<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) {
        int n = s.size(); 
        int m = p.size(); 

        vector<int> hash(26); 
        for (auto ch: p) {
            hash[ch - 'a']++; 
        }

        int dp[26][n]; 
        memset(dp, 0, sizeof(dp)); 

        for (int i = 0; i < n; i++) {
            dp[s[i] - 'a'][i]++; 
        }

        for (int i = 0; i < 26; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] += dp[i][j - 1]; 
            }
        }

        vector<int> ans; 

        for (int i = 0; i < n - m + 1; i++) {
            vector<int> curr(26); 
            int l = i; 
            int r = i + m - 1; 
            for (int j = 0; j < 26; j++) {
                curr[j] = dp[j][r]; 
                if (l - 1 >= 0) {
                    curr[j] -= dp[j][l - 1]; 
                }
            }
            if (curr == hash) {
                ans.push_back(i); 
            }
        }
        return ans; 
    }
};
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