LeetCode in Kotlin

880. Decoded String at Index

Medium

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 10^th letter in the string is “o”.

Example 2:

Input: s = “ha22”, k = 5

Output: “h”

Explanation: The decoded string is “hahahaha”. The 5^th letter is “h”.

Example 3:

Input: s = “a2345678999999999999999”, k = 1

Output: “a”

Explanation: The decoded string is “a” repeated 8301530446056247680 times. The 1^st 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 <= 109
• 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

@Suppress("NAME_SHADOWING")
class Solution {
    fun decodeAtIndex(s: String, k: Int): String {
        var k = k
        var i = 0
        var count: Long = 0
        while (i < s.length && count <= k) {
            val c = s[i]
            count = if (Character.isDigit(c)) count * (c.code - '0'.code) else count + 1
            i++
        }
        i--
        while (i < s.length) {
            val c = s[i]
            if (Character.isDigit(c)) {
                count /= (c.code - '0'.code).toLong()
                k %= count.toInt()
            } else {
                if (k % count == 0L) {
                    break
                }
                --count
            }
            i--
        }
        return s[i].toString()
    }
}

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