StringBuilder II
Very often if you're getting stuck into TLE due to string manipulation, especially concatenation, switch to StringBuilder and your problems will be gone. This is an example of such switch. The TLE came from my initial implementation using String manipulation. Code is down below, cheers, ACC.
Sort Vowels in a String - LeetCode
2785. Sort Vowels in a String
Medium
Given a 0-indexed string s
, permute s
to get a new string t
such that:
- All consonants remain in their original places. More formally, if there is an index
i
with0 <= i < s.length
such thats[i]
is a consonant, thent[i] = s[i]
. - The vowels must be sorted in the nondecreasing order of their ASCII values. More formally, for pairs of indices
i
,j
with0 <= i < j < s.length
such thats[i]
ands[j]
are vowels, thent[i]
must not have a higher ASCII value thant[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.
public string SortVowels(string s) { string vowels = "AEIOUaeiou"; int[] countVowels = new int[vowels.Length]; foreach (char c in s) if (vowels.Contains(c)) countVowels[vowels.IndexOf(c)]++; int index = 0; StringBuilder retVal = new StringBuilder(s); int indexRetVal = 0; foreach(char c in s) { if (!vowels.Contains(c)) retVal[indexRetVal] = c; else { while (index < countVowels.Length && countVowels[index] == 0) index++; retVal[indexRetVal] = vowels[index]; countVowels[index]--; } indexRetVal++; } return retVal.ToString(); }
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