<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;
        if (p.size() > s.size()) {
            return res;
        }

        map<char, int> m;
        for (char c : p) {
            m[c] -= 1;
        }

        for (int i = 0; i < p.size(); i++) {
            m[s[i]] += 1;
            if (m[s[i]] == 0) {
                m.erase(s[i]);
            }
        }

        if (m.size() == 0) {
            res.push_back(0);
        }

        for (int r = p.size(); r < s.size(); r++) {
            int l = r - p.size();
            m[s[r]] ++;
            m[s[l]] --;
            if (m[s[r]] == 0) {
                m.erase(s[r]);
            }
            if (m[s[l]] == 0) {
                m.erase(s[l]);
            }

            if (m.size() == 0) {
                res.push_back(l+1);
            }
        }
        
        return res;
    }
};

// 1. sort strings
// 2. map[counts] O(n * k)
// 3. map[counts] O(n)
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