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Topic: interview questions (Read 15088 times) |
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kens
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interview questions
« on: Feb 3rd, 2008, 6:22am » |
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Q.1 "jack jill" will be broken in tokens and then a document id list for 'jack' will be retrieved and similarly an document id list for 'jill' will be retrieved separately. Now how can we find those id's which contain both jack and jill?.
Q.2 there are random arrays on ‘n’ computer. Each of size of RAM of it’s computer. The arrays are not sorted. The problem was to sort the arrays.
Q.2 Imagine you have 10000 computer network and you are the admin of this network. how would you monitor the performance and how would you rectify the problem.
Q.3 If the mean faliure hour is 10,000 and 20 is the mean repair hour.If the printer is used by 100 customer, then find the availability.
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| « Last Edit: Feb 3rd, 2008, 6:39am by kens » |
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kens
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Re: interview questions
« Reply #1 on: Feb 4th, 2008, 12:02pm » |
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5)Given an array of red, green and blue balls arrange them in groups of all red together, greens together and blue together. Do in a single scan of the array.
6)Merge two linked lists without using extra memory.
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R
Senior Riddler
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Re: interview questions
« Reply #2 on: Apr 14th, 2009, 10:49am » |
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on Feb 4th, 2008, 12:02pm, kens wrote:5)Given an array of red, green and blue balls arrange them in groups of all red together, greens together and blue together. Do in a single scan of the array.
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has been repeated here a lot many times. search for it.
Quote:
6)Merge two linked lists without using extra memory.
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Merge two linked list  They are sorted or not?
For sorted case the following code will work and for unsorted case, just append one to the other.
Code:
node* merge(node *head1, node *head2)
{
if(head1 == NULL)
return head2;
if(head2 == NULL)
return head1;
node* temp3 = null, temp1 = head1, temp2 = head2;
if(temp1->data < temp2->data )
{
temp3 = temp1;
temp1 = temp1->next;
}
else
{
temp3 = temp2;
temp2 = temp2->next;
}
node *head = temp3;
while(temp1 != NULL && temp2 != NULL)
{
if(temp1->data < temp2->data)
{
temp3->next = temp1;
temp1 = temp1->next;
temp3 = temp3->next;
}
else
{
temp3->next = temp2;
temp2 = temp2->next;
temp3 = temp3->next;
}
}
if(temp1 == NULL)
{
if(temp2 == NULL)
return head;
else
temp3->next = temp2;
}
else
{
temp3->next = temp1;
}
return head;
}
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| « Last Edit: Apr 14th, 2009, 10:56am by R » |
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The first experience seems like Magic, but the second tells you the Trick behind it.
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R
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Re: interview questions
« Reply #3 on: Apr 14th, 2009, 11:00am » |
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on Feb 3rd, 2008, 6:22am, kens wrote:Q.2 there are random arrays on ‘n’ computer. Each of size of RAM of it’s computer. The arrays are not sorted. The problem was to sort the arrays.
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If am getting it right, there are n very very large arrays (each one is of size of a RAM) we have to sort them.
If they had been sorted separately, then an n-way merge would have done. Still we can sort each array in its own computer and then apply n-way merge for all of them.
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The first experience seems like Magic, but the second tells you the Trick behind it.
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R
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Re: interview questions
« Reply #4 on: Apr 14th, 2009, 11:03am » |
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on Feb 3rd, 2008, 6:22am, kens wrote:Q.1 "jack jill" will be broken in tokens and then a document id list for 'jack' will be retrieved and similarly an document id list for 'jill' will be retrieved separately. Now how can we find those id's which contain both jack and jill?.
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You have now two arrays of IDs, the problem is just finding the intersection of the two arrays. Those will be the documents having both "jack" and "jill"
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The first experience seems like Magic, but the second tells you the Trick behind it.
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master2k10
Newbie
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Re: interview questions
« Reply #5 on: Dec 31st, 2009, 4:47am » |
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1) solution1: merge two arrays and sort.
Traverse the sorted array . if if you find two id's in sequence then print it... continue till the end of the sorted array......
N = n1 + n2; (Combined length of the two arrays)
o(NlogN) for sort.
o(N) for traversal .
complexity o(NlogN)
2) solution2: create a hashtable of doc_id->count pair.
Traverse both array one at the same time fill/update the hashtable.
Traverse hashtable and look for count>1 and print those.
Time Complexity : o(N=(n1+n2));
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bmudiam
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Re: interview questions
« Reply #6 on: Dec 31st, 2009, 8:32am » |
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on Feb 3rd, 2008, 6:22am, kens wrote:
Q.2 Imagine you have 10000 computer network and you are the admin of this network. how would you monitor the performance and how would you rectify the problem.
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At that scale, you could only monitor the machines by their runtime statistics displayed on graphs because a picture is worth 1000 words.
Which problem are you asking about? network or computer?
If its a network problem I will be at the center of all servers and all routers beside me..so I will restart router.
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“Nobody can go back and start a new beginning, but anyone can start today and make a new ending.” - Maria Robinson
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chu082011
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Re: interview questions
« Reply #7 on: Jan 29th, 2012, 5:50pm » |
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Hi,
Thank very much for your reply. It help me to think about for my ideals.
Tks again and pls keep posting.
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jassmee
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Re: interview questions
« Reply #8 on: Apr 6th, 2012, 10:35pm » |
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you provide the better information question related in interview such a good question . it is very helpful in my interview..
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arun gupta
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hello friends :)
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Re: interview questions
« Reply #9 on: Apr 9th, 2012, 10:02pm » |
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One of the most common question for interview is "Tell about your self"? Should i wait for interviewer to ask or i start by self?
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LarryCurtis
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Re: interview questions
« Reply #10 on: Jun 29th, 2013, 12:34pm » |
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intresting questions i hope no one ask me this while in a interview 
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ConnieBrocket
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How to start our introduction
« Reply #11 on: Aug 18th, 2014, 10:52pm » |
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:)
I want to know how to start our introduction during interview.
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alien2
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Re: interview questions
« Reply #13 on: May 16th, 2016, 7:54am » |
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on Apr 9th, 2012, 10:02pm, arun gupta wrote:| One of the most common question for interview is "Tell about your self"? Should i wait for interviewer to ask or i start by self? |
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I think therefore I am. I am a thinker rather than a stinker. I cannot me more or less than myself and I happen to be a beautiful person. So, when do I get started?
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gitanas
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Re: interview questions
« Reply #14 on: Jun 21st, 2016, 3:46am » |
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Anybody works in Microsoft?
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