Reputation: 11
i, j, N, sum is all integer type. N is input.
( Code1 )
i = N;
while(i > 1)
{
i = i / 2;
for (j = 0; j < 1000000; j++)
{
sum = sum + j;
}
}
( Code2 )
sum = 0;
d = 1;
d = d << (N-1);
for (i = 0; i < d; i++)
{
for (j = 0; j < 1000000; j++)
{
sum = sum + i;
}
}
How to calculate step count and time complexity for a Code1, Code2?
Upvotes: 0
Views: 2564
Reputation: 11577
to calculate the time complexity try to understand what takes how much time, and by what n
are you calculating.
if we say the addition ("+") takes O(1) steps then we can check how many time it is done in means of N
.
the first code is dividing i
in 2 each step, meaning it is doing log(N) steps. so the time complexity is
O(log(N) * 1000000)= O(log(N))
the second code is going form 0 to 2 in the power of n-1 so the complexity is:
O(s^(N-1) * 1000000)= O(2^(N-1))
but this is just a theory, because d has a max of 2^32/2^64 or other number, so it might not be O(2^(N-1)) in practice
Upvotes: 1