## IBM Interview Question

Developer Program Engineers**Team:**ISL

**Country:**India

**Interview Type:**In-Person

One way to prove it mathematically:

Assume there is a loop in the list, we have:

a = index of pointer P

b = index of pointer Q

start = index of the loop's starting point

len = loop's length (how many elements in the loop)

t = how many times we have moved pointers of P and Q since both of them are in the loop

We get:

a = start + t % len

b = start + 2t % len

So the distance between P and Q will be:

distance = | a - b | = | t % len |

When there is a loop, it is always possible that

t % len == 0

When there is not a loop, distance will never be 0.

So this approach works.

There is the no loop in your example :). I explained this, Interviewer was expecting some other answer.

let this be address of list

100->200->300->400->500->600->300(loop back to 3rd node)

let two pointers p and q

initailly p,q are @ 100

then p @200

q @300

then p @300

q @500

then p @400

q @300

then p @500

q @500

slow n fast pointer will point to node whoose

next node is making loop

slow move with x speed

fast will move with 2x speed

they will meet at point where actually loop

tend to start

but there is a assumption that head node must not

create a loop.

100->200->300->400->500->600->100(loop back at 1st node)

then above calculations fails

Algorithm is here :

Complexity : O(Length of linked list)

- sonesh March 05, 2013Prove is here :

It is easy to see that if the linked list has a loop then after some time both the pointer will be in the loop.

Consider the time when they are in the loop. Now let First pointer is at Ith node, and Second pointer is at Jth node in linked list. Lets assume that node at Ith place is faster one. So Ith node require J-I+1 move in order to reach to Jth node, and we know that First pointer is moving 1 node relative to second one.which mean it move one node nearer to second in every step. so in J-I+1 step's it will reaches to node second. Hence proved.