## Problem Statement

Given a linked list. If the linked list contains a loop, return True and remove the loop.

A linked list contains a cycle if it consists of a node that can be reached again by continuously following the next pointer.

**Examples:****Input:**

**Output: **1 -> 2 -> 3 -> 4 -> 5**Explanation:** The linked list consists of a loop, where the last node connects to the second node. Hence, remove the loop

### Confused about your next job?

**Output: **1 -> 2

## Approach: Using HashSet

The most straightforward approach to solve this problem is to check whether a node in the linked list has been visited before. To perform this operation, a hashmap can be used. If a node has already occurred before, simply set the current pointer to **NULL**.

### Algorithm

- Initialise a hashmap.
- Traverse the linked list till the
**head**pointer isn’t**NULL:**- If the current node is already present in the hashmap, it ensures that the linked list contains a loop. Hence, set the node to
**NULL**. - Else, continue traversing and continue inserting the node into the HashSet.

- If the current node is already present in the hashmap, it ensures that the linked list contains a loop. Hence, set the node to
- If no node satisfies the above conditions, then the linked list does not contain any cycle.

### C++ Implementation

void hashAndRemove(Node * head) { unordered_map < Node * , int > node_map; Node * last = NULL; while (head != NULL) { if (node_map.find(head) == node_map.end()) { node_map[head]++; last = head; head = head -> next; } else { last -> next = NULL; break; } } }

### Java Implementation

static boolean removeLoop(Node h) { HashSet<Node> s = new HashSet<Node>(); Node prev = null; while (h != null) { if (s.contains(h)) { prev.next = null; return true; } else { s.add(h); prev = h; h = h.next; } } return false; }

### Python Implementation

def removeLoop(head): mp = set() prev = NULL while head is not None: if head in mp: prev.next = NULL return True else: mp.add(head) prev = head head = head.next return False

**Time Complexity:**O(N) where N is the number of nodes of the linked list**.****Space Complexity:**O(N), as HashSet is used

## Efficient Approach: Using Floyd’s Cycle Detection Algorithm

This approach uses a two-pointer – a **fast pointer **and a **slow pointer** to determine if there exists a cycle in the loop. The **slow pointer** moves one node ahead at a time, while the **fast pointer **moves two nodes ahead at a time.

If a loop exists in the linked list, the **fast **and **slow** pointers are bound to meet at some point.

### Algorithm

- Initialise two pointers,
**fast**and**slow**to the**head**of the linked list. - Traverse through the linked list until the
**fast**pointer doesn’t reach the end of the linked list. - If the
**fast**pointer reaches the end, the linked list doesn’t contain any cycle. Hence, return False. - Otherwise, move the slow pointer by one node i.e. slow = slow -> next and the
**fast**pointer by two nodes i.e.**fast = fast -> next -> next**. - At any point, if the
**fast**and the**slow**pointers point to the same node, set**node-> next = NULL**and return**True**as a loop has been detected.

### C++ Code

void removeCycle(Node * slow, Node * head) { for (Node * curr = head; curr != nullptr; curr = curr -> next) { Node * ptr = slow; while (ptr -> next != slow && ptr -> next != curr) { ptr = ptr -> next; } if (ptr -> next == curr) { ptr -> next = nullptr; return; } } } Node * identifyCycle(Node * head) { Node * slow = head, * fast = head; while (fast && fast -> next) { slow = slow -> next; fast = fast -> next -> next; if (slow == fast) { return slow; } } return nullptr; }

### Java Code

public static void removeCycle(Node slow, Node head) { for (Node curr = head; curr != null; curr = curr.next) { Node ptr = slow; while (ptr.next != slow && ptr.next != curr) { ptr = ptr.next; } if (ptr.next == curr) { ptr.next = null; return; } } } public static Node identifyCycle(Node head) { Node slow = head, fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) { return slow; } } return null; }

### Python Code

def removeCycle(slow, head): curr = head while curr: ptr = slow while ptr.next is not slow and ptr.next is not curr: ptr = ptr.next if ptr.next == curr: ptr.next = None return curr = curr.next def identifyCycle(head): slow = fast = head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: return slow return None

**Time Complexity:**O(N), where N is the number of nodes of the linked list.**Space Complexity:**O(1), as a map is used.

## FAQs

### Q.1: How do you detect a loop in a linked list?

**Ans.** A loop can be detected efficiently using the fast and slow pointer algorithm, where the fast pointer moves by two nodes and the slow pointer move by one node at a time. Read more here

### Q.2: Will the fast and slow pointer always meet at some point if the list contains a cycle?

**Ans**. Yes, if the linked list contains a cycle, the fast and slow pointer will always meet at some point.