Linked List Cycle II

Problem: Find where a linked list cycle begins.

Approach: $1.$ Fast/slow pointers find meeting point in cycle.

$2.$ Reset one to head, move both at same speed.

$3.$ They meet at cycle start.

Math: Fast travels $a + b + c$, slow travels $a + b$. Since $2(a+b) = a+b+c$, we get $a = c - b$. From meeting point, $c-b$ steps reaches cycle start, same as $a$ steps from head.

Time: $O(n)$. Space: $O(1)$.