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Still trying to find out more about myself

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Alma Mater:

B.E. in Compter Science and Engineering
Savitribai Phule Pune University
2009 to 2013

Experience:

Intern
Bentley Systems
2014 to 2015
Software Developer
Startup
2015 to 2016
Tud
Answer
2 years 2 months ago

I think, it should be c). In of case of static hashing, the number of buckets is static i.e. it does not change. So, I don't think the search time will increase.

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Tud
Answer
2 years 2 months ago

For the given string, consider X=011101, W=1, W^R=1. It is clear that this is a regular language. W^R means reverse of string W. Reverse of 101 cannot be 1011 :) Since, x belongs to (a+b)*, you can consider WXW^R as a language accepting all the strings where first and last alphabets are same.

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Tud
2 years 3 months ago
It should be C) 3 does not have an inverse.
The inverse is an element of a group such that when you apply the operation between the element and the inverse, you get back the identity element. Now, here the identity element is 1. So when you perform (3*7) modulo 10, you get back 1. Hence, 3 has an inverse. In the same way, you can check other choices.
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Tud
2 years 3 months ago
I think it will return the first node in the linked list.
first will be recursively called untill the last node. After that null will be returned to the calling func. In the same way, each node will return to the calling function and ultimately the function will return the first node.
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Tud
2 years 3 months ago
All except 1)
Check 1st option=> Push 1,2,3,4=> Pop 4,3=> Push 5=> Pop 5=> Now you cannot pop 1 in this sequence, hence it is not a valid sequence.
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I think, it should be c). In of case of static hashing, the number of buckets is static i.e. it does not change. So, I don't think the search time will increase.

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5 May 2016 - 4:45pm

For the given string, consider X=011101, W=1, W^R=1. It is clear that this is a regular language. W^R means reverse of string W. Reverse of 101 cannot be 1011 :) Since, x belongs to (a+b)*, you can consider WXW^R as a language accepting all the strings where first and last alphabets are same.

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It should be C) 3 does not have an inverse.
The inverse is an element of a group such that when you apply the operation between the element and the inverse, you get back the identity element. Now, here the identity element is 1. So when you perform (3*7) modulo 10, you get back 1. Hence, 3 has an inverse. In the same way, you can check other choices.
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13 Apr 2016 - 2:29pm
I think it will return the first node in the linked list.
first will be recursively called untill the last node. After that null will be returned to the calling func. In the same way, each node will return to the calling function and ultimately the function will return the first node.
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12 Apr 2016 - 10:56am
All except 1)
Check 1st option=> Push 1,2,3,4=> Pop 4,3=> Push 5=> Pop 5=> Now you cannot pop 1 in this sequence, hence it is not a valid sequence.
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16 Dec 2015 - 12:00pm

Stack Space is required for head and body recursion because they need to do some work after the recursive calls. But in Tail recursion, the recursive call is the last thing, a function does. And after the recursive call, no work is done, so tail recursion does not require any stack space.

Check this link for how tail recursion works:https://www.quora.com/What-is-tail-recursion-Why-is-it-so-bad

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2 Dec 2015 - 1:01pm

Here, do we delete the additional tuples after the tuple 4,3 is deleted(on-delete cascade) or is it the case that in order to delete 4,3 we need to delete the dependent tuples first(as mentioned in the link)?

 

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One doubt about the question, do we also need to consider the context switch which occurs initially when p1 starts executing and at last after p4 has finished executing?

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Thanks for clearing the the concept.

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