Question

Sum <- 0  // 1 Operation 
for i <- 1 to n do // 2n operations
 for j <-1 to n do // 2(n-1) operations
  k <-1 // 1 operation
while k < n do // n-1 operations
k <- k *c // 2 operations 
sum <- sum +1  // 2 operations 

In total the number of operations in the code are :

1+2n+2(n-1)+1+(n-1)+2+2 == 5n+3 total # of operations ,

is this how you calculate it , because i understand the concept of each stmt has 3 portions to it ( Comparison, Assignment, Incrementation )

please feel free to correct me is my observations are incorrect

Answer 1

No, that’s probably incorrect.

First of all: You’re thinking too hard. At least for the purposes of Big-O calculations, you can probably treat each assignment as a single operation, no matter whether it’s a constant assignment or a calculated value.

Second of all: You aren’t thinking hard enough. The fourth line is a single operation, but it’s run n * n times, so it should be counted as n^2, not 1. Similarly for other lines in loops.