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3\. Longest Substring Without Repeating Characters

Medium

Given a string `s`, find the length of the **longest substring** without repeating characters.

**Example 1:**

**Input:** s = "abcabcbb"

**Output:** 3

**Explanation:** The answer is "abc", with the length of 3. 

**Example 2:**

**Input:** s = "bbbbb"

**Output:** 1

**Explanation:** The answer is "b", with the length of 1. 

**Example 3:**

**Input:** s = "pwwkew"

**Output:** 3

**Explanation:** The answer is "wke", with the length of 3. Notice that the answer must be a substring, "pwke" is a subsequence and not a substring. 

**Example 4:**

**Input:** s = ""

**Output:** 0 

**Constraints:**

*   0 <= s.length <= 5 * 10⁴
*   `s` consists of English letters, digits, symbols and spaces.

To solve the Longest Substring Without Repeating Characters problem in Java using a `Solution` class, we'll follow these steps:

1. Define a `Solution` class with a method named `lengthOfLongestSubstring`.
2. Initialize variables to keep track of the starting index of the substring (`start`), the maximum length (`maxLen`), and a hashmap to store characters and their indices.
3. Iterate through the string `s`, and for each character:
   - Check if the character exists in the hashmap and its index is greater than or equal to the `start` index.
   - If found, update the `start` index to the index after the last occurrence of the character.
   - Update the maximum length if necessary.
   - Update the index of the current character in the hashmap.
4. Return the maximum length found.
5. Handle the edge case where the input string is empty.

Here's the implementation:

```java
import java.util.HashMap;

public class Solution {
    
    public int lengthOfLongestSubstring(String s) {
        // Initialize variables
        int start = 0;
        int maxLen = 0;
        HashMap map = new HashMap<>();
        
        // Iterate through the string
        for (int end = 0; end < s.length(); end++) {
            char ch = s.charAt(end);
            // If the character exists in the hashmap and its index is greater than or equal to the start index
            if (map.containsKey(ch) && map.get(ch) >= start) {
                // Update the start index to the index after the last occurrence of the character
                start = map.get(ch) + 1;
            }
            // Update the maximum length if necessary
            maxLen = Math.max(maxLen, end - start + 1);
            // Update the index of the current character in the hashmap
            map.put(ch, end);
        }
        
        return maxLen;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();

        // Test cases
        String s1 = "abcabcbb";
        System.out.println("Example 1 Output: " + solution.lengthOfLongestSubstring(s1));

        String s2 = "bbbbb";
        System.out.println("Example 2 Output: " + solution.lengthOfLongestSubstring(s2));

        String s3 = "pwwkew";
        System.out.println("Example 3 Output: " + solution.lengthOfLongestSubstring(s3));

        String s4 = "";
        System.out.println("Example 4 Output: " + solution.lengthOfLongestSubstring(s4));
    }
}
```

This implementation provides a solution to the Longest Substring Without Repeating Characters problem in Java.