Time complexity mcq

#1

Please explain solution.
Find the Complexity of the given code.

int gcd(int n, int m)
{
if (n%m ==0) return m;
if (n < m) swap(n, m);
while (m > 0)
{
n = n%m;
swap(n, m);
}
return n;
}

ans : Θ(logn)

and this question also : Please explain this mcq question :
lg (n!) = … ?
ans : O(n lg n)

#2

hi poorva
for gcd()
the complexity is logn
because the n is decrease each time by m
like take example
if n is decrease by 2 u have time complexity
log base 2 (n)
here exact complexity will be
log base m(n)

1 Like
#3

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#4

please explain second question also
log(n!) = ?

#5

Can u tell me the recurrence relation of that