<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 sSize = s.length();
        int pSize = p.length();
        
        vector<int> answer;
        
        if(pSize > sSize) {
            return answer;
        }
        
        vector<int> sMap(26, 0);
        vector<int> pMap(26, 0);
        
        for(int i=0; i<pSize; i++) {
            pMap[p[i] - 'a']++;
        }
        
        
        int start = 0;
        int length = 0;
        
        int i = 0;
        int increment = 0;
        
        while(i<sSize) {
            if(pMap[s[i] - 'a'] != 0) {
                // we have already traversed then go forward
                
                // we haven't trabersed
                
                if( pMap[s[i] - 'a'] > sMap[s[i] - 'a']) {
                    sMap[s[i] - 'a']++;
                    length++;
                    cout<<i<<" ";
                    
                    if(length == pSize) {
                        answer.push_back(i-pSize+1);
                    }
                }else {
                    int charac = s[i] - 'a';
                    while(increment<i && sMap[charac] >= pMap[charac]) {
                        cout<<increment<<" ";
                        sMap[s[increment] - 'a']--;
                        length--;
                        increment++;
                    }
                
                    continue;
                    
                }
            }else {
                sMap = vector<int>(26, 0);
                length = 0;
                increment = i+1;
            }
            
            i++;
        }
        
        return answer;
    }
};
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