<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:
    std::vector<int> findAnagrams(const std::string& s, const std::string& p) {
        std::unordered_map<char, int> pChars{};
        for (const auto i: p) {
            ++pChars[i];
        }
        std::vector<int> result{};
        std::unordered_map<char, int> curr{};
        for (int i{}; i < p.size(); ++i) {
            ++curr[s[i]];
        }
        if (curr == pChars) result.push_back(0);
        for (int i{}; (i + p.size() - 1) < s.size(); ++i) {
            ++curr[s[i + p.size()]];
            --curr[s[i]];
            if (curr[s[i]] == 0) curr.erase(s[i]);
            if (curr == pChars) result.push_back(i + 1);
        }
        return result;
    }
};
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