Problem Description

In this problem, we are provided with an immutable linked list through an interface named ImmutableListNode. We cannot directly access or modify the nodes of the linked list. We are only allowed to use two provided API functions:

• ImmutableListNode.printValue(): This method, when called on a node, prints the value of that node.
• ImmutableListNode.getNext(): This method, when called on a node, returns the next node in the linked list.

Given the head of the list, our task is to print the values of all the nodes in the linked list in reverse order, adhering strictly to using the given interface without altering the list.

Intuition

The solution to printing the values of nodes in reverse without modifying the list lies in recursion. The recursive approach allows us to visit each node once and "unwind" the calls such that we print the nodes when the recursion stack starts to collapse, effectively printing them in reverse order.

Here's the thinking process behind the recursive solution:

1. Base Case: If the current node (head) is None, it signifies that we have reached the end of the list, and we don't need to process further.

2. Recursive Case: Assuming the rest of the list can be handled and printed in reverse, we first make a recursive call with the next node head.getNext().

3. Printing Step: Once the recursive call is made, and the recursion starts unwinding, we print the current node's value using the head.printValue() method.

We recursively go deeper into the list until we hit the base case. As we return from each recursive call, we're going back from the end of the list towards the front, allowing us to print the values in reverse order by printing the value of the node after the recursive call to the rest of the list.

Learn more about Stack, Recursion, Linked List and Two Pointers patterns.

Solution Approach

The solution provided uses a simple recursive strategy to print the values of an immutable linked list in reverse. Let's delve into the implementation details:

1. Recursion: The primary algorithm employed here is recursion, which allows us to "delay" the printing of a node's value until all of its subsequent nodes have been processed (hence printing in reverse).

2. No Additional Data Structures: Since the list is immutable, we are not using any extra data structures like stacks or arrays to store values. We are simply leveraging the call stack to hold our place as we recurse through the list.

We start with:

if head:
    self.printLinkedListInReverse(head.getNext())
    head.printValue()

As seen in the code, there's an if condition to check for the base case where head is None. In linked lists, a None node typically signifies the end of the list.

When the if condition is true, indicating there are more nodes to explore, we perform two steps:

• Recursive Call: We call self.printLinkedListInReverse(head.getNext()), which moves one node forward in the list. This line is the crux of the recursive step, causing the function to dive one level deeper until it reaches the end of the list.

• Print Value: After the recursive call returns (for each call stack), the subsequent statement head.printValue() is executed. Here is where the actual printing happens. The nature of the recursion ensures that this line is called in the reverse order of the node traversal, since the deepest node (the last one) will return from the recursion first, printing its value before its predecessors.

The simplicity of the recursion obviates the need for explicitly handling complex iterative control flows or additional data manipulations, making the code both clean and efficient.

It's important to note that the recursion depth could potentially be a problem if the list is very long; a stack overflow might occur with sufficiently large input. However, for the scope of this exercise, we assume that input sizes won't exceed the limits of the system's call stack.

Example Walkthrough

Imagine we have an immutable linked list with the following values: [1, 2, 3, 4]. We need to print these values in reverse order using the recursive approach detailed above. Let's walk through the steps of the solution using this example:

Step 1: Start at the head of the list which is value 1.

• We call self.printLinkedListInReverse on the next node of 1, which is the node with value 2.

Step 2: Now our current node is 2.

• We call self.printLinkedListInReverse on the next node which is the node with value 3.

Step 3: Our current node is now 3.

• We call self.printLinkedListInReverse on the next node which is the node with value 4.

Step 4: We have reached the final node, 4.

• We call self.printLinkedListInReverse on the next node of 4, but there is no next node (it's None), so we hit our base case and do not make any further recursive calls.

Step 5: As the recursion starts to unwind, we are now back at node 4.

• We execute head.printValue() which prints 4.

Step 6: The recursion unwinds one step further, and we are back at node 3.

• We execute head.printValue() which prints 3.

Step 7: The recursion continues to unwind, and we are back at node 2.

• We execute head.printValue() which prints 2.

Step 8: Finally, the recursion completely unwinds to the first node, 1.

• We execute head.printValue() which prints 1.

Through this recursion, we've printed the nodes in the reverse order (4, 3, 2, 1) by leveraging the call stack to delay the print operations until we've encountered all nodes. This approach elegantly uses the system stack in place of an explicit stack data structure to reverse the print order.

Solution Implementation

Time and Space Complexity

The given Python code defines a recursive function printLinkedListInReverse that prints the values of nodes in an immutable linked list in reverse order. Here's an analysis of its time complexity and space complexity:

Time Complexity

The time complexity of the code is O(n), where n is the number of nodes in the linked list. This is because the function must visit each node exactly once to print its value, and the recursive calls do not overlap in terms of the number of nodes they process.

Space Complexity

The space complexity of the code is also O(n), due to the recursive nature of the solution. Each recursive call to printLinkedListInReverse adds a new frame to the call stack. Since the function recurses to the end of the list before printing and backtracking, there will be n recursive calls stacked in the call stack, with n being the number of nodes in the list.

Learn more about how to find time and space complexity quickly using problem constraints.