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# Time Complexity

Can anyone please tell me the time complexity of below code and how? for(int i = 1;i<n;i=i*2){ for(int j=1;j<i;j++){ } }

6 Answers

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NB but the inner loop does execute when i > 1. That starts happening upon the second pass of the outer loop. I tried to reduce the inner loop series into O() terms, but my math is too weak.
[The compiler would optimize such a loop into oblivion because there is no work being done inside. In that case it's O(0)!]

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Oh yes that was error in thinking „each second“ and can be replaced.

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In outer for loop variable i iterates from 1 to n times doubling it's value in each iteration so it will execute log₂(n) times and inner for loop is never executed because condition j<i will always give false,so inner for loop does not add to time complexity.
For eg if value of n is 8 then the above code which is only the outer loop,will execute log₂(8)=3 times.

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Yes Brian you are right I mistakenly assumed j=i instead of j=1 in my head

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The outer loop performs n/2 (each second of n) iterations, and the inner loop performs i iterations for each iteration of the outer loop, where i is the current iteration count of the outer loop.
The total number of iterations performed by the inner loop can be calculated by summing the number of iterations performed in each iteration of the outer loop, which is given by the formula
sum(i) from i=1 to outerRange, which is equal to outerRange * (outerRange + 1) / 2
that equals to
n/2 * ((n/2 + 1) / 2) = time complexity

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JaScript, if you squint a little more you'll notice the outer loop incrementer does not add 2; rather it multiplies by 2. So log2(n) is correct, as NB said, for the number of outer loop iterations. The total inner loop iterations is the sum of 0, 1, 3, 7, 15, ..., n/2. Now figure out the general expression of that series summation and it will lead to the complexity solution.