---
title: '567. Permutation in String'
description: Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise
sidebar:
  label: 'Permutation in String'
  badge: 'Medium'
---

Hash Table

### Example 1:
- Input: `s1 = "ab", s2 = "eidbaooo"`
- Output: `true`
- Explanation: `s2` contains one permutation of `s1` ("ba").

### Example 2:
- Input: `s1 = "ab", s2 = "eidboaoo"`
- Output: `false`

### Constraints:

- `1 <= s1.length, s2.length <= 10^4`
- `s1` and `s2` consist of lowercase English letters.

## Approach

```mermaid
flowchart TD
  S(["checkInclusion(s1, s2)"]) --> G{"k = len(s1) > len(s2)?"}
  G -- yes --> X(["return False"])
  G -- no --> I["need = Counter(s1), window = Counter()"]
  I --> L{"more (right, c) in s2?"}
  L -- no --> E(["return False"])
  L -- yes --> A["window[c] += 1"]
  A --> B{"right >= k? — window is now longer than k"}
  B -- yes --> C["drop s2[right - k] from window, deleting the key at 0"]
  B -- no --> Q
  C --> Q{"window == need?"}
  Q -- yes --> R(["return True"])
  Q -- no --> L
```

## Solution

```py
from collections import Counter


class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        k = len(s1)
        if k > len(s2):
            return False

        need = Counter(s1)
        window = Counter()

        for right, c in enumerate(s2):
            window[c] += 1
            if right >= k:
                left_char = s2[right - k]
                window[left_char] -= 1
                if window[left_char] == 0:
                    del window[left_char]
            if window == need:
                return True

        return False
```

## Explanation

[Permutation in String - Leetcode 567 - Python](https://www.youtube.com/watch?v=UbyhOgBN834)