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Design Browser History

Last Updated : 23 Jul, 2025
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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. The task is to design a data structure and implement the functionality of visiting a URL starting from the homepage and moving back and forward in the history. The following functionalities should be covered:

  • visit(url) : Visits a URL given as string
  • forward(steps) : Takes 'steps' forward.
  • back(steps) : Takes 'steps' backward.

Note: The starting page of the tab will always be the homepage.

Examples: 

Input: 
homepage = "geeksforgeeks.org"
visit("amazon.com");
back(2);
Output: geeksforgeeks.org
Explanation:  We need to move 2 steps back but since only 1 step is available we would land up at the homepage, i.e., geeksforgeeks.org

Input:
homepage = "gfg.org"
visit("google.com");
visit("facebook.com");
visit("youtube.com");
back(1);
back(1);
forward(1);
visit("linkedin.com");
forward(2);
back(2);
back(7);
Output:
facebook.com
google.com
facebook.com
linkedin.com
google.com
gfg.org
Explanation:
visit("google.com") :  We are at google.com 
visit("facebook.com"): Now, we are at facebook.com
visit("youtube.com"): We are at youtube.com
back(1):  We would land up at facebook.com, if we move one step back.
back(1):  Moving one step back, takes us to google.com 
forward(1): Moving a step forward we would be at facebook.com
visit("linkedin.com"):  We are at linkedin.com
forward(2): We are still at linkedin. since visiting clear the forward history . When we are the current URL, there is no URL to move forward to. 
back(2): Moving two steps back, takes us to google.com
back(7):  We need to move 7 steps back, but only 1 url is available. Therefore we would return gfg.org. 

Table of Content

Brute-Force Approach [Using Two Stacks]

The idea is to implement a browser history design by employing two stacks. We need a stack to keep track of the previously visited URLs and another stack to store the current URL on the browser tab.

Follow the steps mentioned below to implement the idea:

Create two stacks:

  • backStack: Stores the current URL and all previous URLs in the backward navigation.
  • forwardStack: Stores URLs for forward navigation.

BrowserHistory(string homepage):

  • Initialize the browser with the given homepage.
  • Push the homepage into backStack to set it as the starting page.

visit(string url):

  • While visiting a new URL: Clear the forwardStack because visiting a new page invalidates forward history.
  • Pop all elements from forwardStack.
  • Push the new url into the backStack as it is now the current page.

back(int steps):

  • To move backward: Run a while loop for steps number of times.
  • Stop early if backStack has only one element, as we can't go back further.
  • In each step: Push the top element of backStack into forwardStack.
  • Pop the top element from backStack.
  • After moving, return the topmost element of backStack, which is now the current page.

forward(int steps):

  • To move forward: Run a while loop for steps number of times.
  • Stop early if forwardStack is empty.
  • In each step: Push the top element of forwardStack into backStack.
  • Pop the top element from forwardStack.
  • After moving, return the topmost element of backStack, which is now the current page.

Follow the below illustration for a better understanding:

Illustration:

input:
homepage = "gfg.org"
visit("google.com");
visit("facebook.com");
visit("youtube.com");
back(1);
back(1);
forward(1);
visit("linkedin.com");
forward(2);
back(2);
back(7);

Operations:

1st operation: Initialising gfg.org as the homepage and pushing it into the backStack. 

2nd, 3rd and 4th operation: Visiting google.com, facebook.com, and youtube.com. So, push all of these into the backStack.

visit("gogle.com"), visit("facebook.com"), visit("youtube.com")
visit("google.com"), visit("facebook.com"), visit("youtube.com")

5th operation: Move one step back in the browser history. Take youtube.com from the backStack and push it into the forwardStack to keep track of it. After moving one step back, we would land on facebook.com. So, the current page is facebook.com. 

back(1)
back(1)

6th operation: Again we will move one step back by popping the topmost element of the backStack and pushing the same into the forwardStack. After moving one step back, we will be on google.com. 

back(1)
back(1)

7th operation: Move one step forward. We moved to facebook.com after visiting google.com. Therefore, from the forwardStack, we will pick its top and push it into the backStack. facebook.com now serves as the current page. 

