Here are a few sample lists. Try drawing them out as linked lists:

  [3, 4, 2, 3, 1, 5]
[3, 1, 2, 2]
[4, 3, 1, 1, 4]

Look for patterns. Then think about how we might use those patterns to find a duplicate number in our list.

When a value is repeated, how will that affect the structure of our linked list?

If two nodes have the same value, their next pointers will point to the same node!

So if we can find a node with two incoming next pointers, we know the position of that node is a duplicate integer in our list.

For example, if there are two 2s in our list, the node in the 2nd position will have two incoming pointers.

Alright, we’re on to something. But hold on—creating a linked list would take $O(n)$ space, and we don’t want to change our space cost from $O(1)$.

No problem—turns out we can just think of the list as a linked list, and traverse it without actually creating a new data structure.

If you're stuck on figuring out how to traverse the list like a linked list, don't sweat it too much. Just use a real linked list for now, while we finish deriving the algorithm.

Ok, so we figured out that the position of a node with multiple incoming pointers must be a duplicate. If we can find a node with multiple incoming pointers in a constant number of walks through our list, we can find a duplicate value in $O(n)$ time.

How can we find a node with multiple incoming pointers?

Let's look back at those sample lists and their corresponding linked lists, which we drew out to look for patterns:

Are there any patterns that might help us find a node with two incoming pointers?

Here’s a pattern: the last node never has any incoming pointers. This makes sense—since the list has a length $n + 1$ and all the values are $n$ or less, there can never be a pointer to the last position. If $n$ is 5, the length of the list is 6 but there can’t be a value 6 so no pointer will ever point to the 6th node. Since it has no incoming pointers, we should treat the last position in our list as the "head" (starting point) of our linked list.

Here's another pattern: there’s never an end to our list. No pointer ever points to None. Every node has a value in the range $1..n$, so every node points to another node (or to itself). Since the list goes on forever, it must have a cycle (a loop). Here are the cycles in our example lists:

Can we use these cycles to find a duplicate value?

If we walk through our linked list, starting from the head, at some point we will enter our cycle. Try tracing that path on the example lists above. Notice anything special about the first node we hit when we enter the cycle?

The first node in the cycle always has at least two incoming pointers!

We're getting close to an algorithm for finding a duplicate value. How can we find the beginning of a cycle?

Again, drawing an example is helpful here:

If we were traversing this list and wanted to know if we were inside a cycle, that would be pretty easy—we could just remember our current position and keep stepping ahead to see if we get to that position again.

But our problem is a little trickier—we need to know the first node in the cycle.

What if we knew the length of the cycle?

If we knew the length of the cycle, we could use the “stick approach” to start at the head of the list and find the first node. We use two pointers. One pointer starts at the head of the list:

Then we lay down a “stick” with the same length as the cycle, by starting the second pointer at the end. So here, for example, the second pointer is starting 4 steps ahead because the cycle has 4 steps:

Then we move the stick along the list by advancing the two pointers at the same speed (one node at a time).

When the first pointer reaches the first node in the cycle, the second pointer will have circled around exactly once. The stick wraps around the cycle, and the two ends are in the same place: the start of the cycle.

We already know where the head of our list is (the last position in the list) so we just need the length of the cycle. How can we find the length of a cycle?

If we can get inside the cycle, we can just remember a position and count how many steps it takes to get back to that position.

How can we make sure we’ve gotten inside a cycle?

Well, there has to be a cycle in our list, and at the latest, the cycle is just the last node we hit as we traverse the list from the head:

So we can just start at the head and walk $n$ steps. By then we'll have to be inside a cycle.

Alright, we’ve pieced together an entire strategy to find a duplicate integer! Working backward:

1. We know the position of a node with multiple incoming pointers is a duplicate in our list because the nodes that pointed to it must have the same value.
2. We find a node with multiple incoming pointers by finding the first node in a cycle.
3. We find the first node in a cycle by finding the length of the cycle and advancing two pointers: one starting at the head of the linked list, and the other starting ahead as many steps as there are steps in the cycle. The pointers will meet at the first node in the cycle.
4. We find the length of a cycle by remembering a position inside the cycle and counting the number of steps it takes to get back to that position.
5. We get inside a cycle by starting at the head and walking $n$ steps. We know the head of the list is at position $n + 1$.

