# Of Intersecting linked lists

## The problem

You are given two linked list. Basically given a node, you can only access its successor. You know that they merge at one point. Find an algorithm to detect the merging point. Optimize for computational time first, and if possible, memory.

The problem was supposed to be solved in Java, but considering there is more pythonistas reading this blog, I will rather translate everything into python.

So a node of our linked list might be implemented as its first node, also named head :

class Node:

def __init__(self,next=None):
self.next = next

def __iter__(self,):
cur = self
while cur is not None:
yield cur
cur = cur.next

def __len__(self,):
return sum(1 for el in self)

Note that accessing the n-th item of this chain requires stepping ahead n-times.

We need to write a function find_intersection that given two linked lists left and right, returns the first common node. For instance in the following example, find_intersection(L1,R1) should return L3.

## A first answer

Let’s call our two lists left and right A first very simple answer would be go through all of the nodes of left and store them in a set. We can then go through the right and stop as soon as we reach a node we have already met in left.

The python implementation is actually probably shorter and clearer than plain english :

def find_intersection_simple(left,right):
visited = set(left)
for cur_right in right:
if cur_right in visited:
return cur_right
return None

Simple isn’t it? Now let’s take a look at the complexity of this algorithm.

Python sets are basically hash sets, we can consider they have a bounded complexity for both inserting or checking if an element is within our set. All the element of the list we choose within the set. Let’s call L and R length of our two lists, and l and r the number of nodes before the intersection.

• time : O(L+r)
• memory : O(L)

Since the algorithm gives a different role to left and right, the complexity is also assymetric. If we had some clue that right is actually much shorter that left we might prefer to inverse their roles, but well, we don’t have that kind of information.

## Walking is the way to go

We can improve this a lot by advancing alternatively on the left and right leg. As soon as both legs reach the intersection, our program can terminate.

def find_intersection_walking(left,right):
visited = set()
for (cur_left, cur_right) in izip_longest(left, right):
if cur_left == cur_right:
return cur_left
if cur_left in visited:
return cur_left
if cur_right in visited:
return cur_right
if cur_left is not None:
if cur_right is not None:
return None

This is version is a huge improvement over the first implementation in the case the two list interestects very quickly. The resulting complexity is :

• time : O(max(l,r))
• memory : O(max(l,r))

## Naaaa, skipping is the way to go!

Obviously we won’t be able to improve the time complexity of this algorithm, as we will definitely need to at least reach the ancester to discover it.

But maybe we can do something for memory without impacting computational time too much.

Rather than storing all the elements we encounter, let’s consider what would happen if we were skipping an exponentially increasing number of elements.

Let’s assume the left leg reachs a node we have marked as visited by the right leg, we cannot know for sure that it is the first common node. What we know however, is that we crossed the intersection somewhere between this right leg checkpoint, and its previous checkpoint.

We need to have extra information to make it possible to step backwards to the last checkpoint. The trick is that instead of only storing visited nodes into a set, we will associate to each of these checkpoints a node to step back to.

That’s it for the right leg, but we need to step backward in the left leg as well don’t we? To be accurate we need to go back at least the number of steps between the two checkpoint of the right leg. This can be done without extra information, by going back to the last checkpoint, and going to its previous checkpoint. Since the step between checkpoints is increasing we know that we went back sufficiently.

And here is the implementation… I’m afraid it won’t help much about understanding the algorithm as it is getting hairy by the need to handle all the edge cases.

I someone find a more elegant way to implement it, I’d be happy to see it!

def steps():
c = 1
while True:
yield c
c = (c+1)*6/5

def find_intersection_skipping(left,right):
visited_left = { left: left }
visited_right = { right: right }
checkpoint_left = left
checkpoint_right = right
if left==right:
return left
walk_it = izip_longest(left, right)
(cur_left, cur_right) = walk_it.next()
for step in steps():
for skip in xrange(step):
try:
(cur_left, cur_right) = walk_it.next()
except StopIteration:
back_left = visited_left[checkpoint_left]
back_right = visited_right[checkpoint_right]
return find_intersection_skipping(
back_left,
back_right)
if cur_left == cur_right:
return cur_left
if cur_left in visited_right:
if step == 1:
return cur_left
else:
back_left = visited_left[checkpoint_left]
back_right = visited_right[cur_left]
return find_intersection_skipping(
back_left,
back_right)
if cur_right in visited_left:
if step == 1:
return cur_right
else:
back_left = visited_left[cur_right]
back_right = visited_right[checkpoint_right]
return find_intersection_skipping(back_left, back_right)
if cur_left not in visited_left:
visited_left[cur_left] = checkpoint_left.next
checkpoint_left = cur_left
if cur_right not in visited_right:
visited_right[cur_right] = checkpoint_right.next
checkpoint_right = cur_right

The time complexity of this algorithm is still linear, but we are logarithmic in memory.

• time : O(max(l,r))
• memory : O(log(max(l,r))

## A simple and efficient solution (Added on March, Thursday the 26)

Several people came up with a elegant solution and told about it on reddit and my blog comments. Thanks especially to Eric VW and also hraban who went as far as posting an implementation of the algorithm!

The principle is simple, measure the length of both list in linear time. Consume the longer list up to the difference in order to align them. We can them just walk in both of them until we meet the same element on both.

Without further due, a slightly modified version of hraban’s code.

from itertools import izip

'''Advance an iterator by a given number of steps'''
for i in xrange(n):
next(it)

def find_intersection_measure(left, right):
llen = len(left)
rlen = len(right)
li = iter(left)
ri = iter(right)
if llen > rlen:
advance(li, llen - rlen)
else:
advance(ri, rlen - llen)
for (l,r) in izip(left, right):
if l == r:
return l
return None

Its complexity of such an algorithm :

• time : O(max(L,R))
• memory : O(1)

Which is pretty neat as well.

## And then I got the best nerd burn ever…

Somebody from Twisted Oak posted on the company’s blog a simpler, better, faster, stronger solution to this puzzle. I will not post the implementation of this algorithm as he made a great job explaining it .

Somebody on reddit wondered (politely) why people (meaning me I guess) would go with such complicated solutions for a problem for which the best solution seemed so intuitive.

First of all, I had no idea of this algorithm and I had no idea that such an algorithm could exist. I might have thought about that solution if I had been thinking about it after reading people’s suggestion about measuring the list length.

But how incredible that may seem, I kind of thought that my solution couldn’t be improved. I fully understood how bold such a statement is and how stupid I may seem a posteriori, but that’s actually the painful and shameful truth! Experimenting that is actually very enlightening. It also tells a lot about how getting other people’s point of view as soon as possible is invaluable.

Now let’s talk about being intuitive . After knowing a problem’s solution, when it is elegant and clean it always seems pretty obvious… Almost shiny. But intuitive is all about having a simple path guiding your thought to the solution. The existence of such a path really depends on your background, education, culture, work experience, etc.

Today is the anniversary of the birth of a mathematician called Paul Erdos. And he happens to have a math theorem to his name which is extremely simple, and has many different proofs. My point is that the original proof is probably the one most people would come up with… But it is actually quite ugly. I would definitely call it the intuitive proof.

The best solution I know however is so elegant it will seem straight obvious when you read it.

Here is the problem :

101 people are in waiting in a line in a random order. Show that it is always possible to choose 11 of them so that if they remain in the very same order, those 11 people are sorted by size increasingly or decreasingly.