<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) {
        
        vector<int> res;
        
        int ns = s.length();
        int np = p.length();
        
        if(ns<np) {
            return res;
        }
        
        unordered_map<char,int> mp;
        unordered_set<char> ms;
        
        for( char ch: p) {
            mp[ch] += 1;
            ms.insert(ch);
        }
        
        int i = 0;
        int j = np-1;
        
        auto upd = [&]( char key, int val)->void {
            mp[key] += val;
            if(mp[key]==0) {
                ms.erase(key);
            } else {
                ms.insert(key);
            }
        };
        
        for( int ii=0; ii<np; ii++) {
            upd(s[ii], -1);
        }
        
        while(true) {
            if(ms.empty()) {
                res.push_back(i);
            } 
            if(j+1>=ns) {
                break;
            }
            j += 1;
            upd( s[j], -1);
            upd( s[i], 1);
            i += 1;
        }
        
        return res;
    }
};
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