A.  it can be used to decide the best algorithm that solves a given problem 
B. 
it determines the maximum size of a problem that can be solved in a given system, in a given amount of time

C.  Both(a) and (b) 
D.  none of the above 
Option: C Explanation : Click on Discuss to view users comments. 
A.  log n 
B.  n 
C.  n^{2} 
D.  n^{n} 
Option: A Explanation : Click on Discuss to view users comments. 
A.  2.15 
B.  3.01 
C.  2.3 
D.  1.78 
Option: A Explanation :
Using Hoffman's algorithm, code for a is 1111; b is 0; c is 110; d is 1110; e is 10. Average code length is
4 x.12 + 1 x .4 + 3 x.15 + 4 x.08 + 2 x.25 = 2.15
Click on Discuss to view users comments. raj said: (6:04am on Wednesday 5th June 2013)
if code for a is 1111 then it must be multiplied by 15 i.e .12x15 refer link http://www.siggraph.org/education/materials/HyperGraph/video/mpeg/mpegfaq/huffman_tutorial.html

A.  n 
B.  log n 
C.  n^{n} 
D.  n^{2} 
Option: A Explanation :
Let us find what is T(4), T(5), T(6) is.
T( 4) = T(3) + T(2)  T(1) = 3 + 2  1 = 4
T(5) = T(4) + T(3)  T(2) = 4 + 3  2 = 5
T(6) = T(5) + T(4)  T(3) = 5 + 4  3 = 6
By induction it can be proved that T(n) = n. Hence order is n.
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A.  constant 
B.  linear 
C.  logarithmic 
D.  exponential 
Option: A Explanation :
Let T(1) = T(2) = T(3) = k (say). Then T(4) = k + k  k = k
T(5) = k+ k k= k.
By mathematical induction it can be proved that T(n) = k, a constant.
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