## 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:3Explanation:The answer is`"abc"`

, with the length of 3.

**Example 2:**

Input:"bbbbb"Output:1Explanation:The answer is`"b"`

, with the length of 1.

**Example 3:**

Input:"pwwkew"Output:3Explanation:The answer is`"wke"`

, with the length of 3. Note that the answer must be asubstring,`"pwke"`

is asubsequenceand 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.

It is a best solution very popular and helpful:

https://www.youtube.com/watch?v=-QPgQJl_CDg&ab_channel=EricProgramming