Web Browser History
Coding & Implementation

Algorithms and Data Structures: TheAlgorist.com

System Design: www.System.Design

Low Level Design: LowLevelDesign.io

Frontend Engineering: FrontendEngineering.io
জয় শ্রী রাম
🕉
Problem Statement:
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 as follows:
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:
BrowserHistory browserHistory = new BrowserHistory("example.com");
browserHistory.visit("google.com"); // You are in "example.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 "example.com". return "example.com"
Solution:
While solving this design problem I would show you how to methodically go from O(n) approach to O(1) approach using your critical thinking ability and clear understanding of data structures. Please pay special attention to how we systematically optimize our solution. This is one of the qualities we are trying to build in you in this course and would greatly help you in your interviews as well as later in your software engineering career to stand out and get promoted and progress towards becoming an architect.O(n) Doubly LinkedList based O(n) Approach and its drawbacks:
This solution is based on
a very simple thought process:
take a linkedlist to store
the visited webpages in sequence (in the order they are visited).
We have the older visited webpages towards the tail, and the
more recently visited webpages towards the head. Since we have both forward() and back() methods,
that means that we need to move on both sides of the linkedlist, this means that
we would need a doubly linkedlist.
This approach would
automatically come to your mind if you have gone through
LRU Cache
and LFU Cache.
The simple code below implements the above algorithm:
Java Code:
Python Code:
The simple code below implements the above algorithm:
Java Code:
// Implementation using Doubly LinkedList
class BrowserHistory {
Node head; // most recently visited webpage
Node tail; // webpage visited longest ago
Node current; // webpage we are currently at
public BrowserHistory(String homepage) {
// when the first webpage is visited
// that webpage becomes head, tail and current
// becuase we have only one webpage as of now
head = new Node(homepage);
tail = head;
current = head;
}
public void visit(String url) {
Node newNode = new Node(url);
// Requiremnets for visit(string url) method: Visits url from the current page. It clears up all the forward history.
newNode.next = current; // the webpage to which current node currently points to goes one step back in history from the currently
// visited webpage which means the webpage which current node is pointing to becomes next node of the currently visited webpage
current.prev = newNode; // the newly visited node becomes previous node of the webpage that is currently being pointed to by the current node
head = newNode; // newly visited webpage becomes head. This clears all forward history
current = head; // newly visited webpage becomes current webpsge
}
public String forward(int steps) {
// Requirements for forward(int steps) method: 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.
Node node = current;
// go towards head
while (node.prev != null && steps > 0) {
node = node.prev;
steps;
}
current = node;
return node.url;
}
public String back(int steps) {
// Requirements of back(int steps) method: 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.
Node node = current;
// go towards tail
while (node.next != null && steps > 0) {
node = node.next;
steps;
}
current = node;
return node.url;
}
class Node {
public String url;
public Node prev;
public Node next;
public Node(String url) {
this.url = url;
}
}
}
Python Code:
class BrowserHistory:
def __init__(self, homepage):
self.head = Node(self, homepage)
self.tail = self.head
self.current = self.head
def visit(self, url):
newNode = Node(self, url)
newNode.next = self.current
self.current.prev = newNode
self.head = newNode
self.current = self.head
def forward(self, steps):
node = self.current
while node.prev is not None and steps > 0:
node = node.prev
steps = 1
self.current = node
return node.url
def back(self, steps):
node = self.current
while node.next is not None and steps > 0:
node = node.next
steps = 1
self.current = node
return node.url
class Node:
def __init__(self, outerInstance, url):
self.url = None
self.prev = None
self.next = None
self._outerInstance = outerInstance
self.url = url
Drawback:
This doubly linkedlist based approach is not a good approach to be implemented in a language which does not
have support inbuilt garbage collection due to the risk of data leak that would
arise when we are clearing the forward history and reassigning the head. The webpages which are cleared from
the forward history should be handled by the developer to make sure there is no data leak if the
above approach is implemented in a language like C++ where there is no garbage collection.
Time Complexity: O(n)
Space Complexity: O(n)
O(n) Approach using Two Stacks:
This approach is one of the simplest and intuitive and a perfect use of stack data structure.
You keep two stacks:
The picture below beautifully illustrates the concept introduces above.

