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leetcode
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‎README.md

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- [**303.** Range Sum Query - Immutable](RangeSumQuery-Immutable/range_sum_query_immutable.dart)
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- [**433.** Minimum Genetic Mutation](MinimumGeneticMutation/minimum_genetic_mutation.dart)
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- [**2131.** Longest Palindrome by Concatenating Two Letter Words](LongestPalindromeByConcatenatingTwoLetterWords/longest_palindrome_by_concatenating_two_letter_words.dart)
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- [**345.** Reverse Vowels of a String](ReverseVowelsOfAString/reverse_vowels_of_a_string.dart)
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## Reach me via
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‎ReverseVowelsOfAString/reverse_vowels_of_a_string.dart

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/*
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-* 345. Reverse Vowels of a String *-
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Given a string s, reverse only all the vowels in the string and return it.
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The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and upper cases, more than once.
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Example 1:
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Input: s = "hello"
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Output: "holle"
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Example 2:
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Input: s = "leetcode"
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Output: "leotcede"
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Constraints:
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1 <= s.length <= 3 * 105
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s consist of printable ASCII characters.
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Accepted
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403,544
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Submissions
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838,146
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*/
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class A {
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String reverseVowels(String s) {
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List<String> char = s.split("");
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int left = 0;
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int right = s.length - 1;
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while (left < right) {
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while (left < right && !isVowels(char[left])) {
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left++;
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}
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while (left < right && !isVowels(char[right])) {
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right--;
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}
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if (left < right) {
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String temp = char[left];
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char[left++] = char[right];
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char[right--] = temp;
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}
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}
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return char.join();
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}
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bool isVowels(String s) {
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return s == 'a' ||
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s == 'e' ||
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s == 'i' ||
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s == 'o' ||
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s == 'u' ||
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s == 'A' ||
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s == 'E' ||
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s == 'I' ||
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s == 'O' ||
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s == 'U';
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}
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}
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class B {
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// Set<String> vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
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String vowels = 'aeiouAEIOU';
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String reverseVowels(String s) {
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List<String> char = s.split("");
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for (int i = 0, j = char.length - 1; i < j; i++, j--) {
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while (!vowels.contains(char[i]) && i < j) {
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i++;
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}
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while (!vowels.contains(char[j]) && i < j) {
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j--;
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}
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// [char[i], char[j]] = [char[j], char[i]];
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char[i] = char[j];
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char[j] = char[i];
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}
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return char.join('');
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}
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}
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class C {
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String reverseVowels(String s) {
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String str = "aeiouAEIOU";
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String dummy = "";
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for (String c in s.split("")) {
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if (str.contains("" + c)) dummy += c;
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}
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List<String> c = s.split("");
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int j = dummy.length - 1;
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for (int i = 0; i < s.length; i++) {
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if (str.contains("" + s[i])) {
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c[i] = dummy[j];
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j--;
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}
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}
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return c.join();
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}
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}
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class D {
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String reverseVowels(String s) {
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// Create A Pointer Which Is Start From End Of The Array.
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int pointer = s.length - 1;
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// Convert String To Array That Will More Easy To Solve This Problem.
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List<String> ansStr = s.split("");
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int i = 0; // Create A I Veritable That Start From Array Index Of 0
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// Create a All Vowels That Will Help To Check The Value Is Vowel Or Not.
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List<String> vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'];
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// You can use regex like: /[a|e|i|o|u]/gi
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while (i < pointer) {
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// Check Condition That I Should Be Less Then Pointer This Condition Divide The Time Complexity To O(N) To O(LogN).
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if (vowels.contains(ansStr[i]) && vowels.contains(ansStr[pointer])) {
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// Here we check the condition that our I position value and Pointer position value should be vowel
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String tmp = ansStr[i];
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// If condition is true then swap this position value respectively you can create a extra swap function.
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ansStr[i++] = ansStr[pointer];
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ansStr[pointer--] = tmp;
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} else {
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if (vowels.contains(ansStr[i])) {
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// If the Ith position value is vowel then we decrement the Pointer Position value to -1;
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pointer--;
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} else {
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i++; // If the Ith position value does not a vowel then increment the Ith position by +1
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}
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}
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}
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return ansStr.join(""); // Here we convert the array to string
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}
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}