Can you implement this? And remember—we won't want to actually create a linked list. Here's how we can traverse our list as if it were a linked list. ↴

Let's take an example list and try walking through it as if it were a linked list:

Remember that our input list is defined as having a length $n + 1$. So we know $n$ is 3 because the list has a length of 4.

The head (starting point) is the 4th node, since it has no incoming pointers. We’ll want to go from the 4th position to the 2nd position to the 1st position to the 3rd position. Or, in terms of indices in our list, we’ll want to go from index 3 to index 1 to index 0 to index 2.

Let’s get set up:

  n = 3
int_list = [3, 1, 2, 2]

# start at the head
current_position = 4

CC#C++JavaJavaScriptObjective-CPHPPython 2.7Python 3.6RubySwift

Now we need to take $n$ steps:

  for _ in range(n):
    # step ahead

CC#C++JavaJavaScriptObjective-CPHPPython 2.7Python 3.6RubySwift

On our first step, current_position is 4 and the value at the 4th position is 2, so we want to update current_position to 2. The only trick is that we need to convert our position to an index. That’s easy—we just subtract 1 (the 1st position of a list is index 0).

  for _ in range(n):

    # subtract 1 from the current position to get
    # the current index
    current_index = current_position - 1

    # take a step, updating the current position
    # to the *value* at its previous position
    current_position = int_list[current_index]

CC#C++JavaJavaScriptObjective-CPHPPython 2.7Python 3.6RubySwift

So if we’re at a current_position, the next position we want to go to is the value at the index current_position - 1. We can refactor this to 1 line and have this general way to take $n$ steps forward in our list as if it were a linked list:

  for _ in range(number_of_steps):
    current_position = int_list[current_position - 1]

CC#C++JavaJavaScriptObjective-CPHPPython 2.7Python 3.6RubySwift

To get inside a cycle (step E above), we identify $n$, start at the head (the node in position $n + 1$), and walk $n$ steps.

  def find_duplicate(int_list):
    n = len(int_list) - 1

    # STEP 1: GET INSIDE A CYCLE
    # Start at position n+1 and walk n steps to
    # find a position guaranteed to be in a cycle
    position_in_cycle = n + 1
    for _ in range(n):
        position_in_cycle = int_list[position_in_cycle - 1]

Now we’re guaranteed to be inside a cycle. To find the cycle’s length (D), we remember the current position and step ahead until we come back to that same position, counting the number of steps.

  def find_duplicate(int_list):
    n = len(int_list) - 1

    # STEP 1: GET INSIDE A CYCLE
    # Start at position n+1 and walk n steps to
    # find a position guaranteed to be in a cycle
    position_in_cycle = n + 1
    for _ in range(n):
        position_in_cycle = int_list[position_in_cycle - 1]

    # STEP 2: FIND THE LENGTH OF THE CYCLE
    # Find the length of the cycle by remembering a position in the cycle
    # and counting the steps it takes to get back to that position
    remembered_position_in_cycle = position_in_cycle
    current_position_in_cycle = int_list[position_in_cycle - 1]  # 1 step ahead
    cycle_step_count = 1

    while current_position_in_cycle != remembered_position_in_cycle:
        current_position_in_cycle = int_list[current_position_in_cycle - 1]
        cycle_step_count += 1

Now we have the head and the length of the cycle. We need to find the first node in the cycle (C). We set up 2 pointers: 1 at the head, and 1 ahead as many steps as there are nodes in the cycle. These two pointers form our "stick."

  # STEP 3: FIND THE FIRST NODE OF THE CYCLE
# Start two pointers
#   (1) at position n+1
#   (2) ahead of position n+1 as many steps as the cycle's length
pointer_start = n + 1
pointer_ahead = n + 1
for _ in range(cycle_step_count):
    pointer_ahead = int_list[pointer_ahead - 1]

Alright, we just need to find to the first node in the cycle (B), and return a duplicate value (A)!