A linked list with a cycle looks like this:

• μ (mu) = number of nodes before the cycle starts (distance from head to cycle entry)
• λ (lambda) = cycle length

Example shape:

head → ... (μ nodes) ... → [CYCLE ENTRY] → ... → (λ nodes loop back)

1. Compute cycle length λ.
2. Put two pointers at head:

• p1 = head
• p2 = head

3. Move p2 ahead by λ steps.
4. Move both p1 and p2 one step at a time. They will meet at the cycle entry.

Why moving ahead by λ makes them meet at cycle start

Key invariant

At all times during step 4:

p2 is exactly λ nodes ahead of p1 along the list.

Because you advanced p2 by λ at the start, and then you move both pointers equally (1 step per iteration), the gap stays constant.

I suggest if you still didn't understand, draw this part

head → ... (μ nodes) ... → [CYCLE ENTRY] → ... → (λ nodes loop back)

This part is constant for no matter how many nodes are in μ and λ, that's why this works. Why authors couldn't clarify this though - this is the key to understanding this algorithm, right? They were pretty good at explaining previous tasks, but this one imo needs more thought put into it.