forward(1)
forward(1)

8th operation: Now, for visiting another URL, we will push linkedin.com into the backStack. Since this is the most recent URL and there is nothing beyond this, we would clear the forwardStack.

visit("linkedin.com")
visit("linkedin.com")

9th operation: We cannot move 2 steps forward since there is no URL beyond the current page. We will return linkedin.com. 

forward(2)
forward(2)

10th operation: To move 2 steps back, pop linkedin.com and facebook.com and push them into the forwardStack. Now the current page turns out to be google.com.

back(2)
back(2)

11th operation: We need to move 7 steps back, but we only have one URL after the homepage. We cannot move back in history beyond the homepage. Therefore, the homepage is the current page. We will return gfg.org. 

back(7)
back(7)

Below is the implementation of the above approach:

C++ 
// C++ implementation of browser history
// using 2 stacks
#include <bits/stdc++.h>
using namespace std;

class BrowserHistory {
public:
    stack<string> backStack, forwardStack;

    // Constructor to initialize object with homepage
    BrowserHistory(string homepage) {
        backStack.push(homepage);
    }

    // Visit current url
    void visit(string url) {
        while (!forwardStack.empty()) {
            forwardStack.pop();
        }
        backStack.push(url);
    }

    // 'steps' move backward in history and return current page
    string back(int steps) {
        while (backStack.size() > 1 && steps--) {
            forwardStack.push(backStack.top());
            backStack.pop();
        }
        return backStack.top();
    }

    // 'steps' move forward and return current page
    string forward(int steps) {
        while (!forwardStack.empty() && steps--) {
            backStack.push(forwardStack.top());
            forwardStack.pop();
        }
        return backStack.top();
    }
};

int main() {
  
    // Input case :
    string homepage;
    homepage = "gfg.org";

    // Initialise the object of Browser History
    BrowserHistory obj(homepage);

    string url = "google.com";
    obj.visit(url);

    url = "facebook.com";
    obj.visit(url);

    url = "youtube.com";
    obj.visit(url);

    cout << obj.back(1) << endl;

    cout << obj.back(1) << endl;

    cout << obj.forward(1) << endl;

    obj.visit("linkedin.com");

    cout << obj.forward(2) << endl;

    cout << obj.back(2) << endl;

    cout << obj.back(7) << endl;

    return 0;
}
Java 
// Java implementation of browser history
// using 2 stacks
import java.io.*;
import java.util.*;

class GFG {

    static class BrowserHistory {
        Stack<String> backStack = new Stack<>();
        Stack<String> forwardStack = new Stack<>();

        // Constructor to initialize object with homepage
        BrowserHistory(String homepage) {
            backStack.push(homepage);
        }

        // Visit current url
        void visit(String url) {
            while (!forwardStack.isEmpty()) {
                forwardStack.pop();
            }
            backStack.push(url);
        }

        // 'steps' move backward in history and return current page
        String back(int steps) {
            while (backStack.size() > 1 && steps-- > 0) {
                forwardStack.push(backStack.peek());
                backStack.pop();
            }
            return backStack.peek();
        }

        // 'steps' move forward and return current page
        String forward(int steps) {
            while (!forwardStack.isEmpty() && steps-- > 0) {
                backStack.push(forwardStack.peek());
                forwardStack.pop();
            }
            return backStack.peek();
        }
    }

    public static void main(String[] args) {
      
        // Input case :
        String homepage = "gfg.org";

        // Initialise the object of Browser History
        BrowserHistory obj = new BrowserHistory(homepage);

        String url = "google.com";
        obj.visit(url);

        url = "facebook.com";
        obj.visit(url);

        url = "youtube.com";
        obj.visit(url);

        System.out.println(obj.back(1));
        System.out.println(obj.back(1));
        System.out.println(obj.forward(1));
        obj.visit("linkedin.com");
        System.out.println(obj.forward(2));
        System.out.println(obj.back(2));
        System.out.println(obj.back(7));
    }
}
Python 
# Python implementation of browser history
# using 2 stacks
class BrowserHistory:
    
