LeetCode Q 880 - Decoded String at Index

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.

Solution

Solution 1: Recursion Method

Code

public String decodeAtIndex(String S, int K) {
  int len = 0;
  for (char ch: S.toCharArray()) {
    if (Character.isDigit()) {
      int num = ch - '0', count = 1;
      for (; count < num && K > len; count++) {
        K -= len;
      }
      if (count != num) return decodeAtIndex(S, K);
      else len \*= num;
    } else {
        len++; K--;
        if (K == 0) return ch + "";
    }
  }
}

Solution 2: Recursion Method (More Efficient)

Code

public String decodeAtIndex(String S, int K) {
  int len = 0;
  for (char ch: S.toCharArray()) {
    if (Character.isDigit()) {
      int times = ch - '0';
      int prelen = len;
      len \*= times;
      
      if(len >= K)
        return decodeAtIndex(S,(K-1) % (int)prelen + 1);
                

    } else {
        len++;
        if (len == K) return ch + "";
    }
  }
}

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