Remove Duplicates from sorted list solution added

dsa-solutions/lc-solutions/0000-0099/0083-remove-duplicates-from-sorted-list.md

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+---
+id: remove-duplicates-from-sorted-list
+title: Remove Duplicates from Sorted List Solution
+sidebar_label: 0083 Remove Duplicates from Sorted List
+tags:
+  - Linked List
+  - Two Pointers
+  - LeetCode
+  - Python
+  - Java
+  - C++
+description: "This is a solution to the Remove Duplicates from Sorted List problem on LeetCode."
+---
+
+## Problem Description
+
+Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.
+
+### Examples
+
+**Example 1:**
+```
+Input: head = [1,1,2]
+Output: [1,2]
+```
+
+**Example 2:**
+
+```
+Input: head = [1,1,2,3,3]
+Output: [1,2,3]
+```
+
+### Approach for Removing Duplicates from Sorted List
+
+**Intuition:**
+The problem requires removing duplicates from a sorted singly-linked list. The approach involves iterating through the list and removing duplicates while maintaining the sorted order.
+
+**Approach:**
+- Initialize a pointer `current` to the head of the linked list to traverse the list.
+- Start a `while` loop that continues until `current` reaches the end of the list or `current.next` reaches null.
+- Inside the loop, compare the value of the current node `current.val` with the value of the next node `current.next.val`.
+- If the values are equal, it indicates a duplicate node. In this case, update the next pointer of the current node `current.next` to skip the next node (remove the duplicate).
+- If the values are not equal, move the `current` pointer to the next node, continuing the traversal.
+- Repeat the loop until the end of the list is reached, ensuring that all duplicates are removed while maintaining the sorted order of the remaining nodes.
+- After the loop, return the modified linked list, which contains no duplicates.
+
+
+### Code in Different Languages
+
+#### Java (code):
+
+```java
+class Solution {
+   public ListNode deleteDuplicates(ListNode head) {
+        ListNode current = head;
+
+        while (current != null && current.next != null) {
+            if (current.val == current.next.val) {
+                current.next = current.next.next;
+            } else {
+                current = current.next;
+            }
+        }
+
+        return head;
+    }
+}
+```
+
+#### Python (code) 
+
+```python
+class Solution:
+    def deleteDuplicates(self, head: ListNode) -> ListNode:
+        current = head
+
+        while current and current.next:
+            if current.val == current.next.val:
+                current.next = current.next.next
+            else:
+                current = current.next
+
+        return head
+```
+
+#### CPP (code) ;
+
+```cpp
+class Solution {
+public:
+    ListNode* deleteDuplicates(ListNode* head) {
+        ListNode* current = head;
+
+        while (current && current->next) {
+            if (current->val == current->next->val) {
+                current->next = current->next->next;
+            } else {
+                current = current->next;
+            }
+        }
+
+        return head;
+    }
+};
+```
+
+#### Complexity Analysis
+
+- **Time Complexity:** O(n)
+
+  - The algorithm iterates through the linked list once, where n is the number of nodes in the list. Each node is examined once to identify and remove duplicates.
+
+- **Space Complexity:** O(1)
+  - The algorithm uses a constant amount of additional memory space for variables, regardless of the size of the input linked list, making its space complexity constant.
+
+## References
+
+- **LeetCode Problem**: [Balanced Binary Tree](https://leetcode.com/problems/balanced-binary-tree/)
+- **Solution Link**: [LeetCode Solution](https://leetcode.com/problems/balanced-binary-tree/solution/)
+- **Authors GeeksforGeeks Profile:** [Vipul lakum](https://leetcode.com/u/vipul_lakum_02/)