LeetCode in Kotlin
1309. Decrypt String from Alphabet to Integer Mapping
Easy
You are given a string s formed by digits and '#'. We want to map s to English lowercase characters as follows:
- Characters (
'a'to'i')are represented by ('1'to'9') respectively. - Characters (
'j'to'z')are represented by ('10#'to'26#') respectively.
Return the string formed after mapping.
The test cases are generated so that a unique mapping will always exist.
Example 1:
Input: s = “10#11#12”
Output: “jkab”
Explanation: “j” -> “10#” , “k” -> “11#” , “a” -> “1” , “b” -> “2”.
Example 2:
Input: s = “1326#”
Output: “acz”
Constraints:
1 <= s.length <= 1000sconsists of digits and the'#'letter.swill be a valid string such that mapping is always possible.
Solution
class Solution {
fun freqAlphabets(s: String): String {
val map: MutableMap<String, String> = HashMap()
map["1"] = "a"
map["2"] = "b"
map["3"] = "c"
map["4"] = "d"
map["5"] = "e"
map["6"] = "f"
map["7"] = "g"
map["8"] = "h"
map["9"] = "i"
map["10#"] = "j"
map["11#"] = "k"
map["12#"] = "l"
map["13#"] = "m"
map["14#"] = "n"
map["15#"] = "o"
map["16#"] = "p"
map["17#"] = "q"
map["18#"] = "r"
map["19#"] = "s"
map["20#"] = "t"
map["21#"] = "u"
map["22#"] = "v"
map["23#"] = "w"
map["24#"] = "x"
map["25#"] = "y"
map["26#"] = "z"
val sb = StringBuilder()
var i = 0
while (i < s.length) {
if ((("" + s[i]).toInt() == 1 || ("" + s[i]).toInt() == 2) &&
i + 1 < s.length && i + 2 < s.length &&
s[i + 2] == '#'
) {
sb.append(map[s.substring(i, i + 3)])
i += 3
} else {
sb.append(map["" + s[i]])
i++
}
}
return sb.toString()
}
}