Stack for storing back history: Lower the position of a webpage in the stack,
the older it is in history.
 Stack to store forward history: Lower the position of a webpage in the stack, the more forward in the history it is.
The picture below beautifully illustrates the concept introduces above.
Look at the simple code below and everything we discussed so far would become crystal clear:
Code:
Java:
// Two Stacks Approach
class BrowserHistory {
Deque<String> backHistory; // stack to contain current webpage and all its back pages, which means webpage on top of this stack is the current webpage we are at
Deque<String> forwardHistory; // stack to store all forward webpages of the current webpage we are at
public BrowserHistory(String homepage) {
backHistory = new ArrayDeque<>();
forwardHistory = new ArrayDeque<>();
backHistory.push(homepage);
}
public void visit(String url) {
backHistory.push(url);
forwardHistory.clear(); // clear all forward history
}
public String back(int steps) {
while (!backHistory.isEmpty() && steps > 0) {
forwardHistory.push(backHistory.pop());
steps;
}
if (backHistory.isEmpty()) {
backHistory.push(forwardHistory.pop());
}
return backHistory.peek();
}
public String forward(int steps) {
while (!forwardHistory.isEmpty() && steps > 0) {
backHistory.push(forwardHistory.pop());
steps;
}
return backHistory.peek();
}
}
Python:
import collections
class BrowserHistory:
def __init__(self, homepage):
self.backHistory = None
self.forwardHistory = None
self.backHistory = collections.deque([])
self.forwardHistory = collections.deque([])
self.backHistory.insert(len(self.backHistory), homepage)
def visit(self, url):
self.backHistory.insert(len(self.backHistory), url)
self.forwardHistory.clear()
def back(self, steps):
while (not len(self.backHistory) == 0) and steps > 0:
self.forwardHistory.insert(len(self.backHistory), self.backHistory.pop())
steps = 1
if len(self.backHistory) == 0:
self.backHistory.insert(len(self.backHistory), self.forwardHistory.pop())
return self.backHistory.index(0)
def forward(self, steps):
while (not len(self.forwardHistory) == 0) and steps > 0:
self.backHistory.insert(len(self.backHistory), self.forwardHistory.pop())
steps = 1
return self.backHistory.index(0)
Time Complexity: O(n)
Space Complexity: O(n)
O(n) solution using One List :
Instead of keeping two data structures, can we just use one ? Yes we can, we just
need to keep track of the index of the current webpage at all time. Please see the code below:
Java Code:
Python Code:
Time Complexity: O(n)
Space Complexity: O(n)
Java Code:
class BrowserHistory {
List<String> visitedWebpages;
int currentIndex;
public BrowserHistory(String homepage) {
visitedWebpages = new ArrayList<String>();
visitedWebpages.add(homepage);
currentIndex = 0;
}
public void visit(String url) { // O(n)
currentIndex++;
// do not do visitedWebpages.add(currentIndex, url), instead do as shown below: it optimizes complexity from O(n) to O(1)
if (currentIndex <= visitedWebpages.size()  1) {
// update
visitedWebpages.set(currentIndex, url); // replace. This optimizes time complexity rather than doing add(index, value) operation since add(index, value) is O(n) but set() is O(1)
}
else {
// add at the end of the list
visitedWebpages.add(url); // O(1)
}
visitedWebpages.subList(currentIndex + 1, visitedWebpages.size()).clear(); // clear all forward history // clear() is O(n) operation
}
public String back(int steps) { // O(1)
currentIndex = Math.max(0, currentIndex  steps);
return visitedWebpages.get(currentIndex); // O(1)
}
public String forward(int steps) { // O(1)
currentIndex = Math.min(visitedWebpages.size()  1, currentIndex + steps);
return visitedWebpages.get(currentIndex); // O(1)
}
}
The above implementation is not O(1) because in
visitedWebpages.subList(currentIndex + 1, visitedWebpages.size()).clear(); // clear all forward history
