# ---
# title: 1472. Design Browser History
# id: problem1472
# author: Tian Jun
# date: 2020-10-31
# difficulty: Medium
# categories: Design
# link: <https://leetcode.com/problems/design-browser-history/description/>
# hidden: true
# ---
# 
# 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`.
# 
# 
## @lc code=start
using LeetCode

## add your code here:
## @lc code=end