‎ReverseVowelsOfAString/reverse_vowels_of_a_string.go

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package main
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import "unicode"
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// Time Complexity: O(N)
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// Space Complexity: O(N)
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// fun fact:
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// `Y` and `y` can be a vowel as well sometimes.
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// glad the problem statement defines it well
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func isVowel(c rune) bool {
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// alternatively, we can just check
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// return c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U' ||
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// c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
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c = unicode.ToLower(c)
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return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
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}
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func reverseVowels(s string) string {
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ss := []rune(s)
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n := len(ss)
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l, r := 0, n-1
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for l < r {
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// find the index of the first vowel in the given range
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for l < r && !isVowel(ss[l]) {
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l += 1
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}
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// find the index of last vowel in the given range
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for r > l && !isVowel(ss[r]) {
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r -= 1
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}
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// swap ss[l] and ss[r]
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ss[l], ss[r] = ss[r], ss[l]
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// since we've processed the character ss[l],
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// we move the left pointer to the right
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l += 1
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// since we've processed the character ss[r],
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// we move the right pointer to the left
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r -= 1
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}
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return string(ss)
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}

‎ReverseVowelsOfAString/reverse_vowels_of_a_string.md

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# 🔥 Reverse Vowels of a String 🔥 || 3 Approaches || Simple Fast and Easy || with Explanation
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## Solution -1
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```dart
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class Solution {
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String reverseVowels(String s) {
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List<String> char = s.split("");
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int left = 0;
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int right = s.length - 1;
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while (left < right) {
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while (left < right && !isVowels(char[left])) {
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left++;
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}
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while (left < right && !isVowels(char[right])) {
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right--;
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}
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if (left < right) {
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String temp = char[left];
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char[left++] = char[right];
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char[right--] = temp;
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}
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}
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return char.join();
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}
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bool isVowels(String s) {
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return s == 'a' ||
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s == 'e' ||
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s == 'i' ||
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s == 'o' ||
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s == 'u' ||
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s == 'A' ||
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s == 'E' ||
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s == 'I' ||
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s == 'O' ||
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s == 'U';
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}
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}
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```
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## Solution - 2
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```dart
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class Solution {
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String reverseVowels(String s) {
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String str = "aeiouAEIOU";
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String dummy = "";
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for (String c in s.split("")) {
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if (str.contains("" + c)) dummy += c;
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}
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List<String> c = s.split("");
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int j = dummy.length - 1;
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for (int i = 0; i < s.length; i++) {
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if (str.contains("" + s[i])) {
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c[i] = dummy[j];
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j--;
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}
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}
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return c.join();
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}
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}
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```
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## Solution - 3
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```dart
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class Solution {
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String reverseVowels(String s) {
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// Create A Pointer Which Is Start From End Of The Array.
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int pointer = s.length - 1;
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// Convert String To Array That Will More Easy To Solve This Problem.
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List<String> ansStr = s.split("");
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int i = 0; // Create A I Veritable That Start From Array Index Of 0
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// Create a All Vowels That Will Help To Check The Value Is Vowel Or Not.
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List<String> vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'];
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// You can use regex like: /[a|e|i|o|u]/gi
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while (i < pointer) {
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// Check Condition That I Should Be Less Then Pointer This Condition Divide The Time Complexity To O(N) To O(LogN).
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if (vowels.contains(ansStr[i]) && vowels.contains(ansStr[pointer])) {
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// Here we check the condition that our I position value and Pointer position value should be vowel
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String tmp = ansStr[i];
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// If condition is true then swap this position value respectively you can create a extra swap function.
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ansStr[i++] = ansStr[pointer];
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ansStr[pointer--] = tmp;
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} else {
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if (vowels.contains(ansStr[i])) {
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// If the Ith position value is vowel then we decrement the Pointer Position value to -1;
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pointer--;
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} else {
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i++; // If the Ith position value does not a vowel then increment the Ith position by +1
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}
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}
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}
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return ansStr.join(""); // Here we convert the array to string
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}
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}
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```

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