Reverse Vowels of a String - Given a string s, reverse only all the vowels in the string and return it. The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and upper cases, more than once.

In-depth solution and explanation for LeetCode 345. Reverse Vowels of a String in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis. Better than official and forum solutions.

March 17, 2025 - Input: "programming" Output: "prigrammong" Explanation: The vowels 'o', 'a', 'i' are reversed, resulting in "prigrammong". ... The idea is to extract all vowels from the given string s and store them in vowelStr while maintaining their order.

May 23, 2024 - The vowels are 'a', 'e', 'i', 'o', ... more than once. ... “Reverse Words In A String” is a LeetCode medium, with both a clever two pointer technique solution and a fairly straightforward stack solution....

class Solution: def reverseVowels(self, s: str) -> str: vowels = "aeiouAEIOU" i, j = 0, len(s) - 1 cs = list(s) while i < j: while i < j and cs[i] not in vowels: i += 1 while i < j and cs[j] not in vowels: j -= 1 if i < j: cs[i], cs[j] = cs[j], cs[i] i, j = i + 1, j - 1 return "".join(cs) class Solution { public String reverseVowels(String s) { boolean[] vowels = new boolean[128]; for (char c : "aeiouAEIOU".toCharArray()) { vowels[c] = true; } char[] cs = s.toCharArray(); int i = 0, j = cs.length - 1; while (i < j) { while (i < j && !vowels[cs[i]]) { ++i; } while (i < j && !vowels[cs[j]]) { --j; } if (i < j) { char t = cs[i]; cs[i] = cs[j]; cs[j] = t; ++i; --j; } } return String.valueOf(cs); } }

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We iterate over the string and check if the characters at both indices are vowels. In case it is, we swap them and move on with the loop. At last, we join back the array and return it.

January 31, 2025 - Leetcode : Reverse Vowels of a String Given a string s, reverse only all the vowels in the string and return it. The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and …

It finds and swaps the leftmost and rightmost vowels, moving the pointers inward. This continues until the pointers cross each other. Time complexity is O(n) for a single traversal; space complexity is O(1) due to a single temporary variable.

The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both cases. ... public class Solution { private boolean isVowel(char c) { return c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'I' || c == 'o' || c == 'O' || c == 'u' || c == 'U'; } public String reverseVowels(String str) { int i = 0; int j = str.length() - 1; char[] str1 = str.toCharArray(); while (i < j) { if (!isVowel(str1[i])) { i++; } else if (!isVowel(str1[j])) { j--; } else { // swapping char t = str1[i]; str1[i] = str1[j]; str1[j] = t; i++; j--; } } return String.copyValueOf(str1); } }

November 28, 2023 - 4. After swapping, just increase ‘ i ‘ by ( i++) decrease ( j – -) The loop continues until ‘i’ is incremented and ‘j’ is decremented to continue searching for other pairs of vowels. And it will run until ‘ i ‘ is no longer less than ‘j’ which means all the possible pairs or vowels have been swapped. 5. finally we will come outside the loop and the string ‘ s’ is returned .

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August 13, 2023 - We are tasked with reversing the ... transformed to 'holle'. Similarly, if the string has the value 'leetcode', it should be changed to 'leotcede'....