<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.length();
        int m = s2.length();

        int freq[26];

        for(int i = 0 ; i <n ; i++)
        {
            freq[s1[i] - 'a']++;
        }

        map<vector<int>, int> mp;
        vector<int> freq2(26);
        for(int i = 0 ; i < 26 ; i++)
            freq2[i] = 0;
        bool ans = false;
        mp[freq2]++;
        for(int i = 0 ; i < m; i++)
        {
            
            vector<int> freq3(26);
            freq2[s2[i] - 'a']++;
            for(int j =0 ; j < 26 ; j++)
            {
                freq3[j] = freq2[j] - freq[j];
            }
            if(mp[freq3])
                ans = true;
            
            mp[freq2]++;
            
        }
        return ans;
    }
};
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