Time complexity of the code containing condition

问题

foo(int n)
{
   int s=0;
   for(int i=1;i<=n;i++)
     for(int j=1;j<=i*i;j++)
        if(j%i==0)
          for(k=1;k<=j;k++)
            s++;
}

What is the time complexity of the above code?

I am getting it as O(n^5) but it is not correct.

回答1:

The complexity is O(n^4).

Innermost loop will be executed i times for each i. (i multiples of i within 0..i*i)

It will be like the inner loop will run for

j = 0 1 2...i i+1 ...2*i ....3*i .... 4*i .... 5*i... i*i
            x         x       x        x        x      x
    \------/\--------/\-------/                 \------/

These x denotes the execution of the innermost for loop with complexity j. Rest of the time this is not touched and just the test is done and it fails.

So now check the thing, these \-----/ has i*j (j = 1,2,3...i) looping and i checks.

And now we do i times precisely.

So total work = i*(1+1+1+...1) + i*(1+2+3+..i)
              = i*i+ i*i*(i+1)/2 ~ i^3

With the outer loop it will be n^4.

Now what is the meaning of it. The whole work can be divided in like this

O(i*j+i)
  ^^^ ^
   |   The other cases when it simply skips
  The innermost loop executed

Now if we iterate over j then it will have complexity O(n^3).

With added external loop it will be O(n^4).

回答2:

Your function computes 4-dimensional pyramidal numbers (A001296). The number of increments to s can be computed using this formula:

a(n) = n*(1+n)*(2+n)*(1+3*n)/24

Therefore, the complexity of your function is O(n⁴).

The reason it is not O(n⁵) is that if (j%i == 0) proceeds with the "payload" loop only for multiples of i, of which we have exactly i among all js in the range from 1 to i², inclusive.

Hence, we add one for the outermost loop, one for the loop in the middle, and two for the innermost loop, because it iterates up to i², for the total of 4.

Why only one for middle (j) ? It runs up to i² right?

Perhaps it would be easier to see if we rewrite the code to exclude the condition:

int s=0;
for(int i=1;i<=n;i++)
    for(int j=1;j<=i;j++)
        for(int k=1;k<=i*j;k++)
            s++;
return s;

This code produces the same number of "payload loop" iterations, but rather than "filtering out" the iterations that skip the inner loop it removes them from consideration by computing the terminal value of k in the innermost loop as i*j.

来源：https://stackoverflow.com/questions/47836794/time-complexity-of-the-code-containing-condition

标签

algorithm

time-complexity