Question

I want to know the complexity of this algorithm. In my case both in good, medium and worst is O(n^2)

public char getModa(char[] a){
   int ii[] =new int[a.length];
   char[] t= new char[a.length];

   for(int i=0;i<a.length;i++){
    for(int j=0;j<a.length;j++){
       if(a[j]==a[i]){ 
          ii[i]++;
          t[i]=a[j];
       }
     }
   }
   int cc=0;
   for(int i=0;i<ii.length;i++){
     if(ii[i]>ii[cc]) cc=i;
   }
   return a[cc];
 }

Answer 1

The complexity of all cases (best/average/worst) is O(n^2), the bottle-neck is the double iteration over (i,j) in range([0,a.length),[0,a.length)).

To clarify why it is indeed O(n^2) and not O(n^3) as one might think – because the last loop is not ‘nested’ in side the bottle-neck, so the complexity of the 3 loops is basically O(n^2+n), but since O(n^2+n) = O(n^2), this is the answer.

In fact – there is no variation at all, for the same length arrays – the algorithm has the same number of iterations, regardless of what the input exactly is.