GATE | GATE-CS-2009 | Question 35

The running time of an algorithm is represented by the following recurrence relation:

    if  n <= 3  then   T(n) = n
    else T(n) = T(n/3) + cn

Which one of the following represents the time complexity of the algorithm? <pre>
(A) \theta(n)
(B) \theta(n log n)
(C) \theta(n^2)
(D) \theta(n^2log n) </pre>

(A) A
(B) B
(C) C
(D) D

Answer: (A)



T(n) = cn + T(n/3)
     = cn + cn/3 + T(n/9)
     = cn + cn/3 + cn/9 + T(n/27)
Taking the sum of infinite GP series. The value of T(n) will
be less than this sum.
T(n) <= cn(1/(1-1/3))
     <= 3cn/2

or we can say 
cn <= T(n) <= 3cn/2
Therefore T(n) = \theta(n)

This can also be solved using Master Theorem for solving recurrences. The given expression lies in Case 3 of the theorem.

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