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# Permutation in String

ID: 567

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`.

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

## Idea

{% hint style="info" %}
Bitmap, not brute force
{% endhint %}

If s1 is longer than s2, cannot contain permutation

If character frequency in s1 equals to character frequency in a consecutive order in s2, is permutation, else false

Use int\[26] to record character frequency of 26 characters

## Code

```java
public boolean checkInclusion(String s1, String s2) {
        if (s1.length() > s2.length())
            return false;
        int[] s1map = new int[26];
        for (int i = 0; i < s1.length(); i++)
            s1map[s1.charAt(i) - 'a']++;
        for (int i = 0; i <= s2.length() - s1.length(); i++) {
            int[] s2map = new int[26];
            for (int j = 0; j < s1.length(); j++) {
                s2map[s2.charAt(i + j) - 'a']++;
            }
            if (matches(s1map, s2map))
                return true;
        }
        return false;
    }
    
    public boolean matches(int[] s1map, int[] s2map) {
        for (int i = 0; i < 26; i++) {
            if (s1map[i] != s2map[i])
                return false;
        }
        return true;
    }
```