<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> ans;
    if (p.size() > s.size()) return ans;

    // Frequency map for characters in p
    map<char, int> pCount;
    for (char ch : p) {
        pCount[ch]++;
    }

    // Frequency map for characters in the current window in s
    map<char, int> sCount;
    int i = 0, j = 0;
    int pLength = p.size();
    int sLength = s.size();
    int required = pCount.size();  // Number of unique characters in p
    int matched = 0;  // Number of unique characters matched

    while (j < sLength) {
        // Add character at j to the window
        sCount[s[j]]++;

        // Check if this character is part of p and its count is correct
        if (pCount.count(s[j]) && sCount[s[j]] == pCount[s[j]]) {
            matched++;
        }

        // If window size is equal to p's size
        if (j - i + 1 == pLength) {
            // If matched all character counts, record the start index
            if (matched == required) {
                ans.push_back(i);
            }

            // Remove character at i from the window
            if (pCount.count(s[i])) {
                if (sCount[s[i]] == pCount[s[i]]) {
                    matched--;
                }
                sCount[s[i]]--;
                if (sCount[s[i]] == 0) {
                    sCount.erase(s[i]);
                }
            }

            i++;  // Slide the window
        }

        j++;  // Expand the window
    }

    return ans;
    }
};
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