<p>You are given a string <code>s</code> and an array of strings <code>words</code>. All the strings of <code>words</code> are of <strong>the same length</strong>.</p>

<p>A <strong>concatenated string</strong> is a string that exactly contains all the strings of any permutation of <code>words</code> concatenated.</p>

<ul>
	<li>For example, if <code>words = [&quot;ab&quot;,&quot;cd&quot;,&quot;ef&quot;]</code>, then <code>&quot;abcdef&quot;</code>, <code>&quot;abefcd&quot;</code>, <code>&quot;cdabef&quot;</code>, <code>&quot;cdefab&quot;</code>, <code>&quot;efabcd&quot;</code>, and <code>&quot;efcdab&quot;</code> are all concatenated strings. <code>&quot;acdbef&quot;</code> is not a concatenated string because it is not the concatenation of any permutation of <code>words</code>.</li>
</ul>

<p>Return an array of <em>the starting indices</em> of all the concatenated substrings in <code>s</code>. You can return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;barfoothefoobarman&quot;, words = [&quot;foo&quot;,&quot;bar&quot;]</span></p>

<p><strong>Output:</strong> <span class="example-io">[0,9]</span></p>

<p><strong>Explanation:</strong></p>

<p>The substring starting at 0 is <code>&quot;barfoo&quot;</code>. It is the concatenation of <code>[&quot;bar&quot;,&quot;foo&quot;]</code> which is a permutation of <code>words</code>.<br />
The substring starting at 9 is <code>&quot;foobar&quot;</code>. It is the concatenation of <code>[&quot;foo&quot;,&quot;bar&quot;]</code> which is a permutation of <code>words</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;wordgoodgoodgoodbestword&quot;, words = [&quot;word&quot;,&quot;good&quot;,&quot;best&quot;,&quot;word&quot;]</span></p>

<p><strong>Output:</strong> <span class="example-io">[]</span></p>

<p><strong>Explanation:</strong></p>

<p>There is no concatenated substring.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;barfoofoobarthefoobarman&quot;, words = [&quot;bar&quot;,&quot;foo&quot;,&quot;the&quot;]</span></p>

<p><strong>Output:</strong> <span class="example-io">[6,9,12]</span></p>

<p><strong>Explanation:</strong></p>

<p>The substring starting at 6 is <code>&quot;foobarthe&quot;</code>. It is the concatenation of <code>[&quot;foo&quot;,&quot;bar&quot;,&quot;the&quot;]</code>.<br />
The substring starting at 9 is <code>&quot;barthefoo&quot;</code>. It is the concatenation of <code>[&quot;bar&quot;,&quot;the&quot;,&quot;foo&quot;]</code>.<br />
The substring starting at 12 is <code>&quot;thefoobar&quot;</code>. It is the concatenation of <code>[&quot;the&quot;,&quot;foo&quot;,&quot;bar&quot;]</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words.length &lt;= 5000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
	<li><code>s</code> and <code>words[i]</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        unordered_map<string, int> m;
        unordered_map<string, int> freq;
        vector<int> res;
        int n = words[0].size();
        for(int i=0; i<words.size(); i++){
            m[words[i]] = i + 1;
            freq[words[i]]++;
        }
        for(int start = 0; start < n; start++)
        {
            vector<int> visit(words.size()+1, 0);
            int found = 0;
            queue<string> q;
            for(int i=start; i+n-1<s.size(); i+=n)
            {
                string sub_string;
                for(int j=0; j<n; j++)
                    sub_string.push_back(s[i+j]);

                int new_found = m[sub_string], max_freq = freq[sub_string];


                if(new_found && visit[new_found] < max_freq)
                {
                    visit[new_found]++;
                    found++;
                    q.push(sub_string);
                }else if(new_found && visit[new_found] >= max_freq)
                {
                    visit[new_found]++;
                    found++;
                    q.push(sub_string);
                    while(visit[new_found] > max_freq)
                    {
                        string front = q.front();
                        q.pop();
                        int idx = m[front];
                        visit[idx]--;
                        found--;
                    }
                }else
                {
                    found = 0;
                    visit = vector<int>(words.size()+1, 0);
                    while(!q.empty())
                        q.pop();
                }

                if(found == words.size())
                {
                    string front = q.front();
                    q.pop();
                    int idx = m[front];
                    visit[idx]--;
                    found--;
                    res.push_back( i - n * words.size() + n);
                }
            }


        }
        return res;

    }
};
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