java codility training Genomic-range-query

Here is the solution that got 100 out of 100 in codility.com. Please read about prefix sums to understand the solution:

public static int[] solveGenomicRange(String S, int[] P, int[] Q) {
        //used jagged array to hold the prefix sums of each A, C and G genoms
        //we don't need to get prefix sums of T, you will see why.
        int[][] genoms = new int[3][S.length()+1];
        //if the char is found in the index i, then we set it to be 1 else they are 0
        //3 short values are needed for this reason
        short a, c, g;
        for (int i=0; i<S.length(); i++) {
            a = 0; c = 0; g = 0;
            if ('A' == (S.charAt(i))) {
                a=1;
            }
            if ('C' == (S.charAt(i))) {
                c=1;
            }
            if ('G' == (S.charAt(i))) {
                g=1;
            }
            //here we calculate prefix sums. To learn what's prefix sums look at here https://codility.com/media/train/3-PrefixSums.pdf
            genoms[0][i+1] = genoms[0][i] + a;
            genoms[1][i+1] = genoms[1][i] + c;
            genoms[2][i+1] = genoms[2][i] + g;
        }

        int[] result = new int[P.length];
        //here we go through the provided P[] and Q[] arrays as intervals
        for (int i=0; i<P.length; i++) {
            int fromIndex = P[i];
            //we need to add 1 to Q[i], 
            //because our genoms[0][0], genoms[1][0] and genoms[2][0]
            //have 0 values by default, look above genoms[0][i+1] = genoms[0][i] + a; 
            int toIndex = Q[i]+1;
            if (genoms[0][toIndex] - genoms[0][fromIndex] > 0) {
                result[i] = 1;
            } else if (genoms[1][toIndex] - genoms[1][fromIndex] > 0) {
                result[i] = 2;
            } else if (genoms[2][toIndex] - genoms[2][fromIndex] > 0) {
                result[i] = 3;
            } else {
                result[i] = 4;
            }
        }

        return result;
    }