GATE CS 2024 Set 1 Q.42: Asymptotic Analysis and Time/Space Complexity

Consider the following recurrence relation:

T(n) = \begin{cases} \sqrt{n}T(\sqrt{n}) + n & \text{for } n \ge 1, \\ 1 & \text{for } n = 1. \end{cases}

Which one of the following options is CORRECT?

$T(n) = \Theta(n \log \log n)$

$T(n) = \Theta(n \log n)$

$T(n) = \Theta(n^2 \log n)$

$T(n) = \Theta(n^2 \log \log n)$

Official answer and solution

Solution

Divide both sides of the recurrence by $n$ and define $S(n) = T(n)/n$ to reduce it to $S(n) = S(\sqrt{n}) + 1$ .

Dividing both sides of $T(n) = \sqrt{n}T(\sqrt{n}) + n$ by $n$ gives:

\frac{T(n)}{n} = \frac{T(n^{1/2})}{n^{1/2}} + 1

Let $S(n) = \frac{T(n)}{n}$ . Then $S(n) = S(n^{1/2}) + 1$ . Substituting $n = 2^{2^k}$ (so $k = \log_2 \log_2 n$ and $n^{1/2} = 2^{2^{k-1}}$ ):

S(2^{2^k}) = S(2^{2^{k-1}}) + 1 = \dots = S(2) + k = \Theta(\log \log n)

Multiplying back by $n$ gives:

T(n) = n \cdot S(n) = \Theta(n \log \log n)

Final answer

Option (A)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

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