1472. Design Browser History

Method 1

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 1

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"

Method 1

【O(1) time | O(n) space】

package Leetcode.Array;

import java.util.ArrayList;

/**
 * @author zhengxingxing
 * @date 2025/02/26
 */
public class BrowserHistory {

    private ArrayList<String> history; // Store browser history
    private int current;              // Current position index
    private int size;                 // Valid history size

    /**
     * Initialize browser history with homepage
     * @param homepage Initial homepage URL
     */
    public BrowserHistory(String homepage) {
        history = new ArrayList<>();
        history.add(homepage);
        current = 0;
        size = 1;
    }

    /**
     * Visit a new URL
     * @param url URL to visit
     */
    public void visit(String url) {
        current++;
        // If current exceeds ArrayList size, add new element
        if (current == history.size()) {
            history.add(url);
        } else {
            history.set(current, url);
        }
        size = current + 1; // Update valid history size
    }

    /**
     * Move back in history
     * @param steps Number of steps to move back
     * @return URL after moving back
     */
    public String back(int steps) {
        current = Math.max(0, current - steps);
        return history.get(current);
    }

    /**
     * Move forward in history
     * @param steps Number of steps to move forward
     * @return URL after moving forward
     */
    public String forward(int steps) {
        current = Math.min(size - 1, current + steps);
        return history.get(current);
    }
    
    public static void main(String[] args) {
        BrowserHistory browserHistory = new BrowserHistory("leetcode.com");
        browserHistory.visit("google.com");
        browserHistory.visit("facebook.com");
        browserHistory.visit("youtube.com");

        System.out.println(browserHistory.back(1));    // Returns "facebook.com"
        System.out.println(browserHistory.back(1));    // Returns "google.com"
        System.out.println(browserHistory.forward(1)); // Returns "facebook.com"

        browserHistory.visit("linkedin.com");
        System.out.println(browserHistory.forward(2)); // Returns "linkedin.com"
        System.out.println(browserHistory.back(2));    // Returns "google.com"
        System.out.println(browserHistory.back(7));    // Returns "leetcode.com"
    }
}

Method 1

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