Medium
Given a 0-indexed string s
, permute s
to get a new string t
such that:
i
with 0 <= i < s.length
such that s[i]
is a consonant, then t[i] = s[i]
.i
, j
with 0 <= i < j < s.length
such that s[i]
and s[j]
are vowels, then t[i]
must not have a higher ASCII value than t[j]
.Return the resulting string.
The vowels are 'a'
, 'e'
, 'i'
, 'o'
, and 'u'
, and they can appear in lowercase or uppercase. Consonants comprise all letters that are not vowels.
Example 1:
Input: s = “lEetcOde”
Output: “lEOtcede”
Explanation: ‘E’, ‘O’, and ‘e’ are the vowels in s; ‘l’, ‘t’, ‘c’, and ‘d’ are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places.
Example 2:
Input: s = “lYmpH”
Output: “lYmpH”
Explanation: There are no vowels in s (all characters in s are consonants), so we return “lYmpH”.
Constraints:
1 <= s.length <= 105
s
consists only of letters of the English alphabet in uppercase and lowercase.class Solution {
fun sortVowels(s: String): String {
val vowelCount = IntArray(11)
val countIndexMap = IntArray(128)
val result = s.toCharArray()
val charMap = "AEIOUaeiou".toCharArray()
run {
var i = 0
while (i < charMap.size) {
countIndexMap[charMap[i].code] = ++i
}
}
for (c in result) vowelCount[countIndexMap[c.code]]++
var j = 1
var i = 0
while (j < vowelCount.size) {
if (vowelCount[j] > 0) {
while (i < result.size) {
if (countIndexMap[result[i++].code] == 0) continue
vowelCount[j]--
result[i - 1] = charMap[j - 1]
break
}
} else {
j++
}
}
return String(result)
}
}