<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 s2, string s1) {
        int n1=s1.length();
        int n2=s2.length();
        vector<vector<int>>v(n1+1,vector<int>(26,0));
        for(int i=0;i<n1;i++)
        {
            v[i+1][s1[i]-'a']++;
        }
        for(int i=0;i<26;i++)
        {
            for(int j=1;j<=n1;j++)
            {
                v[j][i]=v[j-1][i]+v[j][i];
            }
        }
        vector<int>temp(26,0);
        for(int i=0;i<n2;i++)
        {
            temp[s2[i]-'a']++;
        }
        for(int i=0;i+n2-1<n1;i++)
        {
            int flag=1;
            int l=i+1;
            int r=i+1+n2-1;
            for(int j=0;j<26;j++)
            {
                if(v[r][j]-v[l-1][j]!=temp[j])
                {
                    flag=0;break;
                }
            }
            if(flag)
            return true;
        }
        return false;
    }
};
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