DecodedStringAtIndex884.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".
 * 
 * Note:
 * 2 <= S.length <= 100
 * S will only contain lowercase letters and digits 2 through 9.
 * S starts with a letter.
 * 1 <= K <= 10^9
 * The decoded string is guaranteed to have less than 2^63 letters.
 */

public class DecodedStringAtIndex884 {
 public String decodeAtIndex(String S, int K) {
 char[] chars = S.toCharArray();
 int N = chars.length;
 long[] lens = new long[N];
 long len = 0;
 for (int i=0; i<N; i++) {
 char c = chars[i];
 if (c >= 'a' && c <= 'z') {
 len++;
 } else {
 int re = Character.getNumericValue(c);
 len *= re;
 }
 lens[i] = len;
 }

 for (int i=N-1; i>=0; i--) {
 char c = chars[i];
 if (c >= 'a' && c <= 'z') {
 if (lens[i] == K) return Character.toString(c);
 } else {
 K %= lens[i-1];
 if (K == 0) {
 int j = i-1;
 while (!(chars[j] >= 'a' && chars[j] <= 'z')) {
 j--;
 }
 return Character.toString(chars[j]);
 }
 }
 }

 return Character.toString(chars[0]);
 }

}