<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) {
        // create a hashmap to store the frequency of each word in words
        unordered_map<string, int> hash;
        for (auto word : words) {
            if (hash.find(word) == hash.end()) {
                hash[word] = 1;
            } else {
                hash[word]++;
            }
        }

        // create a sliding window to check against the hashmap
        int N = s.size();
        int n = words.size();
        int m = words[0].size();
        
        vector<int> res;
        int start;
        int end;
        unordered_map<string, int> wordCount;
        int wordsFound;
        
        // iterate start through entire s
        for (int i = 0; i < m; i++) {
            start = i;
            end = start;
            wordCount = hash;
            wordsFound = 0;            

            while (start <= N - m * n) {
                // get the word between end and end + m
                string sub = s.substr(end, m);

                // if match
                if (wordCount.count(sub) > 0) {
                    
                    // if not exceed
                    if (wordCount[sub] > 0) {
                        wordCount[sub]--;
                        wordsFound++;

                        // if all words found, record start, then start += m, end += m        
                        if (wordsFound == n) {
                            res.push_back(start);
                            wordCount[s.substr(start, m)]++;
                            wordsFound--;
                            start += m;
                            end += m;
                        }
                        
                        // if not all words found, go to next, end += m
                        else {
                            end += m;
                        }
                    }
                    // else if match but exceed, move start += m
                    else {
                        string startSub;
                        do {
                            startSub = s.substr(start, m);
                            wordCount[startSub]++;
                            wordsFound--;
                            start += m;
                        } while (startSub != sub);
                    }
                }
                
                // else if not match, move start to end + m, end = start
                else {
                    start = end + m;
                    end = start;
                    wordCount = hash;
                    wordsFound = 0;
                }
            }
        }

        return res;
    }
};
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