Leetcode #1472: Design Browser History
In this guide, we solve Leetcode #1472 Design Browser History in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Design, Array, Linked List, Data Stream, Doubly-Linked List
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
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"
Python Solution
class BrowserHistory:
def __init__(self, homepage: str):
self.stk1 = []
self.stk2 = []
self.visit(homepage)
def visit(self, url: str) -> None:
self.stk1.append(url)
self.stk2.clear()
def back(self, steps: int) -> str:
while steps and len(self.stk1) > 1:
self.stk2.append(self.stk1.pop())
steps -= 1
return self.stk1[-1]
def forward(self, steps: int) -> str:
while steps and self.stk2:
self.stk1.append(self.stk2.pop())
steps -= 1
return self.stk1[-1]
# Your BrowserHistory object will be instantiated and called as such:
# obj = BrowserHistory(homepage)
# obj.visit(url)
# param_2 = obj.back(steps)
# param_3 = obj.forward(steps)
Complexity
The time complexity is . The space complexity is , where is the length of the browsing history.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.