UGC NET CSE | June 2014 | Part 3 | Question: 63

Question

The solution of the recurrence relation of $ T(n)=3T\left(floor \left(\frac{n}{4}\right)\right)+n$ is

• A. $O(n^{2})$
• B. $O(n/g n)$
• C. $O(n)$ 
• D. $O(l g n)$ 

Answer 1

T(n) = 3T(floor (n/4))+ n  ~ 3T(n/4)+n

Apply Masters theorem

n ^ (log ₄ 3)  < n

Hence T(n) = Θ(n)

If T(n) = Θ(n) then T(n) = O(n)

If T(n) = Θ(n) then T(n) = O(n ² )

answer can be A and C

Correct me if i am wrong

Comments

why O(n²) ?

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f(n) = O(g(n)) means that if there exist positive constants c and n0 such that 0 ≤ f(n) ≤ c g(n) for all n> n0.

Simply speaking, for a value n, f(n) will be always less than or equal to g(n).

In our case f(n) =n; then g(n) can be any thing that has value greater than equal to f(n)

Example:

• n= O(n)
• n= O(n²)
• n/1000= O(n²)
• n=  O(n ³)
• n² = O(n²)
• n²+n = O(n²)