1472. Design Browser History LeetCode Solution

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Problem Statement
Complexity Analysis
Design Browser History solution in C++
Design Browser History solution in Java
Design Browser History solution in Python
Additional Resources

1472. Design Browser History LeetCode Solution image

Problem Statement of 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.

Example not found

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.

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(n)

1472. Design Browser History LeetCode Solution in C++

class BrowserHistory {
 public:
  BrowserHistory(string homepage) {
    visit(homepage);
  }

  void visit(string url) {
    if (++index < urls.size())
      urls[index] = url;
    else
      urls.push_back(url);
    lastIndex = index;
  }

  string back(int steps) {
    index = max(0, index - steps);
    return urls[index];
  }

  string forward(int steps) {
    index = min(lastIndex, index + steps);
    return urls[index];
  }

 private:
  vector<string> urls;
  int index = -1;
  int lastIndex = -1;
};
/* code provided by PROGIEZ */

1472. Design Browser History LeetCode Solution in Java

class BrowserHistory {
  public BrowserHistory(String homepage) {
    visit(homepage);
  }

  public void visit(String url) {
    if (++index < urls.size())
      urls.set(index, url);
    else
      urls.add(url);
    lastIndex = index;
  }

  public String back(int steps) {
    index = Math.max(0, index - steps);
    return urls.get(index);
  }

  public String forward(int steps) {
    index = Math.min(lastIndex, index + steps);
    return urls.get(index);
  }

  private List<String> urls = new ArrayList<>();
  private int index = -1;
  private int lastIndex = -1;
}
// code provided by PROGIEZ

1472. Design Browser History LeetCode Solution in Python

class BrowserHistory:
  def __init__(self, homepage: str):
    self.urls = []
    self.index = -1
    self.lastIndex = -1
    self.visit(homepage)

  def visit(self, url: str) -> None:
    self.index += 1
    if self.index < len(self.urls):
      self.urls[self.index] = url
    else:
      self.urls.append(url)
    self.lastIndex = self.index

  def back(self, steps: int) -> str:
    self.index = max(0, self.index - steps)
    return self.urls[self.index]

  def forward(self, steps: int) -> str:
    self.index = min(self.lastIndex, self.index + steps)
    return self.urls[self.index]
 # code by PROGIEZ