<p>Given two strings <code>s1</code> and <code>s2</code>, return <code>true</code><em> if </em><code>s2</code><em> contains a permutation of </em><code>s1</code><em>, or </em><code>false</code><em> otherwise</em>.</p>

<p>In other words, return <code>true</code> if one of <code>s1</code>&#39;s permutations is the substring of <code>s2</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidbaooo&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> s2 contains one permutation of s1 (&quot;ba&quot;).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;ab&quot;, s2 = &quot;eidboaoo&quot;
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
</ul>

class Solution
{
public:
    bool checkInclusion(string p, string s)
    {
        int n = s.size(), m = p.size();
        vector<int> res;
        vector<int> pChar(26, 0);
        vector<vector<int>> arr(n, vector<int>(26, 0));
        for (char ch : p)
            pChar[ch - 'a']++;
        for (int i = 0; i < n; i++)
        {
            arr[i] = i - 1 >= 0 ? arr[i - 1] : vector<int>(26, 0);
            arr[i][s[i] - 'a']++;
            if (i >= m - 1)
            {
                vector<int> prev = (i - m >= 0 ? arr[i - m] : vector<int>(26, 0));
                vector<int> curr = arr[i];
                for (int i = 0; i < 26; i++)
                    curr[i] -= prev[i];
                if (curr == pChar)
                    return true;
            }
        }
        return false;
    }
};

static int fastIO = []
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 0;
}();
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