<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 iszero(vector<int>freq){
            for(int i=0;i<freq.size();i++){
                if(freq[i] !=0){
                    return false;
                }
            }
            return true;
        }

    vector<int> findAnagrams(string s, string p) {
        
        int k=p.size();
        // window size =k

        int n=s.size();
        int cnt=0;
        vector<int>soln;
        if(p.size()>s.size()){return soln;}

        vector<int>freq(26,0);
        for(int i=0;i<k;i++){
            // increased freq of p
            freq[p[i]-'a']++;
        }

        int i=0;
        int j=k-1;

        // window size is from i to j

        for(int l=0;l<=j;l++){
            freq[s[l]-'a']--;
            // decreased freq of first window
        }

        if(iszero(freq)==true){
                cnt++;
                soln.push_back(0);
                // window initialy starts from 0;
            }

            j++;

        while(j<n){

            char to_exclude=s[i];
            i++;
            freq[to_exclude-'a']++;
        
            
            freq[s[j]-'a']--;

            if(iszero(freq)==true){
                cnt++;
                soln.push_back(i);
            }

            // now we will increment j and include jTh element in window
            // in next iteration
            j++;

        }

        return soln;

    }
};
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