Added solution for leetcode 149

dsa-solutions/lc-solutions/0100-0199/0141-linked-list-cycle.md

@@ -0,0 +1,157 @@
+---
+id: linked-list-cycle
+title: Linked List Cycle
+sidebar_label: 0141- Linked List Cycle
+tags:
+  - DSA
+  - Leetcode
+  - Linked List
+
+description: "This is a solution to the Linked List cycle  on LeetCode."
+---
+
+## Problem Statement
+
+Given head, the head of a linked list, determine if the linked list has a cycle in it.
+
+There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.
+
+Return true if there is a cycle in the linked list. Otherwise, return false.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: head = [3,2,0,-4], pos = 1
+Output: true
+Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).
+```
+
+**Example 2:**
+
+```
+Input: head = [1,2], pos = 0
+Output: true
+Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.
+```
+
+**Example 3:**
+
+```
+Input: head = [1], pos = -1
+Output: false
+Explanation: There is no cycle in the linked list.
+```
+
+### Constraints:
+
+- The number of the nodes in the list is in the range $[0, 10^4].$
+- $-105 <= Node.val <= 105$
+- $pos is -1 or a valid index in the linked-list.$
+
+## Algorithm
+
+1. **Initialization**:
+
+   - Initialize two pointers, `slow` and `fast`, both pointing to the head of the linked list.
+
+2. **Traversal**:
+
+   - Move the `slow` pointer one step at a time.
+   - Move the `fast` pointer two steps at a time.
+
+3. **Cycle Detection**:
+
+   - If there is a cycle, the `fast` pointer will eventually meet the `slow` pointer within the cycle.
+   - If there is no cycle, the `fast` pointer will reach the end of the list (NULL).
+
+4. **Return Result**:
+   - If the `fast` pointer meets the `slow` pointer, return `true` indicating a cycle is detected.
+   - If the `fast` pointer reaches the end of the list, return `false` indicating no cycle is detected.
+
+## Code Implementation
+
+### C++
+
+```cpp
+class Solution {
+public:
+    bool hasCycle(ListNode *head) {
+        ListNode* fast = head;
+        ListNode* slow = head;
+
+        while (fast != NULL && fast->next != NULL) {
+            fast = fast->next->next;
+            slow = slow->next;
+            if (fast == slow) {
+                return true;
+            }
+        }
+
+        return false;
+    }
+};
+```
+
+### Python
+
+```python
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution:
+    def hasCycle(self, head: ListNode) -> bool:
+        fast = head
+        slow = head
+
+        while fast is not None and fast.next is not None:
+            fast = fast.next.next
+            slow = slow.next
+            if fast == slow:
+                return True
+
+        return False
+```
+
+### Java
+
+```java
+public class Solution {
+    public boolean hasCycle(ListNode head) {
+        ListNode fast = head;
+        ListNode slow = head;
+
+        while (fast != null && fast.next != null) {
+            fast = fast.next.next;
+            slow = slow.next;
+            if (fast == slow) {
+                return true;
+            }
+        }
+
+        return false;
+    }
+}
+```
+
+### JavaScript
+
+```javascript
+var hasCycle = function (head) {
+  let fast = head;
+  let slow = head;
+
+  while (fast !== null && fast.next !== null) {
+    fast = fast.next.next;
+    slow = slow.next;
+    if (fast === slow) {
+      return true;
+    }
+  }
+
+  return false;
+};
+```

dsa-solutions/lc-solutions/0100-0199/0142-linked-list-cycle-II.md

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+---
+id: linked-list-cycle--II
+title: Linked List Cycle II
+sidebar_label: 0142- Linked List Cycle II
+tags:
+  - DSA
+  - Leetcode
+  - Linked List
+
+description: "This is a solution to the Linked List cycle II on LeetCode."
+---
+
+## Problem Statement
+
+Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.
+
+There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.
+
+Do not modify the linked list.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: head = [3,2,0,-4], pos = 1
+Output: tail connects to node index 1
+Explanation: There is a cycle in the linked list, where tail connects to the second node.
+```
+
+**Example 2:**
+
+```
+Input: head = [1,2], pos = 0
+Output: tail connects to node index 0
+Explanation: There is a cycle in the linked list, where tail connects to the first node.
+```
+
+**Example 3:**
+
+```
+Input: head = [1], pos = -1
+Output: no cycle
+Explanation: There is no cycle in the linked list.
+```
+
+### Constraints:
+
+- The number of the nodes in the list is in the range $[0, 10^4].$
+- $-10^5 <= Node.val <= 10^5$
+- pos is `-1` or a valid index in the linked-list.
+
+## Algorithm for Detecting the Start of a Cycle in a Linked List
+
+1. **Initialization**:
+
+   - Initialize two pointers, `slow` and `fast`, both pointing to the head of the linked list.
+
+2. **Cycle Detection**:
+
+   - Move the `slow` pointer one step at a time.
+   - Move the `fast` pointer two steps at a time.
+   - If there is a cycle, the `fast` pointer will eventually meet the `slow` pointer within the cycle.
+
+3. **Finding the Cycle Start**:
+
+   - If the `fast` pointer meets the `slow` pointer, reset the `slow` pointer to the head of the list.
+   - Move both pointers one step at a time until they meet again.
+   - The node where they meet is the start of the cycle.
+
+4. **Return Result**:
+   - If the `fast` pointer meets the `slow` pointer, return the meeting node.
+   - If there is no cycle, return `NULL`.
+
+## Code Implementations
+
+### C++
+
+```cpp
+class Solution {
+public:
+    ListNode *detectCycle(ListNode *head) {
+        ListNode* fast = head;
+        ListNode* slow = head;
+
+        while (fast != NULL && fast->next != NULL) {
+            slow = slow->next;
+            fast = fast->next->next;
+            if (slow == fast) {
+                slow = head;
+                while (slow != fast) {
+                    slow = slow->next;
+                    fast = fast->next;
+                }
+                return slow;
+            }
+        }
+        return NULL;
+    }
+};
+```
+
+### Python
+
+```python
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution:
+    def detectCycle(self, head: ListNode) -> ListNode:
+        fast = head
+        slow = head
+
+        while fast is not None and fast.next is not None:
+            slow = slow.next
+            fast = fast.next.next
+            if slow == fast:
+                slow = head
+                while slow != fast:
+                    slow = slow.next
+                    fast = fast.next
+                return slow
+        return None
+```
+
+### Java
+
+```java
+public class Solution {
+    public ListNode detectCycle(ListNode head) {
+        ListNode fast = head;
+        ListNode slow = head;
+
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+            if (slow == fast) {
+                slow = head;
+                while (slow != fast) {
+                    slow = slow.next;
+                    fast = fast.next;
+                }
+                return slow;
+            }
+        }
+        return null;
+    }
+}
+```
+
+### JavaScript
+
+```javascript
+var detectCycle = function (head) {
+  let fast = head;
+  let slow = head;
+
+  while (fast !== null && fast.next !== null) {
+    slow = slow.next;
+    fast = fast.next.next;
+    if (slow === fast) {
+      slow = head;
+      while (slow !== fast) {
+        slow = slow.next;
+        fast = fast.next;
+      }
+      return slow;
+    }
+  }
+  return null;
+};
+```