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 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.

Logic

Take care of edge case of nil pointers while iterating
keep code simple to understand
Read the problem carefully.
Dry run it for entire example

Golang Code

/**
 * 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);
 */

type Node struct {
	url  string
	next *Node
	prev *Node
}

type BrowserHistory struct {
	current *Node
}

func Constructor(homepage string) BrowserHistory {
	return BrowserHistory{
		current: &Node{
			url: homepage,
		},
	}
}

func NewNode(url string) *Node {
	return &Node{
		url: url,
	}
}

func (this *BrowserHistory) Visit(url string) {
	newNode := NewNode(url)
	this.current.next = nil
	this.current.next = newNode
	newNode.prev = this.current
	this.current = newNode
	return
}

func (this *BrowserHistory) Back(steps int) string {
	if this.current != nil && this.current.prev == nil {
		return this.current.url
	}

	prev := this.current
	temp := this.current.prev
	for i := 0; i < steps && temp != nil; i++ {
		prev = temp
		temp = temp.prev
	}

	this.current = prev
	return this.current.url
}

func (this *BrowserHistory) Forward(steps int) string {
	if this.current != nil && this.current.next == nil {
		return this.current.url
	}

	next := this.current
	temp := this.current.next
	for i := 0; i < steps && temp != nil; i++ {
		next = temp
		temp = temp.next
	}

	this.current = next
	return this.current.url
}

Output

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"]

Optimised Solution

/**
 * 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);
 */

type Node struct {
	url  string
	next *Node
	prev *Node
}

type BrowserHistory struct {
	current *Node
}

func Constructor(homepage string) BrowserHistory {
	return BrowserHistory{
		current: &Node{
			url: homepage,
		},
	}
}

func NewNode(url string) *Node {
	return &Node{
		url: url,
	}
}

func (this *BrowserHistory) Visit(url string) {
	newNode := NewNode(url)
	this.current.next = nil
	this.current.next = newNode
	newNode.prev = this.current
	this.current = newNode
	return
}

func (this *BrowserHistory) Back(steps int) string {
	// pay attention to for loop
	for steps > 0 && this.current.prev != nil {
		this.current = this.current.prev
		steps--
	}

	return this.current.url
}

func (this *BrowserHistory) Forward(steps int) string {
	// pay attention to for loop
	for steps > 0 && this.current.next != nil {
		this.current = this.current.next
		steps--
	}

	return this.current.url
}

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CODE & ALGO

Design Browser History

Table of Contents

Problem Statement

Logic

Golang Code

Output

Optimised Solution

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