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+---
+id: remove-linked-list-elements
+title: Remove Linked List Elements
+sidebar_label: 203-Remove-Linked-List-Elements
+tags:
+- Linked List
+- Java
+- C++
+- Python
+- Data Structures
+description: "This document provides solutions for removing all nodes from a linked list that have a specific value."
+---
+
+## Problem
+
+Given the head of a linked list and an integer `val`, remove all the nodes of the linked list that have `Node.val == val`, and return the new head.
+
+### Examples
+
+**Example 1:**
+
+**Input:** head = [1,2,6,3,4,5,6], val = 6
+
+**Output:** [1,2,3,4,5]
+
+**Example 2:**
+
+**Input:** head = [], val = 1
+
+**Output:** []
+
+**Example 3:**
+
+**Input:** head = [7,7,7,7], val = 7
+
+**Output:** []
+
+### Constraints
+
+- The number of nodes in the list is in the range `[0, 10^4]`.
+- `1 <= Node.val <= 50`
+- `0 <= val <= 50`
+
+---
+
+## Approach
+
+The approach involves iterating through the linked list and removing nodes that have the specified value. This can be achieved using a dummy node to handle edge cases where the head itself needs to be removed.
+
+### Steps:
+
+1. **Dummy Node:**
+   - Create a dummy node that points to the head of the list. This helps to easily handle the case where the head node itself needs to be removed.
+  
+2. **Iterate and Remove:**
+   - Use a pointer to iterate through the list, checking each node's value.
+   - If a node's value equals `val`, adjust the pointers to bypass this node.
+
+3. **Return New Head:**
+   - Return the next node of the dummy node as the new head of the list.
+
+## Solution for Remove Linked List Elements
+
+### Java Solution
+
+```java
+class ListNode {
+    int val;
+    ListNode next;
+    ListNode() {}
+    ListNode(int val) { this.val = val; }
+    ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+}
+
+class Solution {
+    public ListNode removeElements(ListNode head, int val) {
+        // Create a dummy node to handle edge cases
+        ListNode dummy = new ListNode(0);
+        dummy.next = head;
+        
+        // Use a pointer to iterate through the list
+        ListNode current = dummy;
+        
+        while (current.next != null) {
+            if (current.next.val == val) {
+                // Bypass the node with the value equal to val
+                current.next = current.next.next;
+            } else {
+                // Move to the next node
+                current = current.next;
+            }
+        }
+        
+        // Return the new head
+        return dummy.next;
+    }
+}
+```
+### C++ solution
+
+```cpp
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode() : val(0), next(nullptr) {}
+    ListNode(int x) : val(x), next(nullptr) {}
+    ListNode(int x, ListNode *next) : val(x), next(next) {}
+};
+
+class Solution {
+public:
+    ListNode* removeElements(ListNode* head, int val) {
+        // Create a dummy node to handle edge cases
+        ListNode* dummy = new ListNode(0);
+        dummy->next = head;
+        
+        // Use a pointer to iterate through the list
+        ListNode* current = dummy;
+        
+        while (current->next != nullptr) {
+            if (current->next->val == val) {
+                // Bypass the node with the value equal to val
+                current->next = current->next->next;
+            } else {
+                // Move to the next node
+                current = current->next;
+            }
+        }
+        
+        // Return the new head
+        return dummy->next;
+    }
+};
+```
+### Python Solution
+
+```python
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+class Solution:
+    def removeElements(self, head: ListNode, val: int) -> ListNode:
+        # Create a dummy node to handle edge cases
+        dummy = ListNode(0)
+        dummy.next = head
+        
+        # Use a pointer to iterate through the list
+        current = dummy
+        
+        while current.next is not None:
+            if current.next.val == val:
+                # Bypass the node with the value equal to val
+                current.next = current.next.next
+            else:
+                # Move to the next node
+                current = current.next
+        
+        # Return the new head
+        return dummy.next
+```
+### Complexity Analysis
+**Time Complexity:** O(n)
+
+>Reason: The algorithm involves a single pass through the linked list, resulting in a time complexity of $O(n)$, where $n$ is the number of nodes in the list.
+
+**Space Complexity:** O(1)
+>Reason: The space complexity is $O(1)$ because the algorithm uses a constant amount of extra space.
