About

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

Role

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 6 days 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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Ok, will try to do this asap.

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Let T(N) be the number of distinct ordered trees that can be constructed from a fixed set of N (labeled) nodes.

Base Case: N = 1. If we have exactly one node we can construct only one tree where that node is the root. So T(1) = 1.

Second Case: N = 2. There are two choices for the root node. The remaining node is necessarily the first and only child of the root. So T(2) = 2.

Third Case: N = 3. There are now three choices for the root node. Once we've selected a root node, we again have two cases:

  • Case A: The root node has two children, each of which is an ordered tree with two elements. There are two ways we could order the two remaining nodes. So there are 3*2 = 6 possible ordered trees with three nodes given that the root node has two children.

  • Case B: The root node has one child, which is necessarily an ordered tree with two elements. There are T(2) = 2 different ways we could construct an ordered tree from the remaining two elements, so there are a total of 3*2 = 6 possible ordered trees with three nodes given that the root node has only one child.

These two subcases cover all the possibilities and they don't overlap (they partition the possible ordered trees with three nodes), so we can just add them: T(3) = 6 + 6 = 12.

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Ordered trees are the binary tree having some labels on the nodes.

Consider 2 nodes.

For unordered trees, you can make only two structure of trees.

(Root + left child) or (Root + right child)

Now consider Ordered trees case,

Suppose we label nodes as A,B.

Now in the two possible structures we can put A and B in 4 ways.

1. A(root) + B(left child)

2. A(root) + B ( right child)

3. B(root) + A( left child)

4. B(root) + A(right child)

I hope got the difference.

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I will suggest to put this part at the least precedence. Try to complete technical part first. Vocab part is very unpredictable. Sometimes, all the questions asked are very esay and sometimes those are hard. But it depend on individuals. 

I will say to solve vocab questions from virtual gate tests and previous yr GATE. Few websites like, Campus gate, indiabix are also helpful.

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The outer for loop iterates log n times.
For first iteration of the outer for loop, inner for loop iterates n times.
For second iteration of the outer for loop, inner for loop iterates n/2 times.
For third iteration of the outer for loop, inner for loop iterates n/4 times...

Therefore, the total number of iterations of the inner for loop or the total number of times the statement count += 1executes equals to
n + n/2 + n/4 + ... + 1 (total log n number of terms) = 2n - 1 = O(n)
Therefore, time complexity is O(n).

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