Time Complexity of an Algorithm (Nested Loops)

Question

I'm trying to figure out the time complexity of this pseudocode given algorithm:

sum = 0;
for (i = 1; i <= n; i++)
    for (j = 1; j <= n / 6; j++)
        sum = sum + 1;

I know that the first line runs

n times

But I'm not sure about the second line.

Answer 1

Using Sigma notation, we can find the asymptotic bounds of your algorithm as follows:

Answer 2

Here you have a simple double loop:

for i=1;i<=n;i++
   for j=1; j<=n/6; j++

so if you count how many times the body of the loop will be executed (ie how many times this line of code sum = sum + 1; will be executed), you will see it's:

n*n/6 = n²/6

which in terms of big-O notation is:

O(n²)

because we do not really care for the constant term, because as n grows, the constant term makes no (big) difference if it's there or not!

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When and only when you fully realize what I am saying, you can go deeper with this nice question: Big O, how do you calculate/approximate it?

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However, please notice that such questions are more appropriate for the Theoretical Computer Science , rather than SO.

Answer 3

您进行n * n / 6次操作，因此时间复杂度为O（n ^ 2/6）= O（n ^ 2）。