<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) {
        int wordSize = words[0].size();
        int numWords = words.size();
        int subStrSize = wordSize * numWords;
        unordered_map<string, long> wordToFreq;
        if(s.size() < wordSize)
            return {};

        for(auto word: words) {
            wordToFreq[word]++;
        }

        int l = 0, r = 0;
        //Keep track of words in window at each point
        vector<unordered_map<string, long>> window(wordSize);
        vector<int> leftBounds(wordSize, 0);
        for(int i = 0; i < wordSize; ++i) {
            window[i] = wordToFreq;
            leftBounds[i] = i;
        }

        vector<int> answer;
        cout << subStrSize << endl;
        for(; r <= s.size()- wordSize; ++r) {
            int i = r % wordSize;
            string sWord = s.substr(r, wordSize);
            if(window[i].find(sWord) != window[i].end() && window[i][sWord]-- >0) {
                if(r + wordSize - leftBounds[i] == subStrSize) {
                    answer.push_back(leftBounds[i]);
                    //Move the left pointer one unit to the right
                    window[i][s.substr(leftBounds[i], wordSize)]++;
                    leftBounds[i] += wordSize;
                }
            } else if(window[i].find(sWord) == window[i].end()) { //reset window entirely
                while(leftBounds[i] < r) {
                    window[i][s.substr(leftBounds[i], wordSize)]++;
                    leftBounds[i] += wordSize;
                }
                leftBounds[i] += wordSize;
                //include the current word if its legit
            } else { //move left pointer until legit again
                while(window[i][sWord] < 0) {
                    window[i][s.substr(leftBounds[i], wordSize)]++;
                    leftBounds[i] += wordSize;
                }
            }
        }
        return answer;

    }
};
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