<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:
bool isAllZero(vector<int> vec){
    for(int &i : vec){
        if(i != 0){
            return false;
        }
    }
    return true;
}
    vector<int> findAnagrams(string txt, string pat) {
         int n = txt.length();
	    
	    vector<int> vec(26,0);
	    
	    
	    
	    for(int i = 0; i<pat.length(); i++){
	        char ch = pat[i];
	        
	        vec[ch-'a']++;
	    }
	    
	   int i=0,j=0;
	   vector<int> res;
	   int k = pat.length();
	   
	   while(j<n){
	       vec[txt[j] - 'a']--;
	       
	       if(j-i+1 == k){
	           if(isAllZero(vec)){
	               res.push_back(i);
	           }
	           vec[txt[i] - 'a']++;
	           i++;
	       }
	       
	       
	       j++;
	   }
	   
	   return res;
    }
};
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