Time complexity - recursive call

Question

I am trying to understand how to compute the time complexity of an algorithm .

I have this piece of code: This is the whole method:

public void solve(int i) {
    if(i < 2) {
        return;
    }
    solve(i-1); //recursive call
    int x = v[n-i];
    for(int j = n-i+1; j < n; j++) {
        if(x > v[j]) {
            count++;
        }
    }
    return;
}

I think the complexity is O(n). Am I right?

Thanks

Okay at least now we have the source of `solve`. What's `n`? What's `v`? — T.J. Crowder, Dec 01 '13 at 12:05
Thanks. n is a number read from a file and v is an array with n elements. — user3054293, Dec 01 '13 at 12:09

Alexey A. · Accepted Answer · 2013-12-01T12:27:20.817

0

The complexity should be O(N^2) as you will have N + (N-1) + (N-2) + 1 = N ( N + 1 ) / 2 iterations in worst case.

edited Dec 01 '13 at 12:27

answered Dec 01 '13 at 12:03

Alexey A.

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Thanks, i'll check the article – user3054293 Dec 01 '13 at 13:06

score 0 · Answer 2 · answered Dec 01 '13 at 12:14

0

The complexity is O(i^2), n here no matter.

The function will run i times (until i<2). each iteration will run i times (n-(n-i+1)=i-1).

We can call it O(N^2) when N mention to i.

Note! N and n are not the same here!

answered Dec 01 '13 at 12:14

MeNa

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Time complexity - recursive call

2 Answers2