Python 3   ||   6 lines,  iteration   ||   T/S: 80%  / 95%
\nclass Solution:\n    def sortVowels(self, s: str) -> str:\n        \n        vowels, ans = \'AEIOUaeiou\', \'\'\n\n        sVowels = deque(sorted(filter(\n   
```\nclass Solution:\n    def sortVowels(self, s: str) -> str:\n        \n        vowels, ans = \'AEIOUaeiou\', \'\'\n\n        sVowels = deque(sorted(filter(\n                        lambda x: x in vowels, s)))\n\n        for ch in s:\n            if ch in vowels: ans+= sVowels.popleft()\n            else           : ans+= ch\n\n        return ans\n```\n[https://leetcode.com/problems/sort-vowels-in-a-string/submissions/1297114765/](https://leetcode.com/problems/sort-vowels-in-a-string/submissions/1297114765/)\n\n\nI could be wrong, but I think that time complexity is *O*(*N*+*V*log*V*) and space complexity is *O*(*N*, in which *N* ~ `len(s)`, and *V* ~ the count of vowels in`s`.
['Python3']