# Question

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

Examples:

Given “`abcabcbb`

”, the answer is “`abc`

”, which the length is 3.

Given “`bbbbb`

”, the answer is “`b`

”, with the length of 1.

Given “`pwwkew`

”, 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.

# Solution

In the naive approaches, we repeatedly check a substring to see if it has duplicate character. But it is unnecessary. If a substring \(s_{ij}\) from index \(i\) to \(j - 1\) is already checked to have no duplicate characters. We only need to check if `s[j]`

is already in the substring \(s_{ij}\).

To check if a character is already in the substring, we can scan the substring, which leads to an \(O(n^2)\) algorithm. But we can do better.

By using Set as a sliding window, checking if a character in the current can be done in \(O(1)\).

A sliding window is an abstract concept commonly used in array/string problems. A window is a range of elements in the array/string which usually defined by the start and end indices, i.e. \([i, j)\). A sliding window is a window “slides” its two boundaries to the certain direction. For example, if we slide \([i, j)\) to the right by \(1\) element, then it becomes \([i+1, j+1)\).

Back to our problem. We use HashSet to store the characters in current window \([i, j)\) (\(j = i + 1\) initially). Then we slide the index \(j\) to the right. If it is not in the HashSet, we slide \(j\) further. Doing so until `s[j]`

is already in the HashSet. At this point, we found the maximum size of substrings without duplicate characters start with index \(i\). Next, to include \(j\) in the substring, we need to slide \(i\) further until `s[j]`

is not in the set, or when \(i == j\) (when `s[j]`

is the same as `s[j-1]`

). In the latter case, we slide both \(i\) and \(j\) further and repeat. We maintain \(0 \leq i \lt j \lt len(s)\) relationship.

```
class Solution:
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
if len(s) <= 1:
return len(s)
charSet = set(s[0])
ans = 1
i = 0
j = 1
while i < j < len(s):
if s[j] not in charSet:
charSet.add(s[j])
j += 1
if len(charSet) > ans:
ans = j - i
else:
charSet.remove(s[i])
i += 1
if i == j and i < len(s):
j += 1
charSet.add(s[i])
return ans
```