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Design Browser History

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.

Solution Explanation: Design Browser History

This problem requires designing a data structure to efficiently manage browser history, allowing for back and forward navigation. The optimal approach leverages the properties of stacks to achieve efficient time complexity.

Approach: Using Two Stacks

We employ two stacks:

  • stk1 (main stack): Stores the visited URLs in chronological order. The top of the stack represents the currently viewed page.
  • stk2 (forward stack): Stores URLs that can be accessed by moving forward. This stack is cleared whenever a new page is visited.

1. BrowserHistory(homepage):

  • The constructor initializes stk1 with the homepage. stk2 remains empty.

2. visit(url):

  • Pushes the new url onto stk1.
  • Clears stk2 because moving forward from the new page resets the forward history. This operation is O(1) on average if stk2 uses a dynamic array implementation. In a linked list implementation it would be O(n) in the worst case.

3. back(steps):

  • Iteratively pops elements from stk1 and pushes them onto stk2 for steps times or until stk1 has only one element (homepage). This simulates moving back in history.
  • Returns the top element of stk1 (the currently viewed URL after moving back).

4. forward(steps):

  • Iteratively pops elements from stk2 and pushes them onto stk1 for steps times or until stk2 is empty. This simulates moving forward in history.
  • Returns the top element of stk1 (the currently viewed URL after moving forward).

Time and Space Complexity Analysis

  • Time Complexity:

    • visit: O(1) Average case; O(n) worst-case for linked list
    • back: O(min(steps, n)) where n is the length of the history. In the worst case it will iterate through all elements.
    • forward: O(min(steps, n)) where n is the length of the forward history. In the worst case it will iterate through all elements.

  • Space Complexity: O(n), where n is the total number of URLs visited. This is because we store all visited URLs in stk1 and potentially a subset in stk2.

Code Implementations

The code implementations provided in Python, Java, C++, and Go all follow this two-stack approach. The choice of stack data structure (e.g., list in Python, Deque in Java, stack in C++, a slice in Go) affects minor implementation details, but the core algorithm remains the same. Remember that the average time complexity for visit is O(1) assuming a dynamic array or list based implementation but could be O(n) for a linked list in the worst-case scenario. The other operations have a worst-case time complexity that is linear in the number of steps or the size of the history.