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1472. Design Browser History

Description

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.

Solutions

Solution 1: Two Stacks

We can use two stacks, \(\textit{stk1}\) and \(\textit{stk2}\), to store the back and forward pages, respectively. Initially, \(\textit{stk1}\) contains the \(\textit{homepage}\), and \(\textit{stk2}\) is empty.

When calling \(\text{visit}(url)\), we add \(\textit{url}\) to \(\textit{stk1}\) and clear \(\textit{stk2}\). The time complexity is \(O(1)\).

When calling \(\text{back}(steps)\), we pop the top element from \(\textit{stk1}\) and push it to \(\textit{stk2}\). We repeat this operation \(steps\) times until the length of \(\textit{stk1}\) is \(1\) or \(steps\) is \(0\). Finally, we return the top element of \(\textit{stk1}\). The time complexity is \(O(\textit{steps})\).

When calling \(\text{forward}(steps)\), we pop the top element from \(\textit{stk2}\) and push it to \(\textit{stk1}\). We repeat this operation \(steps\) times until \(\textit{stk2}\) is empty or \(steps\) is \(0\). Finally, we return the top element of \(\textit{stk1}\). The time complexity is \(O(\textit{steps})\).

The space complexity is \(O(n)\), where \(n\) is the length of the browsing history.

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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)

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class BrowserHistory {
    private Deque<String> stk1 = new ArrayDeque<>();
    private Deque<String> stk2 = new ArrayDeque<>();

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

    public void visit(String url) {
        stk1.push(url);
        stk2.clear();
    }

    public String back(int steps) {
        for (; steps > 0 && stk1.size() > 1; --steps) {
            stk2.push(stk1.pop());
        }
        return stk1.peek();
    }

    public String forward(int steps) {
        for (; steps > 0 && !stk2.isEmpty(); --steps) {
            stk1.push(stk2.pop());
        }
        return stk1.peek();
    }
}

/**
 * Your BrowserHistory object will be instantiated and called as such:
 * BrowserHistory obj = new BrowserHistory(homepage);
 * obj.visit(url);
 * String param_2 = obj.back(steps);
 * String param_3 = obj.forward(steps);
 */

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class BrowserHistory {
public:
    stack<string> stk1;
    stack<string> stk2;

    BrowserHistory(string homepage) {
        visit(homepage);
    }

    void visit(string url) {
        stk1.push(url);
        stk2 = stack<string>();
    }

    string back(int steps) {
        for (; steps && stk1.size() > 1; --steps) {
            stk2.push(stk1.top());
            stk1.pop();
        }
        return stk1.top();
    }

    string forward(int steps) {
        for (; steps && !stk2.empty(); --steps) {
            stk1.push(stk2.top());
            stk2.pop();
        }
        return stk1.top();
    }
};

/**
 * Your BrowserHistory object will be instantiated and called as such:
 * BrowserHistory* obj = new BrowserHistory(homepage);
 * obj->visit(url);
 * string param_2 = obj->back(steps);
 * string param_3 = obj->forward(steps);
 */

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type BrowserHistory struct {
    stk1 []string
    stk2 []string
}

func Constructor(homepage string) BrowserHistory {
    t := BrowserHistory{[]string{}, []string{}}
    t.Visit(homepage)
    return t
}

func (this *BrowserHistory) Visit(url string) {
    this.stk1 = append(this.stk1, url)
    this.stk2 = []string{}
}

func (this *BrowserHistory) Back(steps int) string {
    for i := 0; i < steps && len(this.stk1) > 1; i++ {
        this.stk2 = append(this.stk2, this.stk1[len(this.stk1)-1])
        this.stk1 = this.stk1[:len(this.stk1)-1]
    }
    return this.stk1[len(this.stk1)-1]
}

func (this *BrowserHistory) Forward(steps int) string {
    for i := 0; i < steps && len(this.stk2) > 0; i++ {
        this.stk1 = append(this.stk1, this.stk2[len(this.stk2)-1])
        this.stk2 = this.stk2[:len(this.stk2)-1]
    }
    return this.stk1[len(this.stk1)-1]
}

/**
 * Your BrowserHistory object will be instantiated and called as such:
 * obj := Constructor(homepage);
 * obj.Visit(url);
 * param_2 := obj.Back(steps);
 * param_3 := obj.Forward(steps);
 */

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