Thread: Big O - Algorithm Analysis

How do I analyze a for loop that looks like

for (j = 0; j < 2*n; j += 2)

I know that if it was just j++ it would loop 2n times but the += is throwing me off. This loop is nesting in a normal loop of

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

I know this will make it n^2 but I don't know how the j += 2 affects the time.


Thanks for the help,
PeskyToaster

j += 2

just adds 2 to j, it is the same as

 j = j + 2

, and because j is incrementing twice every loop, it is going to reach it's end twice as fast, so it is going to be looping

1/2 * 2n

or

n

times.

And, does the factor of 2x or 0.5x or whatever was worrying you, really matter?

I have another problem. First here's the code snippet. It's just general code for analysis

 
for (j=0; j<2n; j++) {
     i = 1;
    Generic Statement;
    while (j<n) {
          Generic statement;
          Generic statement;
          i = i *2;
   }
}

How does the i = 2n and the j < 2n affect the Big O and execution time. The execution time is just the equation with each statement being one time unit.