Intuition

Slide a window over the string keeping it free of duplicates. When the new character was already inside the window, jump the left edge to just past its previous position — that removes the duplicate in O(1).

The window only ever grows on the right and jumps forward on the left, so each character is handled once.

Approach

Store the last index seen for each character and a left boundary.
For character at right: if it was seen at index ≥ left, set left = lastSeen + 1.
Record its new index and update the best length right - left + 1.

Walkthrough

s = "abcabcbb"
window grows to "abc" (len 3)
at index 3 (a): a seen at 0 -> left=1, window "bca"
... best stays 3 -> answer 3

Solution

def lengthOfLongestSubstring(s):
    seen = {}
    left = best = 0
    for right, c in enumerate(s):
        if c in seen and seen[c] >= left:
            left = seen[c] + 1
        seen[c] = right
        best = max(best, right - left + 1)
    return best

Complexity

Time — O(n): each character is visited once; left only moves forward.
Space — O(min(n, charset)): at most one map entry per distinct character.

Common Pitfalls

The seen[c] >= left check matters — an old occurrence outside the window must be ignored.

Longest Substring Without Repeating Characters

Example 1

Example 2

Constraints