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# LC567 - Permutation in String

## Problem

Given two strings `s1` and `s2`, return `true` *if* `s2` *contains a permutation of* `s1`*, or* `false` *otherwise*.

In other words, return `true` if one of `s1`'s permutations is the substring of `s2`.

### Example

**Input:** s1 = "ab", s2 = "eidbaooo"

**Output:** true

**Explanation:** s2 contains one permutation of s1 ("ba").

## Solution

Time complexity is $$O(n)$$, with constant $$O(1)$$ space used.

```python
def checkInclusion(self, s1: str, s2: str) -> bool:
	if len(s1) > len(s2):
		return False
	# 26 eng chars
	s1Counts, s2Counts = [0] * 26, [0] * 26
	
	for i in range(len(s1)):
	
	# ASCII val of char, to build char count in array. ord('a') is offset of a from zero.
		s1Counts[ord(s1[i]) - ord('a')] += 1
		s2Counts[ord(s2[i]) - ord('a')] += 1
		
	if s1Counts == s2Counts:
		return True
	
	for i in range(len(s1), len(s2)):
		s2Counts[ord(s2[i]) - ord('a')] += 1
		s2Counts[ord(s2[i-len(s1)]) - ord('a')] -= 1
		if s1Counts == s2Counts:
			return True
	
	return False

```