    # Constructor to initialize object with homepage
    def __init__(self, homepage: str): 
        self.backStack = [homepage]
        self.forwardStack = []

    # Visit current url
    def visit(self, url: str): 
        self.forwardStack.clear()
        self.backStack.append(url)

    # 'steps' move backward in history and return current page
    def back(self, steps: int): 
        while len(self.backStack) > 1 and steps > 0: 
            self.forwardStack.append(self.backStack.pop())
            steps -= 1
        return self.backStack[-1]

    # 'steps' move forward and return current page
    def forward(self, steps: int): 
        while len(self.forwardStack) > 0 and steps > 0: 
            self.backStack.append(self.forwardStack.pop())
            steps -= 1
        return self.backStack[-1]
      
if __name__ == "__main__": 

    # Input case
    homepage = "gfg.org"
    obj = BrowserHistory(homepage)

    url = "google.com"
    obj.visit(url)

    url = "facebook.com"
    obj.visit(url)

    url = "youtube.com"
    obj.visit(url)

    print(obj.back(1))
    print(obj.back(1))
    print(obj.forward(1))

    obj.visit("linkedin.com")

    print(obj.forward(2))
    print(obj.back(2))
    print(obj.back(7))
C# 
// C# implementation of browser history
// using 2 stacks
using System;
using System.Collections.Generic;

public class GFG {

    class BrowserHistory {
        Stack<string> backStack = new Stack<string>();
        Stack<string> forwardStack = new Stack<string>();

        // Constructor to initialize object with homepage
        public BrowserHistory(string homepage) {
            backStack.Push(homepage);
        }

        // Visit current url
        public void Visit(string url) {
            while (forwardStack.Count > 0) {
                forwardStack.Pop();
            }
            backStack.Push(url);
        }

        // 'steps' move backward in history and return current page
        public string Back(int steps) {
            while (backStack.Count > 1 && steps-- > 0) {
                forwardStack.Push(backStack.Peek());
                backStack.Pop();
            }
            return backStack.Peek();
        }

        // 'steps' move forward and return current page
        public string Forward(int steps) {
            while (forwardStack.Count > 0 && steps-- > 0) {
                backStack.Push(forwardStack.Peek());
                forwardStack.Pop();
            }
            return backStack.Peek();
        }
    }

    public static void Main() {

        // Input case:
        string homepage = "gfg.org";

        // Initialize the object of BrowserHistory
        BrowserHistory obj = new BrowserHistory(homepage);

        string url = "google.com";
        obj.Visit(url);

        url = "facebook.com";
        obj.Visit(url);

        url = "youtube.com";
        obj.Visit(url);

        Console.WriteLine(obj.Back(1));
        Console.WriteLine(obj.Back(1));
        Console.WriteLine(obj.Forward(1));
        obj.Visit("linkedin.com");
        Console.WriteLine(obj.Forward(2));
        Console.WriteLine(obj.Back(2));
        Console.WriteLine(obj.Back(7));
    }
}
JavaScript 
// JavaScript implementation of browser history
// using 2 stacks
class BrowserHistory {
    constructor(homepage) {
        this.backStack = [];
        this.forwardStack = [];

        // Initialize object with homepage
        this.backStack.push(homepage);
    }

    // Visit current URL
    visit(url) {
        this.forwardStack = [];
        this.backStack.push(url);
    }

    // 'steps' move backward in history and 
    // return current page
    back(steps) {
        while (this.backStack.length > 1 && steps-- > 0) {
            this.forwardStack.push(this.backStack[
                              this.backStack.length - 1]);
                              
            this.backStack.pop();
        }
        return this.backStack[this.backStack.length - 1];
    }

    // 'steps' move forward and return current page
    forward(steps) {
        while (this.forwardStack.length > 0 && steps-- > 0) {
            this.backStack.push(this.forwardStack[
                             this.forwardStack.length - 1]);
                             
            this.forwardStack.pop();
        }
        return this.backStack[this.backStack.length - 1];
    }
}

// Input case
let homepage = "gfg.org";

