Introduction

In this blog post, we will explore how to solve the LeetCode problem "1472" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You have a browser of one tab where you start on the homepage and you can visit another url, get back in the history number of steps or move forward in the history number of steps. Implement the BrowserHistory class: BrowserHistory(string homepage) Initializes the object with the homepage of the browser. void visit(string url) Visits url from the current page. It clears up all the forward history. string back(int steps) Move steps back in history. If you can only return x steps in the history and steps > x, you will return only x steps. Return the current url after moving back in history at most steps. string forward(int steps) Move steps forward in history. If you can only forward x steps in the history and steps > x, you will forward only x steps. Return the current url after forwarding in history at most steps. Example: 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"] Explanation: BrowserHistory browserHistory = new BrowserHistory("leetcode.com"); browserHistory.visit("google.com"); // You are in "leetcode.com". Visit "google.com" browserHistory.visit("facebook.com"); // You are in "google.com". Visit "facebook.com" browserHistory.visit("youtube.com"); // You are in "facebook.com". Visit "youtube.com" browserHistory.back(1); // You are in "youtube.com", move back to "facebook.com" return "facebook.com" browserHistory.back(1); // You are in "facebook.com", move back to "google.com" return "google.com" browserHistory.forward(1); // You are in "google.com", move forward to "facebook.com" return "facebook.com" browserHistory.visit("linkedin.com"); // You are in "facebook.com". Visit "linkedin.com" browserHistory.forward(2); // You are in "linkedin.com", you cannot move forward any steps. browserHistory.back(2); // You are in "linkedin.com", move back two steps to "facebook.com" then to "google.com". return "google.com" browserHistory.back(7); // You are in "google.com", you can move back only one step to "leetcode.com". return "leetcode.com" Constraints: 1 <= homepage.length <= 20 1 <= url.length <= 20 1 <= steps <= 100 homepage and url consist of '.' or lower case English letters. At most 5000 calls will be made to visit, back, and forward.

Explanation

• Use two stacks: One stack to store the history of visited URLs (back stack) and another for the forward history (forward stack).
  • Visit operation: Push the current URL to the back stack and clear the forward stack.
  • Back/Forward operations: Move URLs between the back and forward stacks as needed, ensuring to handle boundary conditions (e.g., moving back when at the beginning of history).

• Runtime Complexity: O(1) for all operations (visit, back, forward).
• Storage Complexity: O(N) where N is the number of visited URLs.

Code

    class BrowserHistory:

    def __init__(self, homepage: str):
        self.back_stack = [homepage]
        self.forward_stack = []

    def visit(self, url: str) -> None:
        self.back_stack.append(url)
        self.forward_stack = []

    def back(self, steps: int) -> str:
        while steps > 0 and len(self.back_stack) > 1:
            self.forward_stack.append(self.back_stack.pop())
            steps -= 1
        return self.back_stack[-1]

    def forward(self, steps: int) -> str:
        while steps > 0 and len(self.forward_stack) > 0:
            self.back_stack.append(self.forward_stack.pop())
            steps -= 1
        return self.back_stack[-1]