<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) {
        unordered_map<char,int> umap;
        vector<int> ans;
        int i=0,j=0;

        for(int i=0;i<p.size();i++){
            umap[p[i]]++;
        }

        int k = p.size();
        int cnt = umap.size();
        while(j<s.size()){
            if( umap.find(s[j]) != umap.end() ) {
                umap[s[j]]--;
                if(umap[s[j]] == 0){
                    cnt--;
                }
            }

            if(j-i+1 <k) {
                j++;
            } else if(j-i+1 == k) {
                if(cnt == 0){
                    ans.push_back(i);
                }
                if(umap.find(s[i]) != umap.end()) {
                    umap[s[i]]++;
                    if(umap[s[i]] == 1) {
                        cnt++;
                    }
                }
                i++;
                j++;
            }
        }

        return ans;
    }
};
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