DecodedString.java

/*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 = "leet2code3", K = 10
Output: "o"
Explanation: 
The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10th letter in the string is "o".
Example 2:

Input: S = "ha22", K = 5
Output: "h"
Explanation: 
The decoded string is "hahahaha".  The 5th letter is "h".
Example 3:

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

Constraints:

2 <= S.length <= 100
S will only contain lowercase letters and digits 2 through 9.
S starts with a letter.
1 <= K <= 10^9
It's guaranteed that K is less than or equal to the length of the decoded string.
The decoded string is guaranteed to have less than 2^63 letters.*/

class Solution {
    public String decodeAtIndex(String str, int K) {
        long len = 0;
        for(int i=0;i<str.length();i++){
            char cur = str.charAt(i);
            if(Character.isLetter(cur)){
                len++;
            }
            else{
                len *= cur-'0';
            }
        }
        for(int i=str.length()-1;i>=0;i--){
            K %= len;
            char cur = str.charAt(i);
            if(K == 0){
                while(Character.isDigit(str.charAt(i))){i--;}
                return str.charAt(i)+"";
            }
            if(Character.isLetter(cur)){len--;}
            else len /= cur-'0';
        }
        return "";
    }
}