---
title: '3. Longest Substring Without Repeating Characters'
description: Given a string s, find the length of the longest substring without duplicate characters
icon: dot
topics:
  - { name: "Hash Table", slug: "hash-table" }
  - { name: "String", slug: "string" }
  - { name: "Sliding Window", slug: "sliding-window" }
video: "wiGpQwVHdE0"
issue: "https://github.com/prdlk/leetcode/issues/51"
companies:
  - { name: "Ola Cabs", period: "Thirty Days", frequency: 100 }
  - { name: "Visa", period: "Thirty Days", frequency: 100 }
  - { name: "Amazon", period: "Thirty Days", frequency: 84.9 }
  - { name: "Microsoft", period: "Thirty Days", frequency: 75.6 }
  - { name: "Bloomberg", period: "Thirty Days", frequency: 69.8 }
  - { name: "Meta", period: "Thirty Days", frequency: 63.8 }
  - { name: "Google", period: "Thirty Days", frequency: 63.7 }
  - { name: "tcs", period: "Thirty Days", frequency: 58.2 }
  - { name: "Cognizant", period: "Three Months", frequency: 100 }
  - { name: "Optum", period: "Three Months", frequency: 100 }
  - { name: "Spotify", period: "Three Months", frequency: 100 }
  - { name: "Deloitte", period: "Three Months", frequency: 89.1 }
  - { name: "Infosys", period: "Three Months", frequency: 87.4 }
  - { name: "Apple", period: "Three Months", frequency: 69 }
  - { name: "Goldman Sachs", period: "Three Months", frequency: 66.3 }
  - { name: "Media.net", period: "Six Months", frequency: 100 }
  - { name: "Wipro", period: "Six Months", frequency: 100 }
  - { name: "Wissen Technology", period: "Six Months", frequency: 100 }
  - { name: "Zoho", period: "Six Months", frequency: 100 }
  - { name: "Qualcomm", period: "Six Months", frequency: 92 }
  - { name: "Turing", period: "Six Months", frequency: 89 }
  - { name: "EPAM Systems", period: "Six Months", frequency: 86.7 }
  - { name: "Netflix", period: "Six Months", frequency: 83.1 }
  - { name: "Tesla", period: "Six Months", frequency: 80.9 }
  - { name: "Yandex", period: "Six Months", frequency: 69.2 }
  - { name: "Agoda", period: "Six Months", frequency: 65 }
  - { name: "IBM", period: "Six Months", frequency: 60.1 }
  - { name: "LinkedIn", period: "Six Months", frequency: 54.1 }
sidebar:
  label: 'Longest Substring Without Repeating Characters'
  badge: 'Medium'
---

### Example 1:
- Input: `s = "abcabcbb"`
- Output: `3`
- Explanation: The answer is "abc", with the length of `3`. Note that "bca" and "cab" are also correct answers.

### Example 2:
- Input: `s = "bbbbb"`
- Output: `1`
- Explanation: The answer is "b", with the length of `1`.

### Example 3:
- Input: `s = "pwwkew"`
- Output: `3`
- Explanation: The answer is "wke", with the length of `3`. Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

### Constraints:

- `0 <= s.length <= 10^5`
- `s` consists of English letters, digits, symbols and spaces.

## Approach

```mermaid
flowchart TD
  S(["lengthOfLongestSubstring(s)"]) --> I["left = 0, ans = 0, window = set()"]
  I --> L{"more (right, c) in s?"}
  L -- no --> E(["return ans"])
  L -- yes --> W{"c already in window?"}
  W -- yes --> P["window.remove(s[left]); left += 1 — shrink until c is free"]
  P --> W
  W -- no --> A["window.add(c)"]
  A --> M["ans = max(ans, right - left + 1)"]
  M --> L
```

## Solution

```py
class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        left = 0
        ans = 0
        window = set()
        for right, c in enumerate(s):
            while c in window:
                window.remove(s[left])
                left += 1
            window.add(c)
            ans = max(ans, right - left + 1)

        return ans
```