Delete Head / Tail / Nth Node

Author: Kondareddy1209

Leetcode/Delete_Head/Tail/Nth_Node.py

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+"""
+Delete a Node at a Given Position in a Singly Linked List
+
+Problem:
+Given the head of a singly linked list and a position `x` (1-based indexing),
+delete the node present at that position and return the updated head.
+
+Example:
+Input:
+    10 -> 20 -> 30 -> 40
+    x = 3
+
+Output:
+    10 -> 20 -> 40
+
+Approach:
+1. Handle the empty linked list.
+2. If x == 1, delete the head node by returning head.next.
+3. Traverse the linked list until reaching the x-th node.
+4. Keep track of:
+   - `temp` : current node
+   - `prev` : node before the current node
+5. Change the previous node's next pointer to skip the node being deleted.
+
+Visualization:
+
+Before deletion (x = 3)
+
+head
+ │
+ ▼
+10 ───► 20 ───► 30 ───► 40 ───► None
+         ▲        ▲
+       prev     temp
+
+After:
+prev.next = temp.next
+
+head
+ │
+ ▼
+10 ───► 20 ─────────► 40 ───► None
+           (30 is disconnected)
+
+How the loop works:
+
+Initial:
+index = 1
+temp = 10
+prev = None
+
+Iteration 1:
+-----------
+prev = 10
+temp = 20
+index = 2
+
+Iteration 2:
+-----------
+prev = 20
+temp = 30
+index = 3
+
+Loop stops because index == x.
+
+Deletion:
+---------
+prev.next = temp.next
+
+20.next = 40
+
+Result:
+10 -> 20 -> 40
+
+Time Complexity:
+----------------
+O(n)
+- In the worst case, we traverse the linked list once.
+
+Space Complexity:
+-----------------
+O(1)
+- Only a few pointer variables are used.
+"""
+
+
+class Node:
+    def __init__(self, data):
+        self.data = data
+        self.next = None
+
+
+class Solution:
+    def deleteNode(self, head, x):
+        # Empty linked list
+        if head is None:
+            return None
+
+        # Delete the first node
+        if x == 1:
+            return head.next
+
+        index = 1
+        prev = None
+        temp = head
+
+        # Traverse until the x-th node
+        while temp and index < x:
+            prev = temp
+            temp = temp.next
+            index += 1
+
+        # If x is greater than the number of nodes
+        if temp is None:
+            return head
+
+        # Delete the node by skipping it
+        prev.next = temp.next
+
+        return head