LeetCode 3. Longest Substring Without Repeating Characters

Description

https://leetcode.com/problems/longest-substring-without-repeating-characters/

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

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. 
             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

Explanation

We use two pointers technique to solve the problem. One slow pointer i, one fast pointer j.

We also add a HashSet to store the characters which have been visited by j pointer to help detect repeating characters.

We keep moving j pointer right further.

• If current s.charAt(j) character is not in the HashSet, we add the character to the HashSet and keep moving j further.
• If current s.charAt(j) character is in the HashSet, we remove the character i is visiting and move i forward. At this point, we found the maximum size of substrings without duplicate characters start with index i. We move i pointer one step further.

When j pointer iterates all the characters of the string, we get the max length of the longest substring without repeating characters.

Java Solution

class Solution {
    public int lengthOfLongestSubstring(String s) {
        int maxLength = 0;
        HashSet<Character> set = new HashSet<>();
        
        int i = 0;
        int j = 0;
        while (j < s.length()) {
            if (!set.contains(s.charAt(j))) {
                set.add(s.charAt(j));
                j++;
                maxLength = Math.max(maxLength, j - i);
            } else {
                set.remove(s.charAt(i));
                i++;
            }
        }
        return maxLength;
    }
}

Python Solution

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        
        counter = {}
        
        j = 0
        
        longest = 0
        
        for i in range(len(s)):            
            while j < len(s) and s[j] not in counter:                
                longest = max(longest, j - i + 1)
                counter[s[j]] = counter.get(s[j], 0) + 1            
                j += 1
                       
            counter[s[i]] -= 1
            
            if counter[s[i]] == 0:
                del counter[s[i]]
                
        return longest

• Time complexity: O(n).
• Space complexity: O(m). m is the size of the charset.