> For the complete documentation index, see [llms.txt](https://grind75-notes.gitbook.io/notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://grind75-notes.gitbook.io/notes/week-3/longest-substring-without-repeating-characters.md).

# Longest Substring Without Repeating Characters

{% embed url="<https://leetcode.com/problems/longest-substring-without-repeating-characters>" %}

### Problem

> Given a string `s`, find the length of the **longest** **substring** without repeating characters.
>
> &#x20;
>
> **Example 1:**
>
> <pre><code>Input: s = "abcabcbb"
> <strong>Output: 3
> </strong><strong>Explanation: The answer is "abc", with the length of 3.
> </strong></code></pre>
>
> **Example 2:**
>
> <pre><code>Input: s = "bbbbb"
> <strong>Output: 1
> </strong><strong>Explanation: The answer is "b", with the length of 1.
> </strong></code></pre>
>
> **Example 3:**
>
> <pre data-overflow="wrap"><code>Input: s = "pwwkew"
> <strong>Output: 3
> </strong><strong>Explanation: The answer is "wke", with the length of 3.
> </strong>Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
> </code></pre>

### Pseudocode

{% code overflow="wrap" %}

```
- iterate through string to add into a Set, using two pointers
    - one to iterate through loop
    - one as a trailing pointer to delete all letters in the set when repeated letters are found
- if letter found to be repeating/not unique
    - delete trailing letters using a second pointer in a while loop
    - increment pointer until all letters are deleted
    - repeated letter is added into set in the following sequence
- if letter is unique
    - add into set
    - set size = longest unique substring
    - use Math.max() to book keep

```

{% endcode %}

### Solution

```javascript
// https://leetcode.com/problems/longest-substring-without-repeating-characters/
var lengthOfLongestSubstring = function (s) {
  let maxLength = 0;
  let set = new Set();
  let left = 0;

  for (let i = 0; i < s.length; i++) {
    let curr = s[i];

    while (set.has(curr)) {
      set.delete(s[left]);
      left++;
    }

    set.add(curr);
    maxLength = Math.max(maxLength, set.size);
  }

  return maxLength;
};

```

### Time and Space Complexity

#### Time

* What did the code do
* Total -

#### Space

* What did the code do
* Total -