-
Consider the following segment of C-code
int j, n;
j = 1;
while (j < = n)
j = j*2;
The number of comparisons made in the execution of the loop for any n > 0 is
-
-
- n
-
-
-
Correct Option: D
From the statement j = j*2 in the code we get to know that j increases in power of 2’s. Let us say that this statement execute x times then, according to the question for while loop
2^x < = n
Therefore, x < = log2 n
And also for termination of while loop there will be an extra comparison required. Thus, total number of comparisons = x + 1 
