Design Browser History Leetcode Problem 1472 [Python Solution]

In today's exploration of LeetCode problems, we're going to dive into the Design Browser History problem.

This problem falls under the category of Linked Lists and is considered of medium difficulty.

The task involves simulating the behavior of navigating through browser pages.

You'll start on a homepage and have the ability to visit other URLs, move back in the history a certain number of steps, and move forward in the history.

It's a fascinating problem that allows you to explore the implementation of a browser's history functionality.

Problem Overview

Let's start by understanding the problem and its requirements.

You are tasked with implementing the BrowserHistory class, which has the following methods:

1. BrowserHistory(string homepage): Initializes the object with the homepage of the browser.
2. void visit(string url): Visits the given URL from the current page.

It clears all forward history.
3. string back(int steps): Moves back in history by the specified number of steps.

If you can only return x steps in the history and steps is greater than x, you return only x steps.

It returns the current URL after moving back in history.
4. string forward(int steps): Moves forward in history by the specified number of steps.

If you can only forward x steps in the history and steps is greater than x, you forward only x steps.

It returns the current URL after moving forward in history.

Example:

Let's consider an example to illustrate the problem:

Input:
["BrowserHistory", "visit", "visit", "visit", "back", "back", "forward", "visit", "forward", "back", "back"]
[["leetcode.com"], ["google.com"], ["facebook.com"], ["youtube.com"], [1], [1], [1], ["linkedin.com"], [2], [2], [7]]

Output:
[null, null, null, null, "facebook.com", "google.com", "facebook.com", null, "linkedin.com", "google.com", "leetcode.com"]

Here's how the operations play out:

• Initialize the browser history with the homepage "leetcode.com."
• Visit "google.com."
• Visit "facebook.com."
• Visit "youtube.com."
• Move back one step to "facebook.com."
• Move back one more step to "google.com."
• Move forward one step to "facebook.com."
• Visit "linkedin.com."
• Move forward two steps (but we can only move one step forward), ending up at "linkedin.com."
• Move back two steps to "facebook.com" and then to "google.com."
• Attempt to move back seven steps, but we can only move back one step, landing at "leetcode.com."

Constraints:

• 1 <= homepage.length <= 20
• 1 <= url.length <= 20
• 1 <= steps <= 100
• homepage and url consist of '.' or lowercase English letters.
• At most 5000 calls will be made to visit, back, and forward.

Now that we have a clear understanding of the problem, let's explore two different solutions to tackle it.

Efficient Python Code Solution:

Linked List Implementation

First, we'll explore a solution using a doubly linked list.

This approach provides an efficient solution to the problem.

class ListNode:
    def __init__(self, val, prev=None, next=None):
        self.val = val
        self.prev = prev
        self.next = next

class BrowserHistory:
    def __init__(self, homepage: str):
        self.cur = ListNode(homepage)

    # Visit a URL
    # Time complexity: `O(1)`
    def visit(self, url: str) -> None:
        self.cur.next = ListNode(url, self.cur)
        self.cur = self.cur.next

    # Move back in history
    # Time complexity: `O(n)`
    def back(self, steps: int) -> str:
        while self.cur.prev and steps > 0:
            self.cur = self.cur.prev
            steps -= 1
        return self.cur.val

    # Move forward in history
    # Time complexity: `O(n)`
    def forward(self, steps: int) -> str:
        while self.cur.next and steps > 0:
            self.cur = self.cur.next
            steps -= 1
        return self.cur.val

In this solution, we use a doubly linked list to keep track of the browser's history.

The ListNode class represents each page visited, and it contains a reference to the previous and next pages.

Here's a breakdown of each method:

1. Constructor (__init__): Initializes the browser history with the homepage provided.

It creates a cur pointer pointing to the current page.

1. visit: This method allows us to visit a URL.

It creates a new ListNode for the visited URL and appends it to the history.

The forward history is automatically cleared by this operation.

It updates the cur pointer to the newly visited page.

The time complexity is O(1).

1. back: Moving back in history is not as straightforward as in a simple array.

We use a while loop to navigate the doubly linked list backward until we reach the desired number of steps or the beginning of the history.

The time complexity is O(n), where n is the number of steps we want to move back.

1. forward: Similar to the back operation, moving forward involves traversing the linked list forward.

We use a while loop to accomplish this, and the time complexity is O(n).

This linked list implementation provides a correct solution to the problem but is not the most efficient one.

To achieve better efficiency, we can explore an alternative array-based solution, as follows:

Array Implementation (Optimized)

The array-based solution is optimized for efficiency, as it simplifies the back and forward operations to constant time.

It makes use of a dynamic array to store the history.

class BrowserHistory:    
    def __init__(self, homepage: str):
        self.i = 0
        self.length = 1
        self.history = [homepage]

    # Visit a URL
    # Time complexity: `O(1)`
    def visit(self, url: str) -> None:
        if len(self.history) < self.i + 2:
            self.history.append(url)
        else:
            self.history[self.i + 1] = url
        self.i += 1
        self.length = self.i + 1

    # Move back in history
    # Time complexity: `O(1)`
    def back(self, steps: int) -> str:
        self.i = max(self.i - steps, 0)
        return self.history[self.i]

    # Move forward in history
    # Time complexity: `O(1)`
    def forward(self, steps: int) -> str:
        self.i = min(self.i + steps, self.length - 1)
        return self.history[self.i]

This optimized array-based solution has the following characteristics:

1. Constructor (__init__): Initializes the browser history with the homepage provided.

It creates a history list,

an index i representing the current position, and a length variable indicating the effective length of the history.

The initial history contains only the homepage.

1. visit: To visit a URL, we add it to the history array.

If the new URL causes the history to extend beyond its current bounds, we append it to the end of the array.

Otherwise, we update the existing entry at the appropriate index.

We increment the index i to point to the newly visited page and update the length variable.

The time complexity is O(1).

1. back: Moving back in history is as simple as decrementing the index i by the specified number of steps.

We use the max function to ensure that i doesn't go below 0, which would indicate going out of bounds.

The time complexity is O(1).

1. forward: Similar to the back operation, moving forward involves incrementing the index i by the specified number of steps.

We use the min function to ensure i doesn't exceed the length of the history, which would indicate going out of bounds.

The time complexity is O(1).

This array-based solution is highly efficient and simplifies the back and forward operations, making them constant-time operations.

It's a great choice for solving the problem optimally.

Reasoning Behind Our Approach

In both the linked list and array-based solutions, we effectively manage the browser's history, ensuring that we can visit URLs, move back and forward efficiently, and handle edge cases, such as not moving out of bounds.

While the linked list solution is a good demonstration of working with data structures, the array-based solution is the more efficient and practical choice for solving the problem.

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Conclusion

In this guide, we explored the LeetCode problem Design Browser History We examined the problem requirements, constraints, and two different solutions.

The first solution employed a doubly linked list to represent the browser history, providing a correct solution but with less efficient back and forward operations.

The second solution, which used a dynamic array, delivered a more efficient and optimized solution, simplifying the back and forward operations to constant time.

As a beginner, it's important to understand that there can be multiple ways to approach a problem.

It's valuable to consider different data structures and algorithms to achieve the desired outcomes.

In this case, both solutions address the problem requirements but with varying levels of efficiency.

If you found this guide helpful, please like and engage to stay updated on more problem-solving techniques and algorithm implementations.

If you have any questions, suggestions, or comments, feel free to share them.

Happy coding!

Question Link

Remember, the best way to learn and improve your coding skills is through practice and exploration.

Keep challenging yourself with interesting problems and keep refining your solutions.