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880. Decoded String at Index

2019/10/24 Leetcode

880. Decoded String at Index

Tags: ‘Stack’

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:

  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.

Solution

Nornmal solution will get Memory Limit Exceeded.

public String decodeAtIndex(String S, int K) {
    long N = 0L;
    int i;
    char[] chs = S.toCharArray();
    for (i = 0; N < K; i++) N = chs[i] >= '0' && chs[i] <= '9' ? N*(chs[i] - '0') : N + 1;
    i--;
    while (true){
        if (chs[i] >= '0' && chs[i] <= '9') {
            N /= chs[i] - '0';
            K %= N;
        } else if (K%N == 0) return "" + chs[i];
        else N--;
        i--;
    }
}

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