Self Doubt

Question

What would be the worst case time complexity of an unreachable code?

Let's say there are two parts to the code, where first part is if (0) with complexity O(n) and the else part with O(1).

 

Something like this:

if (0) {

 Time complexity O(n)

}

else

    return

 

Logically, it would be O(1) because we'll never reach the fragment of code inside if(0).

When we calculate worst case time complexity do we consider only the theoretical aspects or we deduce it logically?

Is there any resource to support this? How is time complexity calculated for unreachable code?

 

Thanks

Registered user 48

 

 

Comments

we donot calculate TC of a part of code

rather, we find TC for whole code

So, here TC will be time complexity of else part

---

let my code is

if()

{

      takes O( n ) times

}

else

{

    takes O( log n ) times

}

what is the time complexity?

TC = either O(n) or O( log n )

let assume this code in for loop, and for loop run 'j' times, but note that 'p' times only if condition satisfies.

then TC =  ( p . O( n ) + (j-p) O( log n ) ) = p . O( n ) + (j-p) . O( log n )

what it internally means?

if loop runs p times, and each time it requires O(n) ===> p . O(n)

else loop runs (j-p) times, and each time it requires O( log n ) ===> (j-p) . O( log n )

 

Now, as per your question, if loop runs 0 times ===> 0 . O(n) = 0

else runs 1 time with O(1) = 1 * O(1) = O(1)

TC = 0 + O(1) = O(1)

---

Thanks srestha and Shaik.
 

Thanks for clearing up things.