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1:  

The running time of an algorithm T(n),where 'n' is the input size, of a recursive algorithm is given as follows.is given by

          T(n) =c + T(n - 1), if n > 1

            d, if n  ≤  1
The order of this algorithm is

 

A.

n2
B.

n
C.

n3
D.

nn
 
Answer Report Discuss
 

Option: B

Explanation :

By recursively applying the relation we finally arrive at
T(n-I) = c(n-I) + T (1) = c(n-I)+d
So, order is n

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bhawna said: (5:47pm on Monday 30th September 2013)
don't understand hat the exact solution what is c and d here
sakshi said: (8:34pm on Tuesday 8th August 2017)
don't understand pls explain in details

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 ≤  1
The order of this algorithm is

 

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2:  
Six files Fl, F2, F3, F4, F5 and F6 have 100,200,50,80, 120, 150 number of records respectively. In what order should they be stored so as to optimize access time? Assume each file is accessed with the same frequency.
 
A.

 F3, F4, FI, F5, F6, F2
B.

 F2, F6, F5, Fl, F4, F3
C.

Fl, F2, F3, F4, F5, F6
D.

Ordering is immaterial as all files are accessed with the same frequency
 
Answer Report Discuss
 

Option: A

Explanation :

Since the access is sequential, greater the distance, greater will be the access time. Since all the files are referenced with equal frequency, overall access time can be reduced by arranging them as in option (a).

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3:  
In  previous question,  the average access time will be
 
A.

268 units
B.

 256 units
C.

293 units
D.

210 units
 
Answer Report Discuss
 

Option: A

Explanation :

Since each file is referenced with equal frequency and each record in a particular file can be referenced with equal frequency, average access time will be (25 + (50 + 40) + (50 + 80 + 50) + ... )/6 = 268 (approximately).

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rama said: (3:28pm on Wednesday 13th February 2013)
from where did u take these values?
sakshi said: (8:41pm on Tuesday 8th August 2017)
from where u get these values pls explain me

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4:  
An algorithm is made up of 2 modules Ml and M2. If order of M1 is f(n) and M2 is g(n) then the order of the algorithm is
 
A.

max (f (n) ,g (n))
B.

min (f (n) ,g (n) )
C.

f (n) + g (n)
D.

f (n) x g (n )
 
Answer Report Discuss
 

Option: A

Explanation :

 

By definition of order, there exists constants c1, ec2, n1, n2 such that
T (n)   ≤ c1 x f (n) , for all n  ≥ n1.
T (n)   ≤ c2 x f (n) , for all n  ≥ n2.
N = max (n1, n2) andC = max (c1, c2). So,
T (n)  ≤ C x f(n) for all n  ≥ N
T (n)  ≤  C x g (n) for all n  ≥  N
T(n) ≤ C/2 x (f(n)+g(n))
Without loss of generality, let max ( f (n), g (n)) =f(n) .
So, T (n)   ≤ C / 2 (f(n) + f (n) ) ≤ C x f(n) .
So, order is  f(n) , which is max (f(n), g (n) ) , by our assumption.

 

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5:  
There are 4 different algorithms AI, A2, A3, A4 to ·solve a given problem with the order Iog(n), log(log(n)), nIog(n) ,n / log(n) respectively. Which is the best algorithm?
 
A.

A1
B.

A2
C.

A4
D.

A3
 
Answer Report Discuss
 

Option: B

Explanation :

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Mayank said: (10:47pm on Friday 8th March 2013)
According to me, The correct answer is: B (A2), because it gives the least value while calculating the log(log(n)) for any value of n.please justify me the reason for selecting the answer as C (A4).

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Syllabus covered in this section is-

  • Abstract data types
  • Algorithms
  • Stacks, Queues
  • Linked Lists
  • Trees, Binary search trees
  • Binary heaps

This Section covers Data Structures Questions Answers using C language .
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