🔁 Importance of "continue" in Solving String Reversal with Vowels 🔁

In this post, I'll demonstrate how the "continue" keyword in Java can greatly simplify and optimize the process of reversing only the vowels in a given string. Let's dive into the solution and see how this powerful keyword enhances the efficiency of the code.

Given a string s, we need to reverse only the vowels ('a', 'e', 'i', 'o', 'u'), regardless of their case, while leaving the non-vowel characters in their original positions.

Example 1:
Input: "hello"
Output: "holle"

Example 2:
Input: "leetcode"
Output: "leotcede"

🔍 The Problem:

public static String reverseVowels(String s) {
    StringBuilder sbs = new StringBuilder(s);
    String vowels = "aeiou";
    int leftPointer = 0;
    int rightPointer = sbs.length()-1;
    while (leftPointer < rightPointer) {
        if (!vowels.contains(Character.toString(sbs.charAt(leftPointer)).toLowerCase())) {
            leftPointer++;
            continue; // Skip to the next iteration if not a vowel
        }
        if (!vowels.contains(Character.toString(sbs.charAt(rightPointer)).toLowerCase())) {
            rightPointer--;
            continue; // Skip to the next iteration if not a vowel
        }
        // Swap the vowels found at leftPointer and rightPointer
        Character rightCharacter = sbs.charAt(rightPointer);
        Character leftCharacter = sbs.charAt(leftPointer);
        sbs.setCharAt(leftPointer++, rightCharacter);
        sbs.setCharAt(rightPointer--, leftCharacter);
    }
    return sbs.toString();
}

🚀 The Solution:
In the above code, we use two pointers, leftPointer and rightPointer, to traverse the string from the start and end simultaneously. Whenever we encounter a non-vowel character, we skip to the next iteration using the "continue" keyword. This means that if a vowel is not found at a particular position, the pointers will advance to the next position without performing any unnecessary operations.

💡 Key Takeaways:

1. Efficient Pointer Movement: By utilizing "continue", we ensure that the pointers move swiftly and only focus on vowels, without repeatedly processing non-vowel characters.

2. Improved Time Complexity: The efficient movement of pointers leads to a reduction in the overall time complexity of the algorithm, making it faster and more optimal.

3. Simplified Logic: The "continue" statement simplifies the code logic by eliminating the need for nested conditionals and flag variables, resulting in cleaner and more readable code.

So, the next time you're dealing with scenarios where you need to skip certain elements while traversing a data structure, remember the power of "continue" and how it can significantly enhance your code's performance and maintainability. Happy coding! 🚀