880. Decoded String at Index

Problem

You are given an encoded string s. To decode the 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 d, the entire current tape is repeatedly written d - 1 more times in total.

Given an integer k, return **the *kth* 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 consists of lowercase English letters and digits 2 through 9.

• s starts with a letter.

• 1 <= k <= 10^9

• It is guaranteed that k is less than or equal to the length of the decoded string.

• The decoded string is guaranteed to have less than 263 letters.

Solution (Java)

class Solution {
    public String decodeAtIndex(String s, int k) {
        long length = 0;
        for (char c : s.toCharArray()) {
            if (c >= 48 && c <= 57) {
                length *= c - '0';
            } else {
                ++length;
            }
        }
        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            k %= length;
            if (c >= 48 && c <= 57) {
                length /= c - '0';
            } else if (k == 0) {
                return String.valueOf(c);
            } else {
                --length;
            }
        }
        return "";
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).