<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 n=p.size();
        //unordered_map<char,int> um;
        vector<int> um(26,0);
        for(int i=0;i<p.size();i++){
            if(um[p[i]-'a']==0){
                um[p[i]-'a']=1;
            }
            else{
                um[p[i]-'a']++;
            }
        }
        vector<int> v;
        for(int i=0;i<s.size();i++){
            if(um[s[i]-'a']!=0){
                int c=0;
                //unordered_map<char,int> um1=um;
                vector<int> um1=um;
                int j=i;
                while(c<n&&j<s.size()){
                    if(um1[(s[j]-'a')]!=0&&um1[s[j]-'a']>0){
                        um1[s[j]-'a']--;
                    }
                    else{
                        break;
                    }
                    j++;
                    c++;
                }
                if(c==n){
                    v.push_back(i);
                }
                
            }
        }
        return v;
    }
};
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