Is the complexity of this code O(n^2) or O(n^2 * n^(1/2))?

Question

I was thinking the complexity would be O(n^2). Am I wrong? If so, could you please explain why?

 public int countXs(char[][] m)


{
    int rows = m.length, cols = m[0].length;
    int r = 0, c = 0, count = 0;
    while (r < rows || c < cols)
    {
      count += r;
      while (r < rows && m[r][c] == 'x')
      {
        count++;
        r++;
      }
      c++;
    }
    return count;
  }

Answer 1

Firstly, your code may crash because of arrayIndexOutofBound with c index.

Next, when you invest the time complexity, firstly, you need to define what is n . Normally n indicates the size of input, in this case n is the size of 2-dimensional array.

So, you need setup the formula to time complexity and specify the worst case of your code to find time complexity. Suppose that array m is pxq (p + q = n) and we denote O(1) = 1 .

T(n) = Sum(i->max(p, q)) {1 + sum(j->p)(1)}
     = max(p,q) + sum(i->max(p,q)){p}
     = max(p, n-p) + sum(i-> max(p, n-p)) {p}
     = max(p, n-p) + p(n-p)

Based on Cauchy inequality :

4*ab <= (a+b)^2 we have: p(n - p) < (p + n - p)^2 /4  = n^2/4

So:

T(n) <= n + n^2/4 = O(n^2) . Q.E.D

Read more: http://en.wikipedia.org/wiki/Time_complexity