How to calculate my method's time complexity

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Java

question background
Hello, I need to calculate my method's time-complexity for school, and they haven't given me the proper tools for it, sadly :(. I have tried googeling but I'm not comfortable enough to do it alone.

The Task
We have a 2-dimensional array that is sorted pretty much like this:
[41] [42] [43] [44] [45] [46] [47] [48] row 5
[33] [34] [35] [36] [37] [38] [39] [40] row 4
[25] [26] [27] [28] [29] [30] [31] [32] row 3
[17] [18] [19] [20] [21] [22] [23] [24] row 2
[9] [10] [11] [12] [13] [14] [15] [16] row 1
[1] [2] [3] [4] [5] [6] [7] [8] row 0

I need to write a method that will get this array as param, and an int val from the user and the method has to search it in the array in a time of no more than O(n).
That is the code I have written:

<pre>
int l = 0;
int r = m.length-1;  
        while (l <= r)
        {
            int middle = l + ((r-l)/2);
            if (m[middle][m[middle].length-1] == val)
                return true;
            if (m[middle][m[middle].length-1] < val)
                l = middle + 1;
            else
            {
                if (middle == 0) {
                    r = m[middle].length-1;
                    l = 0;
                    if (binarySearch(m, l, r, middle, val) == 0) return true;   
                }
                if (middle > 0)
                    if (m[middle-1][m[middle-1].length-1] < val) {
                        r = m[middle].length-1;
                        l = 0;
                        if (binarySearch(m, l, r, middle, val) == 0) return true;
                    }
                r = middle - 1;
            }
        }
return false
}

I start by checking the last column of the middle row of the array. If the value there is val, great.
If smaller, I won't waste my time on the former rows.
If the value is greater, I then check if the row before it(middle -1) has a last column value that is smaller than val. if yes, this means val can only be in the current row. So I proceed into a second binarySearch.
There is another test to check if middle is 0, that proceeds to another binarySearch. Without it, I get exception outOfBounds in my prior middle-1 check.

Basically I have a big BinarySearch method that if it finds a last column that is bigger than val, uses another BinarySearch.
Would you say that it is 2(log n)? n (log n)? I find it hard to call.
Thank you!

What I have tried:

Google, youtube, university book