// Initialize the object of BrowserHistory
let obj = new BrowserHistory(homepage);

let url = "google.com";
obj.visit(url);

url = "facebook.com";
obj.visit(url);

url = "youtube.com";
obj.visit(url);

console.log(obj.back(1));
console.log(obj.back(1));
console.log(obj.forward(1));
obj.visit("linkedin.com");
console.log(obj.forward(2));
console.log(obj.back(2));
console.log(obj.back(7));

Output
facebook.com
google.com
facebook.com
linkedin.com
google.com
gfg.org

Time Complexity: The visit function takes O(f), where f is the size of the forwardStack. The back and forward functions take O(min(steps, b)) and O(min(steps, f)), respectively.
Space Complexity: The backStack and forwardStack together use O(n), where n is the total number of visited URLs. No additional auxiliary space is used beyond the stacks.

Expected Approach [Using a Doubly Linkeded List]

The idea is to use a doubly linked list to keep track of the browser's history, where each node stores a URL. By moving forward or backward through the list, we can simulate navigating through previously visited pages, updating the current page accordingly.

Follow the steps mentioned below to implement the idea:

Create a doubly linked list:

  • Node class represents each URL visited with references to the previous and next nodes.

BrowserHistory(string homepage):

  • Initialize the browser with the given homepage by creating a new node and setting it as the starting page.
  • Set the current pointer (curr) to point to the homepage node.

visit(string url):

  • When visiting a new URL, create a new node for the URL.
  • Set the previous pointer of the new node to the current node.
  • Set the next pointer of the current node to the new node.
  • Move the curr pointer to the new node.

back(int steps):

  • To move backward, iterate the pointer steps times.
  • In each step, move the pointer to the previous node if it's not null.
  • After moving, update the curr pointer to the new node.
  • Return the URL stored in the curr node.

forward(int steps):

  • To move forward, iterate the pointer steps times.
  • In each step, move the pointer to the next node if it's not null.
  • After moving, update the curr pointer to the new node.
  • Return the URL stored in the curr node.

Below is the implementation of the above approach:

C++ 
// C++ implementation of browser history
// using Doubly Linked List
#include <bits/stdc++.h>
using namespace std;

class Node {
public:
    string data;
    Node* prev;
    Node* next;

    Node(string x) {
        data = x;
        prev = nullptr;
        next = nullptr;
    }
};

class BrowserHistory {
public:
  
    // Pointer to the current URL
    Node* curr;

    // Constructor to initialize with the homepage
    BrowserHistory(string homepage) {
        curr = new Node(homepage);
    }

    // Function to visit a new URL
    void visit(string url) {
        
        // Create a node for this visit
        Node* urlNode = new Node(url);

        // Set the previous node of the 
        // new node to current
        urlNode->prev = curr;

        // Update the next of the current 
        // node to the new node
        curr->next = urlNode;

        // Move the current pointer to the new node
        curr = urlNode;
    }

    // Function to move back by 'step' times
    string back(int step) {
        
        // Pointer to traverse backward
        Node* trav = curr;

        // Travel back `step` times if possible
        while (trav->prev != nullptr && step > 0) {
            trav = trav->prev;
            step--;
        }

        // Update current pointer after moving back
        curr = trav;

        return curr->data;
    }

    // Function to move forward by 'step' times
    string forward(int step) {
        
        // Pointer to traverse forward
        Node* trav = curr;

        // Travel forward `step` times if possible
        while (trav->next != nullptr && step > 0) {
            trav = trav->next;
            step--;
        }

        // Update current pointer after moving forward
        curr = trav;

        return curr->data;
    }
};

int main() {
    
    // Initialize with the homepage
    string homepage = "gfg.org";
    BrowserHistory obj(homepage);

    string url = "google.com";
    obj.visit(url);

    url = "facebook.com";
    obj.visit(url);

    url = "youtube.com";
    obj.visit(url);

    cout << obj.back(1) << endl;

    cout << obj.back(1) << endl;

    cout << obj.forward(1) << endl;

    obj.visit("linkedin.com");

    cout << obj.forward(2) << endl;

    cout << obj.back(2) << endl;

    cout << obj.back(7) << endl;

    return 0;
}
Java 
// Java implementation of browser history
// using Doubly Linked List
class Node {
    String data;
    Node prev, next;

