// Complete the maximumSuperiorCharacters function below.
    static long maximumSuperiorCharacters(int[] freq) {
        // number of superior.
        long s = 0;
        // number of non-superior. (l = low)
        long l = 0;
        // add all letters to the set from a to z.
        for (int i = 0; i < freq.length; i++) {
            assert(s == 0 || l > s);
            // number of next letter that is higher than all previous
            long f = freq[i];
            // total number of letters after adding the last one.
            long n = l+s+f;
            // calculate new number of superior:
            if (l == 0) {
                // if current set of letters is empty, they will all be the
                // same, no superiors.
                s = 0;
            } else if (f >= l+s) {
                // if there are more of new letter then all previous, we can
                // put new one between each old and it will be the biggest
                // number of superiors possible
                s = l+s-1;
            } else if (l-s-1 >= f) {
                // there is along sequence of non-superiors on one side of our
                // string, that can be improved by stuffing new letter at every
                // second position.
                s = s + f;
            } else {
                // it's even better, we have enough of new letter to make
                // an optimal string.
                s = (n-1)/2;
            }
            l = n - s ;
        }

        return s;
    }