Day 9 - Remove loop in Linked List.md

Difficulty	Source	Tags
Medium	160 Days of Problem Solving	Linked-List two-pointer-algorithm

🚀 Day 9. Remove loop in Linked List 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given the head of a linked list that may contain a loop. A loop exists when the last node in the linked list connects back to a node in the same list. Your task is to remove the loop from the linked list (if it exists), ensuring the linked list remains intact.

Custom Input Format:

The input consists of:

1. A head node pointing to the start of a singly linked list.
2. A pos value (1-based index) denoting the position of the node to which the last node points:
   • If pos = 0, it means the last node points to null, indicating there is no loop.

Output:

The function returns true if:

1. No loop exists in the list, or
2. The loop is successfully removed, and the linked list is restored.

Otherwise, it returns false.

🔍 Example Walkthrough:

Input:
head = 1 -> 3 -> 4, pos = 2
Output:
true
Explanation:
The linked list looks like this before removing the loop:

After removing the loop, the list becomes:

Input:
head = 1 -> 8 -> 3 -> 4, pos = 0
Output:
true
Explanation:

The linked list has no loop and remains unchanged.

Input:
head = 1 -> 2 -> 3 -> 4, pos = 1
Output:
true
Explanation:
The linked list looks like this before removing the loop:

After removing the loop, the list becomes:

Constraints:

• 1 ≤ size of linked list ≤ $10^5$

🎯 My Approach:

1. Detect Loop in the Linked List:

   • Use Floyd’s Cycle-Finding Algorithm (Tortoise and Hare) to detect a loop in the linked list.
   • If the slow and fast pointers meet, a loop is confirmed.

2. Find the Starting Node of the Loop:

   • Move one pointer back to the head and traverse both pointers (slow and fast) one step at a time.
   • The node where they meet is the starting point of the loop.

3. Break the Loop:

   • Traverse the loop to find the node whose next pointer points to the starting node of the loop.
   • Set this node’s next pointer to null, breaking the loop.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N), where N is the size of the linked list. We traverse the linked list once to detect the loop and another time to remove it.
• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of space (two pointers).

📝 Solution Code

Code (C++)

class Solution {
public:
    void removeLoop(Node* head) {
        Node *slow = head, *fast = head;
        while (fast && fast->next && (slow = slow->next) != (fast = fast->next->next));
        if (!fast || !fast->next) return;
        for (slow = head; slow != fast; slow = slow->next, fast = fast->next);
        while (fast->next != slow) fast = fast->next;
        fast->next = nullptr;
    }
};

👨‍💻 Alternative Approaches

Using Hashing

Approach:

1. Use a hash table (or a set in Python) to keep track of visited nodes.
2. Traverse the linked list:
   • If a node is revisited (i.e., it's already in the hash table), then a loop exists.
   • Break the loop by setting the next pointer of the previous node to null.
3. If the end of the list is reached without revisiting any node, there is no loop.

Drawback: This method uses additional space and is less efficient.

Complexity:

• Expected Time Complexity: O(n), as we traverse each node once and perform constant-time operations to check for presence in the hash table.
• Expected Auxiliary Space Complexity: O(n), as we use a hash table to store visited nodes.

Code (C++)

class Solution {
public:
    void removeLoop(Node* head) {
        std::unordered_set<Node*> visited;
        Node* prev = nullptr;
        while (head) {
            // If the node is already visited, break the loop
            if (visited.find(head) != visited.end()) {
                prev->next = nullptr;
                break;
            }
            visited.insert(head);  // Mark the current node as visited
            prev = head;
            head = head->next;
        }
    }
};

Code (Java)

import java.util.HashSet;

class Solution {
    public static void removeLoop(Node head) {
        HashSet<Node> visited = new HashSet<>();
        Node prev = null;

        while (head != null) {
            // If the node is already visited, break the loop
            if (visited.contains(head)) {
                prev.next = null;
                break;
            }
            visited.add(head);  // Mark the current node as visited
            prev = head;
            head = head.next;
        }
    }
}

Code (Python)

class Solution:
    def removeLoop(self, head):
        visited = set()
        prev = None

        while head:
            # If the node is already visited, break the loop
            if head in visited:
                prev.next = None
                break
            visited.add(head)  # Mark the current node as visited
            prev = head
            head = head.next

This hashing-based approach is simple to implement but less space-efficient compared to the Floyd’s cycle detection method.

Code (Java)

class Solution {
    public static void removeLoop(Node head) {
        Node slow = head, fast = head;
        while (fast != null && fast.next != null && (slow = slow.next) != (fast = fast.next.next));
        if (fast == null || fast.next == null) return;
        for (slow = head; slow != fast; slow = slow.next, fast = fast.next);
        while (fast.next != slow) fast = fast.next;
        fast.next = null;
    }
}

Code (Python)

class Solution:
    def removeLoop(self, head):
        slow, fast = head, head
        while fast and fast.next and (slow := slow.next) != (fast := fast.next.next): pass
        if not fast or not fast.next: return
        slow = head
        while slow != fast: slow, fast = slow.next, fast.next
        while fast.next != slow: fast = fast.next
        fast.next = None

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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