    Node(String x) {
        data = x;
        prev = null;
        next = null;
    }
}

public class GfG {
  
  static class BrowserHistory {

      // Pointer to the current URL
      private Node curr;

      // Constructor to initialize with the homepage
      public BrowserHistory(String homepage) {
          curr = new Node(homepage);
      }

      // Function to visit a new URL
      public void visit(String url) {

          // Create a node for this visit
          Node urlNode = new Node(url);

          // Set the previous node of the
          // new node to current
          urlNode.prev = curr;

          // Update the next of the current
          // node to the new node
          curr.next = urlNode;

          // Move the current pointer to the new node
          curr = urlNode;
      }

      // Function to move back by 'step' times
      public String back(int step) {

          // Pointer to traverse backward
          Node trav = curr;

          // Travel back `step` times if possible
          while (trav.prev != null && step > 0) {
              trav = trav.prev;
              step--;
          }

          // Update current pointer after moving back
          curr = trav;

          return curr.data;
      }

      // Function to move forward by 'step' times
      public String forward(int step) {

          // Pointer to traverse forward
          Node trav = curr;

          // Travel forward `step` times if possible
          while (trav.next != null && step > 0) {
              trav = trav.next;
              step--;
          }

          // Update current pointer after moving forward
          curr = trav;

          return curr.data;
       }
   }

    public static void main(String[] args) {

        // Initialize with the homepage
        String homepage = "gfg.org";
        BrowserHistory obj = new BrowserHistory(homepage);

        String url = "google.com";
        obj.visit(url);

        url = "facebook.com";
        obj.visit(url);

        url = "youtube.com";
        obj.visit(url);

        System.out.println(obj.back(1));

        System.out.println(obj.back(1));

        System.out.println(obj.forward(1));

        obj.visit("linkedin.com");

        System.out.println(obj.forward(2));

        System.out.println(obj.back(2));

        System.out.println(obj.back(7));
    }
}
Python 
# Python implementation of browser history
# using Doubly Linked List
class Node:
    def __init__(self, x):
        self.data = x
        self.prev = None
        self.next = None

class BrowserHistory:

    # Constructor to initialize with the homepage
    def __init__(self, homepage):
      
        # Pointer to the current URL
        self.curr = Node(homepage)

    # Function to visit a new URL
    def visit(self, url):

        # Create a node for this visit
        url_node = Node(url)

        # Set the previous node of the
        # new node to current
        url_node.prev = self.curr

        # Update the next of the current
        # node to the new node
        self.curr.next = url_node

        # Move the current pointer to the new node
        self.curr = url_node

    # Function to move back by 'step' times
    def back(self, step):

        # Pointer to traverse backward
        trav = self.curr

        # Travel back `step` times if possible
        while trav.prev is not None and step > 0:
            trav = trav.prev
            step -= 1

        # Update current pointer after moving back
        self.curr = trav

        return self.curr.data

    # Function to move forward by 'step' times
    def forward(self, step):

        # Pointer to traverse forward
        trav = self.curr

        # Travel forward `step` times if possible
        while trav.next is not None and step > 0:
            trav = trav.next
            step -= 1

        # Update current pointer after moving forward
        self.curr = trav

        return self.curr.data

if __name__ == "__main__":

    # Initialize with the homepage
    homepage = "gfg.org"
    obj = BrowserHistory(homepage)

    url = "google.com"
    obj.visit(url)

    url = "facebook.com"
    obj.visit(url)

    url = "youtube.com"
    obj.visit(url)

    print(obj.back(1))

    print(obj.back(1))

    print(obj.forward(1))

    obj.visit("linkedin.com")

    print(obj.forward(2))

    print(obj.back(2))

    print(obj.back(7))
C# 
// C# implementation of browser history
// using Doubly Linked List
using System;

class Node {
    public string data;
    public Node prev, next;

    public Node(string x) {
        data = x;
        prev = null;
        next = null;
    }
}

class GfG {
  
    class BrowserHistory {

        // Pointer to the current URL
        private Node curr;

