<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>

struct bor {
	struct node {
		map<char, int> mp;
		int c;
	};
	vector<node> a;
	int sm = 0;
	bor() {
		a.resize(1);
	};
	void set(string s) {
		int p = 0;
		for (auto i : s) {
			if (a[p].mp.count(i) == 0) {
				a[p].mp[i] = a.size();
				a.push_back({});
			}
			p = a[p].mp[i];
		}
		a[p].c++;
		sm++;
	};
	void add(string s) {
		int p = 0;
		for (auto i : s) {
			if (a[p].mp.count(i) == 0)
				return;
			p = a[p].mp[i];
		}
		if (a[p].c > 0) 
			sm--;
		a[p].c--;
	};
	void rem(string s) {
		int p = 0;
		for (auto i : s) {
			if (a[p].mp.count(i) == 0)
				return;
			p = a[p].mp[i];
		}
		if (a[p].c >= 0)
			sm++;
		a[p].c++;
	};
};

class Solution {
public:
	vector<int> findSubstring(string s, vector<string>& w) {
		int n = w.size();
		int m = w[0].size();
		vector<int> ans;
		for (int ii = 0; ii < m; ii++) {
			bor b;
			for (auto i : w)
				b.set(i);
			int k = 0;
			for (int j = ii; j < s.size(); j += m) {
				b.add(s.substr(j, m));
				k++;
				if (k > n)
					b.rem(s.substr(j - n * m, m));
				if (b.sm == 0)
					ans.push_back(j - n * m + m);
                k = min(k, n);
			}
		}
		return ans;
	}
};
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