[algogeeks] Re: Amazon Question

2012-09-23 Thread xyz
have a look http://amnwidfrenz-thinkalways.blogspot.in/2012/09/string- reduction.html -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email

[algogeeks] Re: THANX ALGOGEEKS !!!!!!

2011-10-24 Thread XYZ
Hey please post the written test questions ! Thanks On Sep 22, 8:29 am, saurabh sah.saurab...@gmail.com wrote: I sincerely thank this group as i got selected in MicrosoftIDConly because of this group . It was a wonderful experience for me at the interviews because the some of questions

[algogeeks] Re: Urgent...plz help

2011-08-17 Thread XYZ
@Dave How did you get the complexity as O(n log k) On Aug 18, 8:14 am, Dave dave_and_da...@juno.com wrote: @Aditi: Form a max-heap of the first 100 numbers. Then as you read the remaining numbers, if a number is less than the root of the max-heap, replace the root with it and restore the heap

[algogeeks] Re: Expression evaluation in C

2011-07-18 Thread XYZ
I was also thinking about the order of expression evaluation in gcc! No idea though! -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/KVN1Omyl4UIJ. To post to

[algogeeks] Re: Google interview question

2011-07-16 Thread XYZ
If the there are problems with hashing method, What about simply sorting the numbers then comparing the adjacent numbers ! Time complexity O(nlogn)+O(n)=O(nlogn) Cheers! -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this

[algogeeks] Re: Interview Question

2011-07-03 Thread XYZ
Either you will have to use hashmaps which means extra storage or compromise on time complexity as nlogn I dont think there is any other possible workaround! -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web