Time complexity of loop multiplying the value by two or three

Question

I found the below for-loop in an SO question.

I found that the time complexity is O(n log n) .

How would I find the time complexity if we change k *= 2 to k *= 3 ?

// 'int n' is defined somewhere
int var = 0;
for (int k = 1; k <= n; k *= 2)
   for (int j = 1; j <= n; j++)
      var++;

Answer 1

The time complexity would still be O(n log n) .

For k *= 2 , the log would be base 2.

For k *= 3 , the log would be base 3.

But change in the base of log only affects the result by a constant factor (this can be derived from the fact that log b a = log c a / log c b , for any base c ), which is ignored in big-O notation, thus they have the same time complexity.

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Where do we get log 2 n from anyway?

Well, the values of k goes like this:

1, 2, 4, 8, 16, ..., 2m  for some m, where 2m <= n < 2m+1
= 20 + 21 + 22 + ... + 2m

And of course there are just m+1 ( 0 to m ) terms above, so the loop runs m+1 times.

Now if we can use some basic logarithms to get m in terms of n :

2m = c.n   for some constant c, where 1 <= c < 2
log22m = log2(c.n)
m log22 = log2(c.n)
m.1 = log2c + log2n
m = O(log2c + log2n)
  = O(log2n)               // constant terms can be ignored

We can also follow the exact same approach as above for k *= 3 by just replacing 2 by 3 .

Answer 2

The answer is N×log ₃ N.

To see why, you need to figure out why the answer to the original problem is N×log ₂ N. How many times (let's call it k ) will it execute? It will execute as many times as needed to multiply 2 by itself so that the result would exceed N . In other words, 2 ^k > N. Now apply logarithm to both sides of the expression (the definition of logarithm can be found here ) to see that k > log ₂ N.

The reason that we used 2 as the base of the logarithm is that the base of the exponent on the left was 2 . It is easy to see that if the base of exponen is 3 , applying log ₃ yields the answer to the problem that you are solving.

Answer 3

You can proceed formally and methodically using Sigma notation:

Check the last slides of this document .