<p>Given two strings <code>s</code> and <code>p</code>, return <em>an array of all the start indices of </em><code>p</code><em>&#39;s anagrams in </em><code>s</code>. You may return the answer in <strong>any order</strong>.</p>

<p>An <strong>Anagram</strong> is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.</p>

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

<pre>
<strong>Input:</strong> s = &quot;cbaebabacd&quot;, p = &quot;abc&quot;
<strong>Output:</strong> [0,6]
<strong>Explanation:</strong>
The substring with start index = 0 is &quot;cba&quot;, which is an anagram of &quot;abc&quot;.
The substring with start index = 6 is &quot;bac&quot;, which is an anagram of &quot;abc&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abab&quot;, p = &quot;ab&quot;
<strong>Output:</strong> [0,1,2]
<strong>Explanation:</strong>
The substring with start index = 0 is &quot;ab&quot;, which is an anagram of &quot;ab&quot;.
The substring with start index = 1 is &quot;ba&quot;, which is an anagram of &quot;ab&quot;.
The substring with start index = 2 is &quot;ab&quot;, which is an anagram of &quot;ab&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length, p.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>s</code> and <code>p</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
vector<int> findAnagrams(string s, string p) 
{
    int i=0,n=s.size(),m=p.size(),j;
    vector<int> c(26,0),ans,mc(26,0);
    if (n<m)
    {
        return ans;
    }
    vector<vector<int>> mps(n,vector<int>(26,0));
    while (i<m)
    {
        c[p[i]-'a']++;
        i++;
    }
    i=0;
    while (i<n)
    {
        mc[s[i]-'a']++;
        j=0;
        while (j<26)
        {
            mps[i][j]=mc[j];
            j++;
        }
        i++;
    }
        i=0;
        j=0;
        while (j<26)
        {
            if (mps[m-1][j]!=c[j])
            {
                break;
            }
            j++;
        }
        if (j==26)
        {
            ans.push_back(0);
        }
    while (i+m<n)
    {
        j=0;
        while (j<26)
        {
            if (mps[i+m][j]-mps[i][j]!=c[j])
            {
                break;
            }
            j++;
        }
        if (j==26)
        {
            ans.push_back(i+1);
        }
        i++;
    }
    return ans;
}
};
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