Substring with Concatenation of All Words
<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>
Hash Table; String; Sliding Window
class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        unordered_map<string, int> cnt;
        for (const auto& w : words) {
            cnt[w]++;
        }

        vector<int> ans;
        int m = s.length(), n = words.size(), k = words[0].length();

        for (int i = 0; i < k; ++i) {
            int l = i, r = i;
            unordered_map<string, int> cnt1;
            while (r + k <= m) {
                string t = s.substr(r, k);
                r += k;

                if (!cnt.contains(t)) {
                    cnt1.clear();
                    l = r;
                    continue;
                }

                cnt1[t]++;

                while (cnt1[t] > cnt[t]) {
                    string w = s.substr(l, k);
                    if (--cnt1[w] == 0) {
                        cnt1.erase(w);
                    }
                    l += k;
                }

                if (r - l == n * k) {
                    ans.push_back(l);
                }
            }
        }

        return ans;
    }
};
