A lot of people are familiar with the problem of detecting a loop in a linked list. The problem goes as follows: “Given a linked list, what is the algorithm to determine if it has any cycles (loops)?”

The algorithm is pretty straightforward:

- We start at the beginning of the linked list with two pointers.
- The first pointer is incremented through each node of the list. The second pointer moves twice as fast, and skips every other node.
- If the linked list contains a loop, these two pointers will eventually meet at the same node, thus indicating that the linked list contains a loop.

The algorithm is straightforward and it is relatively easy to create a mental model and get an intuitive sense of why it works.

Now, a slight twist to the same question asks: “Given a circular linked list, what is the algorithm to find the first node of the loop.” For instance, in the circular list A->B->C->D->E->C, the first node of the loop is node C. The first part of the algorithm is identical to the algorithm for finding if there is a loop (above). Once a loop has been found, the following additional steps will give us the starting node of the loop:

- Once a loop as been detected (step-3 above), move one of the pointers to the beginning (head) of the linked list. The second pointer remains where it was at the end of step-3.
- Increment both pointers one node at a time. The node at which the two pointers meet will be the starting node of the loop!

This algorithm isn’t too difficult compared to the algorithm for detecting a loop. However, the mental model seems a bit trickier. Why and how does it always find the start of the loop?

### How does the Algorithm work? An intuitive explanation:

Here’s some explanation which would hopefully help you intuitively understand why the algorithm works, without going into a lot of mathematical detail.

**First, meeting point of two pointers in a loop**

Consider two pointers: a slow pointer S that increments by one node at each step, and a fast pointer F that increments by two nodes at each step (i.e. it is twice as fast as S). Both pointers start at the same time from the beginning of an n-node loop. In the time S covers n nodes. F will have covered 2n nodes and they will both meet at the start of the loop.

Now, let us say that the slow pointer S starts at the beginning of the loop, and the fast pointer F starts at node k (where k < n) of the loop. As these two pointers move along the loop, they will meet at node (n-x).

What we really need to do is figure out x, as it will give us the node at which the two pointers meet inside the loop.

- When S takes n/2 steps, it will be at node n/2. During the same time, F will have taken 2(n/2) = n steps, and it will be at node (k+n). Since the we are inside a loop, F will be effectively back at node k.
- In order for the two pointers to meet at node (n-x), S needs to take a further (n-x-n/2)=(n/2-x) steps and it will end up at node n-x. During the same time, F will have taken 2*(n/2-x)=(n-2x) steps and will be at node k+(n-2x). Given our assumption that both S and F meet at the same node:

n-x = k+n-2x => x = k

This means that if S starts from the start of the loop, and F starts k nodes into the loop, both of them will meet at node (n-k), i.e k nodes from the end of the loop. This is a key insight.

**Circular Linked List**

Now, coming back to the linked list that contains a loop. Suppose the start of the loop is m (e.g. m=3) nodes from the start of the linked list. Both S and F start at the beginning of the linked list [Figure-1].

By the time S gets to node m (i.e. start of loop), F will be at node 2m [Figure-2]. This means that S will be at the start of the loop and F will be m nodes ***into the loop***.

Based on the discussion above, we already know that if S begins from the start of the loop and F starts from node m, they will meet m nodes from the end of the loop (i.e. the orange-node in [Figure-3]).

At this point, keep the pointer F at the orange-node where the two pointers met (i.e. m-nodes from the start of the loop), and move the pointer S to the beginning of the linked list [Figure-4]. Now, if we increment both S and F *one node at a time*, it is obvious that they will meet at ‘Node-m’ (red-node) of the list, which is the start of the loop.

For the curious, here’s the Java code snippets for detecting a loop in a linked list and finding the starting node. The complete source code for my linked list project is at my Github page (https://github.com/umairsd/LinkedList-Java):

/** * Checks if the given linked list is a circular linked list (i.e. it * contains a loop). This means a list in which a node's next pointer points * to an earlier node, so as to make a loop in the linked list. For * instance: * A -> B -> C -> D -> E -> C * * @param linkedList * the linked list to be tested * @return true if there is a loop, false if there isn't */ public static boolean hasLoop(LinkedList linkedList) { if (linkedList == null || linkedList.getHead() == null) { return false; } IntegerNode slow = linkedList.getHead(); IntegerNode fast = linkedList.getHead(); while (true) { slow = slow.getNext(); if (fast.getNext() != null) { fast = fast.getNext().getNext(); } else { return false; } if (slow == null || fast == null) { return false; } if (slow == fast) { return true; } } }

/** * Returns the node at the start of a loop in the given circular linked * list. A circular list is one in which a node's next pointer points * to an earlier node, so as to make a loop in the linked list. For * instance: * * input: A -> B -> C -> D -> E -> C [the same C as earlier] * output: C * * (CCI_0205) * * @param linkedList * list to be tested * @return the node at the start of the loop if there is a loop, null * otherwise */ public static IntegerNode findLoopStart(LinkedList linkedList) { if (linkedList == null || linkedList.getHead() == null) { return null; } IntegerNode loopStartNode = null; IntegerNode slow = linkedList.getHead(); IntegerNode fast = linkedList.getHead(); while (slow != null && fast != null) { slow = slow.getNext(); if (fast.getNext() == null) { loopStartNode = null; break; } fast = fast.getNext().getNext(); // If slow and fast point to the same node, it means that the // linkedList contains a loop. if (slow == fast) { slow = linkedList.getHead(); while (slow != fast) { // Keep incrementing the two pointers until they both // meet again. When this happens, both the pointers will // point to the beginning of the loop slow = slow.getNext(); // Can't be null, as we have a loop fast = fast.getNext(); // Can't be null, as we have a loop } loopStartNode = slow; break; } } return loopStartNode; }