design-browser-history

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Algorithms

You have a browser of one tab where you start on the <code>homepage</code> and you can visit another <code>url</code>, get back in the history number of <code>steps</code> or move forward in the history number of <code>steps</code>.

Implement the <code>BrowserHistory</code> class:

<ul>
	<li><code>BrowserHistory(string homepage)</code> Initializes the object with the <code>homepage</code>&nbsp;of the browser.</li>
	<li><code>void visit(string url)</code>&nbsp;Visits&nbsp;<code>url</code> from the current page. It clears up all the forward history.</li>
	<li><code>string back(int steps)</code>&nbsp;Move <code>steps</code> back in history. If you can only return <code>x</code> steps in the history and <code>steps &gt; x</code>, you will&nbsp;return only <code>x</code> steps. Return the current <code>url</code>&nbsp;after moving back in history at most <code>steps</code>.</li>
	<li><code>string forward(int steps)</code>&nbsp;Move <code>steps</code> forward in history. If you can only forward <code>x</code> steps in the history and <code>steps &gt; x</code>, you will&nbsp;forward only&nbsp;<code>x</code> steps. Return the current <code>url</code>&nbsp;after forwarding in history at most <code>steps</code>.</li>
</ul>

&nbsp;
Example:

<pre>
Input:
[&quot;BrowserHistory&quot;,&quot;visit&quot;,&quot;visit&quot;,&quot;visit&quot;,&quot;back&quot;,&quot;back&quot;,&quot;forward&quot;,&quot;visit&quot;,&quot;forward&quot;,&quot;back&quot;,&quot;back&quot;]
[[&quot;leetcode.com&quot;],[&quot;google.com&quot;],[&quot;facebook.com&quot;],[&quot;youtube.com&quot;],[1],[1],[1],[&quot;linkedin.com&quot;],[2],[2],[7]]
Output:
[null,null,null,null,&quot;facebook.com&quot;,&quot;google.com&quot;,&quot;facebook.com&quot;,null,&quot;linkedin.com&quot;,&quot;google.com&quot;,&quot;leetcode.com&quot;]

Explanation:
BrowserHistory browserHistory = new BrowserHistory(&quot;leetcode.com&quot;);
browserHistory.visit(&quot;google.com&quot;); // You are in &quot;leetcode.com&quot;. Visit &quot;google.com&quot;
browserHistory.visit(&quot;facebook.com&quot;); // You are in &quot;google.com&quot;. Visit &quot;facebook.com&quot;
browserHistory.visit(&quot;youtube.com&quot;); // You are in &quot;facebook.com&quot;. Visit &quot;youtube.com&quot;
browserHistory.back(1); // You are in &quot;youtube.com&quot;, move back to &quot;facebook.com&quot; return &quot;facebook.com&quot;
browserHistory.back(1); // You are in &quot;facebook.com&quot;, move back to &quot;google.com&quot; return &quot;google.com&quot;
browserHistory.forward(1); // You are in &quot;google.com&quot;, move forward to &quot;facebook.com&quot; return &quot;facebook.com&quot;
browserHistory.visit(&quot;linkedin.com&quot;); // You are in &quot;facebook.com&quot;. Visit &quot;linkedin.com&quot;
browserHistory.forward(2); // You are in &quot;linkedin.com&quot;, you cannot move forward any steps.
browserHistory.back(2); // You are in &quot;linkedin.com&quot;, move back two steps to &quot;facebook.com&quot; then to &quot;google.com&quot;. return &quot;google.com&quot;
browserHistory.back(7); // You are in &quot;google.com&quot;, you can move back only one step to &quot;leetcode.com&quot;. return &quot;leetcode.com&quot;
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= homepage.length &lt;= 20</code></li>
	<li><code>1 &lt;= url.length &lt;= 20</code></li>
	<li><code>1 &lt;= steps &lt;= 100</code></li>
	<li><code>homepage</code> and <code>url</code> consist of&nbsp; &#39;.&#39; or lower case English letters.</li>
	<li>At most <code>5000</code>&nbsp;calls will be made to <code>visit</code>, <code>back</code>, and <code>forward</code>.</li>
</ul>