Solutions

# Keep a set of seen characters and two pointers. Move the right pointer until we see a duplicate, then move the left pointer until we remove the duplicate from the set. Keep track of the maximum length at each step.
# sliding-window

# Brute force - O(n^2) time, O(1) space
def lengthOfLongestSubstring(s: str) -> int:
    maxlen = 0
    for l in range(len(s)):
        seen = set()
        for r in range(l, len(s)):
            if s[r] in seen:
                break
            seen.add(s[r])
        maxlen = max(maxlen, r-l)
    return maxlen


# Sliding window - O(n) time, O(m) space (m: unique chars in s)
def lengthOfLongestSubstring(s: str) -> int:
    charSet = set()
    l = 0
    res = 0

    for r in range(len(s)):
        while s[r] in charSet:
            charSet.remove(s[l])
            l += 1
        charSet.add(s[r])
        res = max(res, r - l + 1)
    return res


# Sliding window (optimal) - O(n) time, O(m) space (m: unique chars in s)
def lengthOfLongestSubstring(s: str) -> int:
    if not s:
        return 0

    l, r = 0, 0
    seen = {}
    maxlen = 0

    while r < len(s):
        curr = s[r]
        if curr in seen:
            l = max(l, seen[curr] + 1)
        seen[curr] = r
        maxlen = max(maxlen, r-l +1)
        r += 1

    return maxlen