3 Longest Substring Without Repeating Characters

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.

Solution:

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() == 0){
            return 0;
        }
        Map<Character, Integer> map = new HashMap<Character, Integer>();

        int slow = 0;
        int result = 0;
        for (int fast = 0; fast < s.length(); fast++){
            // Step 1: add fast
            map.put(s.charAt(fast), map.getOrDefault(s.charAt(fast), 0) + 1);

            // Step 2: remove slow
            if (map.size() < fast - slow + 1){
                int count = map.get(s.charAt(slow));
                if (count == 1){
                    map.remove(s.charAt(slow));
                }else{
                    map.put(s.charAt(slow), count - 1);
                }

                slow++;
            }

            // Step 3:
            int curLen = fast - slow + 1;
            if (map.size() == curLen){
                result = Math.max(result, curLen);
            }
        }

        return result;
    }
}

/*
// TC: O(n)
// SC: O(n)


            abcabcbb
            s
               f

    map:
    <a,1>
    <b,1>
    <c,1>


*/

class Solution {
    public int lengthOfLongestSubstring(String s) {
        // base case
        if (s == null ){
            return 0;
        }

        if (s.length() <= 1){
            return s.length();
        }

        Set<Character> c = new HashSet<Character>();
        int longest = 0;
        int slow = 0;
        int fast = 0;
        while(fast < s.length()){
            if (c.contains(s.charAt(fast))){
                // exist
                c.remove(s.charAt(slow));
                slow++;
            }else{
                // not exist
                c.add(s.charAt(fast));
                fast++;
                longest = Math.max(longest, fast - slow);
            }
        }

        return longest;

    }
}

// TC: O(n)

// SC: O(n)



/*
    set:  c
    longest = 3
    a   b   c   a   b   c   b   b
    while(f < s.length)
                                s
                                   f

    1. check set
        1.1 if exit
        remove slow
        slow++
        1.2 if not exit
            :   1.2.1: add.set
                1.2.2: fast++;
                1.2.3: Math(longest, new fast - s);


*/