Comments on: Google Phone Interview (Part 1) http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/ Why am I fighting to live, if I am living to die ? Thu, 12 Nov 2009 22:54:29 +0000 http://wordpress.com/ hourly 1 By: jhon http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-28377 jhon Fri, 26 Jun 2009 00:41:43 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-28377 Hi, I had my interview with Google for associate product manager .. you can read the interview here http://ferozeh.com/Interviews/Google/google.php hope it is useful to others.. Hi,
I had my interview with Google for associate product manager .. you can read the interview here

http://ferozeh.com/Interviews/Google/google.php

hope it is useful to others..

]]> By: Naren http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-27812 Naren Wed, 28 Jan 2009 07:50:18 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-27812 Hi, I'm T. Naren from India. I'm pursuing PG diploma in IT. I want to do MS(IT) and then I want to join in Google. Could you please tell me how to prepare for the Technical round. If possible tell me what kind of questions they will ask. Thank you. Hi, I’m T. Naren from India. I’m pursuing PG diploma in IT. I want to do MS(IT) and then I want to join in Google. Could you please tell me how to prepare for the Technical round. If possible tell me what kind of questions they will ask.
Thank you.

]]> By: Cmonkey http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-25306 Cmonkey Wed, 27 Feb 2008 18:05:00 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-25306 The coin denomination solution that you've posted isn't correct... here is a better version: http://cmonkeyfreeware.blogspot.com/2008/02/coin-denomination-problem.html The coin denomination solution that you’ve posted isn’t correct…
here is a better version:
http://cmonkeyfreeware.blogspot.com/2008/02/coin-denomination-problem.html

]]> By: One Fab Fit Mom - Weight Loss Success! » Want to interview at Google? Check out these unbelievable… http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-24863 One Fab Fit Mom - Weight Loss Success! » Want to interview at Google? Check out these unbelievable… Wed, 23 Jan 2008 22:18:41 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-24863 [...] http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/ [...] [...] http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/ [...]

]]> By: Marcus http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-23716 Marcus Sat, 17 Nov 2007 14:35:42 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-23716 You cant run though both lists at the same time cause there are not ordered. You cant run though both lists at the same time cause there are not ordered.

]]> By: tanay http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-21558 tanay Mon, 10 Sep 2007 14:14:52 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-21558 plz tell me what do they ask on telefone ....i have applied for soft. engg., as a fresher graduate. plz tell me what do they ask on telefone ….i have applied for soft. engg., as a fresher graduate.

]]> By: Michael http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-21315 Michael Mon, 03 Sep 2007 12:38:42 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-21315 Yeah - can't edit. The answer is, assuming the integers are sorted, to walk them at the same time: List l1, l2; List ans = /*new List*/; byte toInc=3; /*Bitmask*/ int lastRead1; int lastRead2; for (int pos1=0,pos2=1,n1=l1.size(), n2=l2.size(); pos1 < n1 && pos2 < n2; ) { if ((1&toInc) != 0) { lastRead1 = l1[pos1++]; } if ((2&toInc) != 0) { lastRead2 = l2[pos2++]; } if (lastRead1 == lastRead2) { ans.push(lastRead1); toInc=3; } else if (lastRead1 < lastRead2) { toInc=1; } else { toInc=2; } } return ans; } Probably better to only store offsets into an existing list too and to count backwards if order is not important. Yeah – can’t edit.

The answer is, assuming the integers are sorted, to walk them at the same time:
List l1, l2;
List ans = /*new List*/;
byte toInc=3; /*Bitmask*/
int lastRead1;
int lastRead2;
for (int pos1=0,pos2=1,n1=l1.size(), n2=l2.size(); pos1 < n1 && pos2 < n2; ) {
if ((1&toInc) != 0) {
lastRead1 = l1[pos1++];
}
if ((2&toInc) != 0) {
lastRead2 = l2[pos2++];
}
if (lastRead1 == lastRead2) {
ans.push(lastRead1);
toInc=3;
} else if (lastRead1 < lastRead2) {
toInc=1;
} else {
toInc=2;
}
}
return ans;
}

Probably better to only store offsets into an existing list too and to count backwards if order is not important.

]]> By: Michael http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-21314 Michael Mon, 03 Sep 2007 12:37:35 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-21314 The answer is, assuming the integers are sorted, to walk them at the same time: List l1, l2; List ans = /*new List*/; byte toInc=3; /*Bitmask*/ int lastRead1; int lastRead2; for (int pos1=0,pos2=1,n1=l1.size(), n2=l2.size(); pos1 < n1 && pos2 < n2; ) { if ((1&toInc) != 0) { lastRead1 = l1[pos1++]; } if ((2&toInc) != 0) { lastRead2 = l2[pos2++]; } if (lastRead1 == lastRead2) { ans.push(lastRead1); toInc=3; } else if (lastRead1 < lastRead2) { toInc=1; } else { The answer is, assuming the integers are sorted, to walk them at the same time:
List l1, l2;
List ans = /*new List*/;
byte toInc=3; /*Bitmask*/
int lastRead1;
int lastRead2;
for (int pos1=0,pos2=1,n1=l1.size(), n2=l2.size(); pos1 < n1 && pos2 < n2; ) {
if ((1&toInc) != 0) {
lastRead1 = l1[pos1++];
}
if ((2&toInc) != 0) {
lastRead2 = l2[pos2++];
}
if (lastRead1 == lastRead2) {
ans.push(lastRead1);
toInc=3;
} else if (lastRead1 < lastRead2) {
toInc=1;
} else {

]]> By: ntrphanikumar http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-8545 ntrphanikumar Thu, 03 May 2007 05:17:15 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-8545 hi is the answer to hash??? hi is the answer to hash???

]]> By: Lee http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-8512 Lee Wed, 02 May 2007 12:43:41 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-8512 Thanks for this post. Many interview questions can be found at http://www.technical-interview.com Thanks for this post. Many interview questions can be found at http://www.technical-interview.com

]]> By: ThanhSon http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-4573 ThanhSon Sat, 07 Apr 2007 00:06:06 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-4573 isn't the problem equal to: given two lists of integers (document id). find the ids that appear in both lists. ? isn’t the problem equal to: given two lists of integers (document id). find the ids that appear in both lists. ?

]]> By: Muhammad Rizwan Asghar http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-3371 Muhammad Rizwan Asghar Wed, 21 Mar 2007 07:13:01 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-3371 It is reply to raster's comment. I think raster is wrong to conclude the complexity O(n1). It may be O(n2) if Saeed has more docs than that of Haroon. So, O(n1+n2) represent the complexity in a right way. It is reply to raster’s comment.

I think raster is wrong to conclude the complexity O(n1). It may be O(n2) if Saeed has more docs than that of Haroon. So, O(n1+n2) represent the complexity in a right way.

]]> By: raster http://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-1394 raster Sat, 06 Jan 2007 09:19:05 +0000 https://haroonsaeed.wordpress.com/2006/07/17/google-phone-interview-part-1-2/#comment-1394 If you iterate through the list of docs containing "Haroon" and generated a subset of docs which has "Saeed" in it, you will get the list of docs containing both "Haroon" and "Saeed". Thats O(n1). Does that sound right? If you iterate through the list of docs containing “Haroon” and generated a subset of docs which has “Saeed” in it, you will get the list of docs containing both “Haroon” and “Saeed”. Thats O(n1). Does that sound right?

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