The Decoded String at Index using Java

The Challenge

An encoded string S is given.  To find and write the decoded string to a tape, the encoded string is read one character at a time and the following steps are taken:

• If the character read is a letter, that letter is written onto the tape.
• If the character read is a digit (say d), the entire current tape is repeatedly written d-1 more times in total.

Now for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the decoded string.

Example 1:

Input: S = "ha22", K = 5
Output: "h"
Explanation: 
The decoded string is "hahahaha".  The 5th letter is "h".

Example 2:

Input: S = "a2345678999999999999999", K = 1
Output: "a"
Explanation: 
The decoded string is "a" repeated 8301530446056247680 times.  The 1st letter is "a".

Note:

1. 2 <= S.length <= 100
2. S will only contain lowercase letters and digits 2 through 9.
3. S starts with a letter.
4. 1 <= K <= 10^9
5. The decoded string is guaranteed to have less than 2^63 letters.

The Code

// Our class
class Solution {
    // The entry method
    public String decodeAtIndex(String S, int K) {
        long size = 0;
        int N = S.length();

        // Find size = length of decoded string
        for (int i = 0; i < N; ++i) {
            char c = S.charAt(i);
            if (Character.isDigit(c))
                size *= c - '0';
            else
                size++;
        }

        for (int i = N-1; i >= 0; --i) {
            char c = S.charAt(i);
            K %= size;
            if (K == 0 && Character.isLetter(c))
                return Character.toString(c);

            if (Character.isDigit(c))
                size /= c - '0';
            else
                size--;
        }

        throw null;
    }
}