Leet Code: - 345. Reverse Vowels of a String

345. Reverse Vowels of a String

Intuition
using two pointers

Approach
we start convert the string to array than check both from first to last and swap if two character is vowel

Complexity
Time complexity: O(n)
Space complexity: O(n)

Code
class Solution { public String reverseVowels(String s) { char ch[] = s.toCharArray(); int i = 0; int j = ch.length - 1; while(i < j){ if((ch[i] == 'a' || ch[i] == 'e' || ch[i] == 'i' || ch[i] == 'o' || ch[i] == 'u' || ch[i] == 'A' || ch[i] == 'E' ||
            ch[i] == 'I' || ch[i] == 'O' || ch[i] == 'U') && (ch[j] == 'a' || ch[j] == 'e' || ch[j] == 'i' || ch[j] == 'o' ||
            ch[j] == 'u' || ch[j] == 'A' || ch[j] == 'E' || ch[j] == 'I' || ch[j] == 'O' || ch[j] == 'U')){ char temp = ch[i]; ch[i] = ch[j]; ch[j] = temp; i++; j--; }else if(ch[i] == 'a' || ch[i] == 'e' || ch[i] == 'i' || ch[i] == 'o' || ch[i] == 'u' ||
            ch[i] == 'A' || ch[i] == 'E' || ch[i] == 'I' || ch[i] == 'O' || ch[i] == 'U'){ j--; }else if(ch[j] == 'a' || ch[j] == 'e' || ch[j] == 'i' || ch[j] == 'o' || ch[j] == 'u' ||
            ch[j] == 'A' || ch[j] == 'E' || ch[j] == 'I' || ch[j] == 'O' || ch[j] == 'U'){ i++; }else{ i++; j--; } } String str = new String(ch); return str; } }

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