<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 s1, string s2) {
        int n = s1.size();
        int m = s2.size();

        if(n == 1 && m == 1){
            return s1[0] == s2[0];
        }

        map<string,int> mp;
        string target = s1;
        sort(target.begin(), target.end());

        mp[target] = 1;

        int l = 0;
        int r = l + n - 1;
        string temp = s2.substr(0,n);

        
        string check = temp;
        sort(check.begin(), check.end());
        mp[check]++;
        if(mp[target] == 2) return true;
        l++;
        r++;
        while(r < m) {
            temp.erase(0,1);
            temp += s2[r];
            string check = temp;
            sort(check.begin(), check.end());
            mp[check]++;
            if(mp[target] == 2) return true;
            l++;
            r++;
        }

        return false;
    }
};
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