+
+### References
+**LeetCode Problem:** Remove Linked List Elements
+
+**Solution Link:** Remove Linked List Elements Solution on LeetCode
+
+
diff --git a/dsa-solutions/lc-solutions/0200-0299/0205-isomorphic-strings.md b/dsa-solutions/lc-solutions/0200-0299/0205-isomorphic-strings.md
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+---
+id: isomorphic-strings
+title: Isomorphic Strings
+sidebar_label: 205-Isomorphic-Strings
+tags:
+- Hash Table
+- String
+- Java
+- C++
+- Python
+description: "This document provides solutions for determining if two strings are isomorphic."
+---
+
+## Problem
+
+Given two strings `s` and `t`, determine if they are isomorphic.
+
+Two strings `s` and `t` are isomorphic if the characters in `s` can be replaced to get `t`.
+
+All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.
+
+### Examples
+
+**Example 1:**
+
+**Input:** s = "egg", t = "add"
+
+**Output:** true
+
+**Example 2:**
+
+**Input:** s = "foo", t = "bar"
+
+**Output:** false
+
+**Example 3:**
+
+**Input:** s = "paper", t = "title"
+
+**Output:** true
+
+### Constraints
+
+- `1 <= s.length <= 5 * 10^4`
+- `t.length == s.length`
+- `s` and `t` consist of any valid ASCII characters.
+
+---
+
+## Approach
+
+The problem can be solved by using two hash maps to track the mappings from characters in `s` to `t` and vice versa. If the mappings are consistent throughout the strings, then the strings are isomorphic.
+
+### Steps:
+
+1. **Initialization:**
+   - Create two hash maps to store the character mappings from `s` to `t` and from `t` to `s`.
+
+2. **Iterate and Map:**
+   - Iterate through the characters of both strings simultaneously.
+   - For each character pair, check if the current mapping exists and is consistent with the previous mappings.
+   - If any inconsistency is found, return false.
+
+3. **Return Result:**
+   - If the iteration completes without finding any inconsistency, return true.
+
+## Solution for Isomorphic Strings
+
+### Java Solution
+
+```java
+import java.util.HashMap;
+import java.util.Map;
+
+class Solution {
+    public boolean isIsomorphic(String s, String t) {
+        if (s.length() != t.length()) return false;
+        
+        Map<Character, Character> mapS = new HashMap<>();
+        Map<Character, Character> mapT = new HashMap<>();
+        
+        for (int i = 0; i < s.length(); i++) {
+            char c1 = s.charAt(i);
+            char c2 = t.charAt(i);
+            
+            if (mapS.containsKey(c1)) {
+                if (mapS.get(c1) != c2) return false;
+            } else {
+                mapS.put(c1, c2);
+            }
+            
+            if (mapT.containsKey(c2)) {
+                if (mapT.get(c2) != c1) return false;
+            } else {
+                mapT.put(c2, c1);
+            }
+        }
+        
+        return true;
+    }
+}
+```
+### C++ Solution
+
+```cpp
+#include <unordered_map>
+#include <string>
+
+class Solution {
+public:
+    bool isIsomorphic(std::string s, std::string t) {
+        if (s.length() != t.length()) return false;
+        
+        std::unordered_map<char, char> mapS;
+        std::unordered_map<char, char> mapT;
+        
+        for (int i = 0; i < s.length(); i++) {
+            char c1 = s[i];
+            char c2 = t[i];
+            
+            if (mapS.count(c1)) {
+                if (mapS[c1] != c2) return false;
+            } else {
+                mapS[c1] = c2;
+            }
+            
+            if (mapT.count(c2)) {
+                if (mapT[c2] != c1) return false;
+            } else {
+                mapT[c2] = c1;
+            }
+        }
+        
+        return true;
+    }
+};
+```
+### Python Solution
+
+```python
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+        
+        mapS = {}
+        mapT = {}
+        
+        for c1, c2 in zip(s, t):
+            if c1 in mapS:
+                if mapS[c1] != c2:
+                    return False
+            else:
+                mapS[c1] = c2
+                
+            if c2 in mapT:
+                if mapT[c2] != c1:
+                    return False
+            else:
+                mapT[c2] = c1
+        
+        return True
+```
+### Complexity Analysis
+**Time Complexity:** O(n)
+> Reason: The algorithm involves a single pass through both strings, resulting in a time complexity of $O(n)$, where $n$ is the length of the strings.
+
+**Space Complexity:** O(n)
+>Reason: The space complexity is $O(n)$ due to the additional hash maps used to store the character mappings.
+
+### References
+**LeetCode Problem:** Isomorphic Strings
+
+**Solution Link:** Isomorphic Strings Solution on LeetCode
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