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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"
The decoded string is "hahahaha".  The 5th letter is "h".

Example 2:

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


  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 = ; int N = S.length(); // Find size = length of decoded string for (int i = ; i < N; ++i) { char c = S.charAt(i); if (Character.isDigit(c)) size *= c - '0'; else size++; } for (int i = N-1; i >= ; --i) { char c = S.charAt(i); K %= size; if (K == && Character.isLetter(c)) return Character.toString(c); if (Character.isDigit(c)) size /= c - '0'; else size--; } throw null; } }
Code language: Java (java)

See also  String polynomial converter in Java
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