Question

I just want to make sure if I am doing this correct. I am trying to count the number of operations performed for the worst case scenario in java

int sum = 0;
    for (int i = 0; i < n; i++ )
    sum++;

Is the number of operations 2+3n or 3+3n?

I got the answer from counting int sum = 0 and int i = 0 for the “2” and i < n, i++, and sum++ as the “3n”. Or is it a 3 rather than a 2 because I have to count i < n before going through the loop?

But either way, is the theta characterization going to be Θ(n)?

Now what if there is a nested for loop like this:

int sum = 0;
    for (int i = 0; i < n; i++ )
        for (int a = 0; a < i; a++)
            sum++;

would it be 3+n*(6a+2) = 6na+2n+3? with Θ(n^2)?

if i change the inner for loop from a < i to a < i*i, would the equation still hold as above or change?

Answer 1

Maybe it’s easier to count the number of executions of each statement if there’s only one per line:

int sum = 0;     // 1 time
int i = 0;       // 1 time
while (i < n) {  // n+1 times
  sum++;         // n times
  i++;           // n times
}

Hence, T(n) = 3*n+3 = Θ(n).

int sum = 0;       // 1 time
int i = 0;         // 1 time
while (i < n) {    // n+1 times
  int a = 0;       // n times
  while (a < i) {  // 1 + 2 + ... + n = n*(n+1)/2 times
    sum++;         // 0 + 1 + ... + n-1 = n*(n-1)/2 times
    a++;           // 0 + 1 + ... + n-1 = n*(n-1)/2 times
  }
  i++;             // n times
}

Hence, T(n) = 3*n+3 + n*(n-1) + n*(n+1)/2 = Θ(n^2).