code
clear()
is O(n).
Python Code:
import math
class BrowserHistory:
def __init__(self, homepage):
self.visitedWebpages = []
self.visitedWebpages.append(homepage)
self.currentIndex = 0
def visit(self, url):
self.currentIndex += 1
if self.currentIndex <= len(self.visitedWebpages)  1:
self.visitedWebpages[self.currentIndex] = url
else:
self.visitedWebpages.append(url)
self.visitedWebpages[self.currentIndex + 1: len(self.visitedWebpages)].clear()
def back(self, steps):
self.currentIndex = max(0, self.currentIndex  steps)
return self.visitedWebpages[self.currentIndex]
def forward(self, steps):
self.currentIndex = min(len(self.visitedWebpages)  1, self.currentIndex + steps)
return self.visitedWebpages[self.currentIndex]
Time Complexity: O(n)
Space Complexity: O(n)
O(1) time complexity solution with O(n) space tradeoff:
Look how the
So if we really want to optimize space we would have to do a tradeoff on space. What if we do not delete the forward history, instead keep where the current foward history ends ? This optimizes the space complexity to O(1) but at a cost of huge space requirements.
The below code implements this O(1) algorithm:
Java code:
Python code:
clear()
method in visit()
method is the root cause of
the time complexity to be O(n) and not O(1). One of the requirements is when we call visit() it should
erase all forward history. If we want to remove the forward history from our data structure it would be
O(n) no matter how much optimization we do. But it saves space since we are deleting the forward history from the data structure altogether.
So if we really want to optimize space we would have to do a tradeoff on space. What if we do not delete the forward history, instead keep where the current foward history ends ? This optimizes the space complexity to O(1) but at a cost of huge space requirements.
No wonder, why some web browsers are memory hog!!
The below code implements this O(1) algorithm:
Java code:
// O (1) solution using one list
class BrowserHistory {
List<String> visitedWebpages;
int currentIndex;
int currentHistorySize; // to keep track of where the forward current history end
public BrowserHistory(String homepage) {
visitedWebpages = new ArrayList<String>();
visitedWebpages.add(homepage);
currentIndex = 0;
currentHistorySize = 1;
}
public void visit(String url) {
currentIndex++;
if (currentIndex <= visitedWebpages.size()  1) {
visitedWebpages.set(currentIndex, url); // replace. This optimizes time complexity rather than doing add() operation since add(index, value) is O(n) but set() is O(1)
}
else {
visitedWebpages.add(currentIndex, url); // O(1)
}
currentHistorySize = currentIndex + 1;
}
public String back(int steps) {
currentIndex = Math.max(0, currentIndex  steps);
return visitedWebpages.get(currentIndex);
}
public String forward(int steps) {
currentIndex = Math.min(currentHistorySize  1, currentIndex + steps);
return visitedWebpages.get(currentIndex);
}
}
Python code:
class BrowserHistory:
def __init__(self, homepage):
self.visitedWebpages = []
self.visitedWebpages.append(homepage)
self.currentIndex = 0
self.currentHistorySize = 1
def visit(self, url):
self.currentIndex += 1
if self.currentIndex <= len(self.visitedWebpages)  1:
self.visitedWebpages[self.currentIndex] = url
else:
self.visitedWebpages.insert(self.currentIndex, url)
self.currentHistorySize = self.currentIndex + 1
def back(self, steps):
self.currentIndex = max(0, self.currentIndex  steps)
return self.visitedWebpages[self.currentIndex]
def forward(self, steps):
self.currentIndex = min(self.currentHistorySize  1, self.currentIndex + steps)
return self.visitedWebpages[self.currentIndex]
Time Complexity: O(1)
Space Complexity: O(n)
Instructor:
If you have any feedback, please use this form: https://thealgorists.com/Feedback.