A linked list is a data structure in which each element in the list, known as a node, contains a value and a reference to the next node in the list. Here’s an example of a linked list:

1 -> 2 -> 3 -> 4 -> 5 -> null

In this list, each node contains a value and a reference to the next node in the list. The value of the first node is 1, the value of the second node is 2, and so on. The last node in the list contains a reference to null, indicating the end of the list.

A loop in a linked list occurs when a node in the list points to a previous node, creating a cycle in the list. For example, the following linked list has a loop:

1 -> 2 -> 3 -> 4 -> 5 -> 2 (loop to node 2)

In this list, the fifth node points to the second node, creating a cycle in the list.

The goal of the problem is to detect whether a linked list contains a loop and remove it if one is found. Removing the loop means breaking the cycle by setting the reference of the node that points to a previous node to null.Leetcode 463 Island Perimeter

To detect and remove a loop in a linked list, we can use the Floyd’s cycle detection algorithm. The algorithm uses two pointers, a slow pointer, and a fast pointer, to traverse the list. The fast pointer moves at twice the speed of the slow pointer. If there is a loop in the list, the fast pointer will eventually catch up to the slow pointer.

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Once the loop is detected, the slow pointer is reset to the head of the list, and both the slow and fast pointers move one step at a time until they meet again at the start of the loop. The fast pointer’s next node is then set to null to break the loop.

Detect and Remove a Loop in a Linked List without using Floyd’s

In this article, we will discuss an alternative method to detect and remove the loop in the list.

We can use a hash set to keep track of the nodes we’ve visited while traversing the linked list.

If we encounter a node that is already in the hash set, it means there is a loop in the list. We can then remove the loop by setting the reference of the last node in the loop to null.

Java Solution: Remove a Loop in a Linked List

public void detectAndRemoveLoop(Node head) {
    if (head == null || head.next == null) {
        return;
    }

    Set<Node> set = new HashSet<>();
    Node curr = head;
    Node prev = null;

    while (curr != null) {
        if (set.contains(curr)) {
            // found a loop, remove it
            prev.next = null;
            break;
        }

        set.add(curr);
        prev = curr;
        curr = curr.next;
    }
}

Python Solution: Remove a Loop in a Linked List

def detect_and_remove_loop(head):
    if not head or not head.next:
        return

    visited = set()
    curr = head
    prev = None

    while curr:
        if curr in visited:
            # found a loop, remove it
            prev.next = None
            break

        visited.add(curr)
        prev = curr
        curr = curr.next

Ruby Solution: Remove a Loop in a Linked List

def detect_and_remove_loop(head)
  return if head.nil? || head.next.nil?

  set = Set.new
  curr = head
  prev = nil

  while curr
    if set.include?(curr)
      # found a loop, remove it
      prev.next = nil
      break
    end

    set.add(curr)
    prev = curr
    curr = curr.next
  end
end

JavaScript Solution: Remove a Loop in a Linked List

function detectAndRemoveLoop(head) {
  if (!head || !head.next) {
    return;
  }

  const set = new Set();
  let curr = head;
  let prev = null;

  while (curr) {
    if (set.has(curr)) {
      // found a loop, remove it
      prev.next = null;
      break;
    }

    set.add(curr);
    prev = curr;
    curr = curr.next;
  }
}

These solutions use a set data structure to keep track of the nodes we’ve visited while traversing the linked list. We start by initializing an empty set and a current node pointer that starts at the head of the list.

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We then traverse the list one node at a time, checking if the current node is already in the set. If it is, we’ve found a loop in the list, and we remove it by setting the next reference of the previous node to null.

If the current node is not in the set, we add it to the set and move on to the next node. We continue this process until we reach the end of the list.

The time complexity of this solution is O(n), where n is the number of nodes in the linked list, since we need to visit each node in the list exactly once.

The space complexity of this solution is also O(n), since we need to store all visited nodes in the set.

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