Compute the "Time Complexity"

Question

I hope you are doing well,

I am computing the Time Complexity of my algorithm which has three nested for, but I did a trick by putting an if in the latest for like:

for (i=0 ; i<n1 ; i++){
    for (j=0 ; j<n2 ; j++){
        for (k=0 ; k<n3 ; k++){
            if (A[i][j][k] == true){
               ...
            }
        }
     }
 }

So, if A[i][j][k] is false then it will be skipped and no computation time will be used.

The question that I have is that: Whether we are skipping some parts, again the complexity of the algorithm is O(n1*n2*n3) and when n1=n2=n3=n, is it O(n^3)?

Thanks for your time.

If we follow you and all entries of A are false, then the running time will be 0. — , Nov 22 '20 at 10:39
It might in theory be possible that, in practice, a smart special compiler can prove that this code segment can never enter (if all is false) and then simplify it by removing. Depends on the language and compiler of course. But this is definitely not the case for most and also not subject to big-o analysis. — Zabuzard, Nov 24 '20 at 08:34

Deepak Tatyaji Ahire · Accepted Answer · 2020-11-19T18:27:48.523

If you are performing constant time operation in the if condition:

The complexity stills remains O(n^3) because the condition and the afterthought parts of the for loop are still getting executed.

The condition checks , i.e., i<n1, j<n2, and k<n3 and the increment operations such as i++, j++, and k++ still take place.

So, these computations are not skipped.

If you are not performing constant time operation in the if condition:

Then time complexity will depend on the operations performed in for loop.

Compute the "Time Complexity"

1 Answers1