So what I am doing is to note 1 instead of > and 0 instead of <. Therefore if I have the configuration 10 then it will become 01. The 01 configuration doesn't change.
Therefore if I take the example from the problem I have

110010
101001
010101
001011
000111

So I see there is a shifting of the zero's to the left and a of one's to the right.
So I am saving in the beginning the positions of zero's in an array (zero[]) and the positions of the one's in another array (one[]).
then I consider k = one[0].

Therefore the all algorithm (after reading the data) looks practically like this

k=one[0];
for (int i=0;i<z;i++){
    if ((zero[i]-k) >=0){
     result+=(zero[i]-k);
     k++;
    }
}

I think the complexity of this is even O(z) where z is the numbers of zero's.

The problem with this is that I obtain TL on test 15. Is there anything faster or is only my coding problme (java)?

Edited by author 29.11.2012 19:22

Edited by author 29.11.2012 19:37

The problem is in your code.
Next solution gets AC with O(n) complexity

// obtaining bool array
int pos = 0, res = 0;
for (int i = 0; i < n; i++) if (!arr[i]) res += i - pos++;