About

Pursuing M.Tech in Computer Science & Engineering at IIT Kharagpur and Editor at Techtud.

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Alma Mater:

B.tech in Compter Science and Engineering
Kamla Nehru Institute of Technology, Sultanpur
2012 to 2016

Experience:

Not updated.
Techtud Editor
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5 months 2 weeks ago
All the Best for GATE 2018 :)
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Techtud Editor
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5 months 3 weeks ago
Tomorrow is the last date for attempting VirtualGATE-2.
http://virtualgate.in/login/index.php
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Techtud Editor
Example text
6 months 1 week ago

c stands for Complement here.

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Techtud Editor
Example text
6 months 1 week ago

@vikash957 at x= -3, function is not defind ( devide by zero).

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Techtud Editor
Multichoice
6 months 1 week ago
@reenakandari
Question is about non-preemptive kernel design, where a running process can be preempted only when it is being forced for it.
Timer interrupts and Disk interrupts will wait untill the process is running. A blocking system call or an exit request are the high priority kernel requests, so these can not be hold on wait.
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@shivani1234 plz edit the image size in the previous comment. It is not visible well.

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4 May 2017 - 2:06pm

It is given in the question that tree is a binary search tree (BST) so we need not to perform sorting to get inorder traversal. There are 'n' elements from 1,2....n, so this sequence is inorder traversal.
Now for you have both postorder and inorder traversals, so you can form unique tree in O(n) time.

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Yes, you are right. See this for better understanding,
http://www.techtud.com/video-lecture/simple-2-phase-locking-protocol

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@jana96 Excellent explanation. 

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You can apply Master's theorem here.
a = 7, b = 2, k = 2, p = 0
Here a > bk.
So, Complexity =  θ(nlog27)

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@shweta1920 Matrix is of order nxn so we will get auxiliary equation of order n and roots of this equation will be eigen value. Here it is given that there are only two eigen values, so you can say that all the other roots(eigen values) will be zero.

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Note two things in the question:
1. Matrix is symmetric.
2. Rank of matrix is 2. 
Matrix is of order nxn so n-2 eigen  values will be zero. 
So eigen values will be, e1, e2, 0, 0, 0......
We have, . = Trace of (A . AT)  = Trace of (A2)  = e12 + e22 +0+0......
Hence e12 + e22 = 50.
So with the above equation, it can be determined that one eigenvalue must be in [−5,5].
Hence B is the correct option.

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