `inorder` array to find the root's position.
-
-8. **Recursively build left and right subtrees**: 
-   - Recursively call the `build` method for the left subtree with updated indices.
-   - Recursively call the `build` method for the right subtree with updated indices.
-   
-9. **Return the root node**: After constructing the left and right subtrees, return the root node.
-
-Here's the Java implementation:
-
-```java
-class Solution {
-    public TreeNode buildTree(int[] preorder, int[] inorder) {
-        if (preorder.length == 0 || inorder.length == 0) return null; // Check for empty arrays
-        return build(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1); // Construct binary tree
-    }
-    
-    // Recursive helper method to construct binary tree
-    private TreeNode build(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) {
-        if (preStart > preEnd || inStart > inEnd) return null; // Base case
-        
-        int rootValue = preorder[preStart]; // Root node value
-        TreeNode root = new TreeNode(rootValue); // Create root node
-        
-        // Find root node's position in inorder array
-        int rootIndex = 0;
-        for (int i = inStart; i <= inEnd; i++) {
-            if (inorder[i] == rootValue) {
-                rootIndex = i;
-                break;
-            }
-        }
-        
-        // Recursively build left and right subtrees
-        root.left = build(preorder, inorder, preStart + 1, preStart + rootIndex - inStart, inStart, rootIndex - 1);
-        root.right = build(preorder, inorder, preStart + rootIndex - inStart + 1, preEnd, rootIndex + 1, inEnd);
-        
-        return root; // Return root node
-    }
-    
-    // TreeNode definition
-    public class TreeNode {
-        int val;
-        TreeNode left;
-        TreeNode right;
-        
-        TreeNode() {}
-        TreeNode(int val) { this.val = val; }
-        TreeNode(int val, TreeNode left, TreeNode right) {
-            this.val = val;
-            this.left = left;
-            this.right = right;
-        }
-    }
-}
-```
-
-This implementation follows the steps outlined above and efficiently constructs the binary tree from preorder and inorder traversals in Java.
+*   `inorder` is **guaranteed** to be the inorder traversal of the tree.

src/main/swift/g0101_0200/s0114_flatten_binary_tree_to_linked_list/readme.md

@@ -32,81 +32,4 @@ Given the `root` of a binary tree, flatten the tree into a "linked list":
 *   The number of nodes in the tree is in the range `[0, 2000]`.
 *   `-100 <= Node.val <= 100`
 
-**Follow up:** Can you flatten the tree in-place (with `O(1)` extra space)?
-
-To solve the "Flatten Binary Tree to Linked List" problem in Java with a `Solution` class, we'll use a recursive approach. Below are the steps:
-
-1. **Create a `Solution` class**: Define a class named `Solution` to encapsulate our solution methods.
-
-2. **Create a `flatten` method**: This method takes the root node of the binary tree as input and flattens the tree into a linked list using preorder traversal.
-
-3. **Check for null root**: Check if the root is null. If so, there's no tree to flatten, so return.
-
-4. **Recursively flatten the tree**: Define a recursive helper method `flattenTree` to perform the flattening.
-   - The method should take the current node as input.
-   - Perform a preorder traversal of the tree.
-   - For each node, if it has a left child:
-     - Find the rightmost node in the left subtree.
-     - Attach the right subtree of the current node to the right of the rightmost node.
-     - Move the left subtree to the right subtree position.
-     - Set the left child of the current node to null.
-   - Recursively call the method for the right child.
-
-5. **Call the helper method**: Call the `flattenTree` method with the root node.
-
-Here's the Java implementation:
-
-```java
-class Solution {
-    public void flatten(TreeNode root) {
-        if (root == null) return; // Check for empty tree
-        flattenTree(root); // Flatten the tree
-    }
-    
-    // Recursive helper method to flatten the tree
-    private void flattenTree(TreeNode node) {
-        if (node == null) return;
-        
-        // Flatten left subtree
-        flattenTree(node.left);
-        
-        // Flatten right subtree
-        flattenTree(node.right);
-        
-        // Save right subtree
-        TreeNode rightSubtree = node.right;
-        
-        // Attach left subtree to the right of the current node
-        node.right = node.left;
-        
-        // Set left child to null
-        node.left = null;
-        
-        // Move to the rightmost node of the flattened left subtree
-        TreeNode current = node;
-        while (current.right != null) {
-            current = current.right;
-        }
-        
-        // Attach the saved right subtree to the right of the rightmost node
-        current.right = rightSubtree;
-    }
-    
-    // TreeNode definition
-    public class TreeNode {
-        int val;
-        TreeNode left;
-        TreeNode right;
-        
-        TreeNode() {}
-        TreeNode(int val) { this.val = val; }
-        TreeNode(int val, TreeNode left, TreeNode right) {
-            this.val = val;
-            this.left = left;
-            this.right = right;
-        }
-    }
-}
-```
-
-This implementation follows the steps outlined above and efficiently flattens the binary tree into a linked list using preorder traversal in Java.
+**Follow up:** Can you flatten the tree in-place (with `O(1)` extra space)?