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Best answer

Please refer this https://gateoverflow.in/1243/gate2007-45

$T(n) = T(\sqrt{n}) +1;$

Let $n = 2^{m} $

$T(2^{m}) = T(\sqrt{2^{m}}) +1;$

$T(2^{m}) = T(2^{m/2}) +1;$

Now let $T(2^{m}) = S(m)$

So $S(m) = S(\frac{m}{2} ) +1$

It is binary search recurrence relation

so $S(m) = O(log m)= O(log log n)$ ($\because n = 2^{m}$)

since $T(n) = T(2^{m}) = S(m)$

T(n) is $O(log log n)$

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