LeetCode: Longest Substring Without Repeating Characters

 The problem statement is copy pasted from leetcode as it is:

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.

My solution:

package madwani.sushil.leetcode.Amazon;

import java.util.*;

public class LongestSubstringWithoutRepeatingCharacters {
    public static void main(String[] args) {
        System.out.println(lengthOfLongestSubstring6("daabcdefabcffg"));
        System.out.println(lengthOfLongestSubstring6("abcabcbb"));
        System.out.println(lengthOfLongestSubstring6("bbbbb"));
        System.out.println(lengthOfLongestSubstring6("pwwkew"));
        System.out.println(lengthOfLongestSubstring6(" abccbaddeabcde"));
        System.out.println(lengthOfLongestSubstring6(" abcdeafbdgcbb"));
        System.out.println(lengthOfLongestSubstring6(" abc__fbdg_bb"));
    }
    // we will solve using sliding window method
    public static int lengthOfLongestSubstring6(String s) {
        if ( s == null ) {
            return 0;
        }
        // windows start index and end index
        int start_index = 0, end_index = 0;

        // current window size
        int window_size = 0;

        // to track different characters
        Set<Character> uniqueChars = new HashSet<>();
        while ( end_index < s.length()) {
            if (!uniqueChars.add(s.charAt(end_index))) {
                // if not able to add, means duplicate character
                // remove the start_index char
                uniqueChars.remove(s.charAt(start_index));
                // and move the start_index till the duplicate character is not removed
                start_index++;
            } else {
                // unique character added to window
                // if window_size is increased
                window_size = Math.max(uniqueChars.size(), window_size);
                // move the window further
                end_index++;
            }
        }
        return window_size;
    }
}