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Amazon Interview Question for Software Engineer / Developers


Team: Hyderabad
Country: India
Interview Type: Phone Interview



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sed?

- Anonymous March 14, 2012 | Flag Reply
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Textpad has some options for doing it bulk.

At the end of the day it is through regular expression find and replace the pattern will be faster :-)

- howaboutthis March 14, 2012 | Flag Reply
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1- have read write pointers on the string to replace space
2- Fill the memory to minimize disk access
3- If block in the memory is not changed dont write it back, keep a clean dirty bit + location info for %5 and just write that part back instead of whole block

- appdev March 14, 2012 | Flag Reply
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Dude was this question asked for Freshers?

- VIneet Setia March 14, 2012 | Flag Reply
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No, this question asked for 5+ exp in java

- Hussain March 14, 2012 | Flag
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Shall we use String.replaceAll method for that?

- flyinocean March 14, 2012 | Flag Reply
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I started my answer with available java api, but the interviewer told not to use api as api always has its own overhead.

- Hussain March 14, 2012 | Flag
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the chance of you writing something that's more efficient than the JDK? very slim. the interviewer should have said "this is an interview question i want you to do it manually."

- jeff April 05, 2012 | Flag
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1. Regular Expression? Here's a Java Snippet

public class RegularExpression
{
	public static void regularExpression(String s)
	{
		Pattern p = Pattern.compile("%20");
		Matcher m = p.matcher(s);
		System.out.println(m.replaceAll(" "));
	}
}

- movingincircles March 15, 2012 | Flag Reply
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I would appreciate if you provide the complexity as well.

- Test March 15, 2012 | Flag
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You write this kind of code when you are at work, but in the interview you are not allowed to use these function

- Hieu March 15, 2012 | Flag
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I believe they will not allow you to use built-in library, for example Replace, Regex. You have to write your own algorithm, you can use KMP to answer this question

- Hieu March 15, 2012 | Flag Reply
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Yes, You are right. They told me to use my own algorithm.

- Hussain April 05, 2012 | Flag


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