        // Constructor to initialize with the homepage
        public BrowserHistory(string homepage) {
            curr = new Node(homepage);
        }

        // Function to visit a new URL
        public void Visit(string url) {

            // Create a node for this visit
            Node urlNode = new Node(url);

            // Set the previous node of the
            // new node to current
            urlNode.prev = curr;

            // Update the next of the current
            // node to the new node
            curr.next = urlNode;

            // Move the current pointer to the new node
            curr = urlNode;
        }

        // Function to move back by 'step' times
        public string Back(int step) {

            // Pointer to traverse backward
            Node trav = curr;

            // Travel back `step` times if possible
            while (trav.prev != null && step > 0) {
                trav = trav.prev;
                step--;
            }

            // Update current pointer after moving back
            curr = trav;

            return curr.data;
        }

        // Function to move forward by 'step' times
        public string Forward(int step) {

            // Pointer to traverse forward
            Node trav = curr;

            // Travel forward `step` times if possible
            while (trav.next != null && step > 0) {
                trav = trav.next;
                step--;
            }

            // Update current pointer after moving forward
            curr = trav;

            return curr.data;
        }
    }

    static void Main(string[] args) {

        // Initialize with the homepage
        string homepage = "gfg.org";
        BrowserHistory obj 
                 = new BrowserHistory(homepage);

        string url = "google.com";
        obj.Visit(url);

        url = "facebook.com";
        obj.Visit(url);

        url = "youtube.com";
        obj.Visit(url);

        Console.WriteLine(obj.Back(1));

        Console.WriteLine(obj.Back(1));

        Console.WriteLine(obj.Forward(1));

        obj.Visit("linkedin.com");

        Console.WriteLine(obj.Forward(2));

        Console.WriteLine(obj.Back(2));

        Console.WriteLine(obj.Back(7));
    }
}
JavaScript 
// JavaScript implementation of browser history
// using Doubly Linked List

class Node {
    constructor(data) {
        this.data = data;
        this.prev = null;
        this.next = null;
    }
}

class BrowserHistory {

    // Constructor to initialize with the homepage
    constructor(homepage) {
        this.curr = new Node(homepage);
    }

    // Function to visit a new URL
    visit(url) {

        // Create a node for this visit
        const urlNode = new Node(url);

        // Set the previous node of the
        // new node to current
        urlNode.prev = this.curr;

        // Update the next of the current
        // node to the new node
        this.curr.next = urlNode;

        // Move the current pointer to the new node
        this.curr = urlNode;
    }

    // Function to move back by 'step' times
    back(step) {

        // Pointer to traverse backward
        let trav = this.curr;

        // Travel back `step` times if possible
        while (trav.prev !== null && step > 0) {
            trav = trav.prev;
            step--;
        }

        // Update current pointer after moving back
        this.curr = trav;

        return this.curr.data;
    }

    // Function to move forward by 'step' times
    forward(step) {

        // Pointer to traverse forward
        let trav = this.curr;

        // Travel forward `step` times if possible
        while (trav.next !== null && step > 0) {
            trav = trav.next;
            step--;
        }

        // Update current pointer after moving forward
        this.curr = trav;

        return this.curr.data;
    }
}

const homepage = "gfg.org";
const obj = new BrowserHistory(homepage);

obj.visit("google.com");
obj.visit("facebook.com");
obj.visit("youtube.com");

console.log(obj.back(1));
console.log(obj.back(1));
console.log(obj.forward(1));

obj.visit("linkedin.com");

console.log(obj.forward(2)); 
console.log(obj.back(2));
console.log(obj.back(7));

Output
facebook.com
google.com
facebook.com
linkedin.com
google.com
gfg.org

Time Complexity: The visit function takes O(1) as it involves adding a node. The back and forward functions take O(min(steps, b)) and O(min(steps, f)) respectively, where b is the number of steps backward and f is the number of steps forward.
Space Complexity: The doubly linked list uses O(n) space, where n is the total number of visited URLs. No additional auxiliary space is used beyond the linked list.
 


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