<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) {
        // convert p to a map of unique letters to counts
        
        std::unordered_map<char, int> letterCounts(p.length());
        
        for (int i = 0; i < p.length(); ++i)
        {
            if (letterCounts.count(p[i]) > 0)
            {
                ++letterCounts[p[i]];
            }
            else
            {
                letterCounts.emplace(p[i], 1);
            }
        }
        
        vector<int> anagramIndicies;
        deque<char> currentCandidate; // use a deque?
        int deficit = 0;

        for (int i = 0; i < s.length(); ++i)
        { 
            if (letterCounts.count(s[i]) == 0)
            {
                // encountered an invalid character

                for (const char& c : currentCandidate)
                {
                    if (letterCounts[c] < 0)
                    {
                        --deficit;
                    }

                    ++letterCounts[c];
                }
                
                currentCandidate.clear();

                continue;
            }
            else if (letterCounts[s[i]] <= 0)
            {
                ++deficit;
            }
            
            --letterCounts[s[i]];
            currentCandidate.emplace_back(s[i]);

            if (currentCandidate.size() == p.length())
            {
                if (deficit == 0)
                {
                    anagramIndicies.emplace_back(i - p.length() + 1);
                }
                
                if (letterCounts[currentCandidate.front()] < 0)
                {
                    --deficit;
                }
                
                ++letterCounts[currentCandidate.front()];
                
                currentCandidate.pop_front();
            }
        }
        
        return anagramIndicies;
    }
};
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