# Leetcode Permutation-In-String

#### 2022-07-14 10:29

> ##### Algorithms:
> #algorithm #sliding_window
> ##### Data structures:
> #DS #string
> ##### Difficulty:
> #coding_problem #difficulty-medium
> ##### Additional tags:
> #leetcode #CS_list_need_understanding 
> ##### Revisions:
> N/A

##### Related topics:
```expander
tag:#sliding_window
```


##### Links:
- [Link to problem](https://leetcode.com/problems/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`.

#### Examples

**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 <= 104`
-   `s1` and `s2` consist of lowercase English letters.

### Thoughts

> [!summary]
> This is a #sliding_window problem.

I tried to use kadane's algorithm, but the problem is a premature string, not set. So I gave up.

Then I found out this is sliding window problem, similar to
[[Leetcode Longest-Substring-Without-Repeating-Characters]]

### Solution

```cpp
class Solution {
public:
  bool checkInclusion(string s1, string s2) {
    // Sliding windows problem.

    // First, register the freqs of s1
    vector<int> freq(26, 0);
    for (char ch : s1) {
      freq[ch - 'a']++;
    }

    // Second, start constructing window and window freq;
    int len = s1.length();
    int l = 0, r = 0;
    vector<int> window(26, 0);

    // Start expanding and sliding!
    while (r < s2.length()) {
      // I add value to window here, to avoid r OOB when r = s2.length
      window[s2[r] - 'a']++;
      if (r - l + 1 < len) {
        // Expand right
        r++;
      } else if (window == freq) { // note that I use else if, to make sure
                                   // every time r is incremented once.
        return true;
      } else {
        // Remove l from the window, slide right
        window[s2[l] - 'a']--;
        l++;
        r++;
      }
    }

    return false;
  }
};
```