<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>

// 438. Find All Anagrams in a String
// https://leetcode.com/problems/find-all-anagrams-in-a-string/description/

class Solution {
public:
	bool equalHashmaps(unordered_map<char, int>& sm, unordered_map<char, int>& pm) {

		for (int i = 0 ; i < 256 ; i++) {

            if (sm[i] != pm[i])
                return false;
        }
			
		return true;
	}

    vector<int> findAnagrams(string s, string p) {

		int pn = p.size();
		int sn = s.size();

		if (sn < pn)
			return {};		// not possible

		vector<int> ans;
		unordered_map<char, int> sm;
		unordered_map<char, int> pm;
		
		for (int i = 0 ; i < pn ; i++)
			pm[p[i]]++;

		for (int i = 0 ; i < pn ; i++)
			sm[s[i]]++;

		if (equalHashmaps(sm, pm))
			ans.push_back(0);

		for (int i = pn ; i < sn ; i++) {

			sm[s[i]]++;
			sm[s[i - pn]]--;

			if (equalHashmaps(sm, pm))
				ans.push_back(i - pn + 1);
		}

		return ans;
    }
};
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