<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:
//using maps and sliding window approach
//If I dont use checker function, it will give me TLE for one case, in which substrings are repeated in whole 's'. So checker will help me to reduce that TC
//TC is O(n)*O(word_len)*O()
    vector<int> findSubstring(string s, vector<string>& words) {
        int n=s.size();
        vector<int> ans;
        int word_arr_len=words.size();
        int word_len=words[0].size();
        if(word_len==0 || word_arr_len==0) return ans;// Edge cases
        unordered_map<string,int> count;//{words[i],count}
        for(auto x:words) count[x]++;
        // Loop through possible starting points of substrings

        //why I am using checker function?-> to decrease time complexity to check the same sliding window word
        unordered_map<string,bool> checker;//{earlier used string, does it match or not with any of permutations}
        for(int i=0;i<=n-word_arr_len*word_len;i++)
        {
            string x= s.substr(i,word_len*word_arr_len);
            if(checker.find(x)!=checker.end())//means this word 'x' has been checked before
            {
                if(checker[x]==true) ans.push_back(i);
                continue;//when in the map but no combination can match
                //or if matched then we have already inserted this in the answer.
            }

            unordered_map<string,int> mp;//{substring,count}
            int j=0;
            while(j<word_arr_len)
            {
                string temp=s.substr(i+j*word_len,word_len);
                mp[temp]++;
                if(mp[temp]>count[temp]) break;// If substring count exceeds required count
                j++;
            }


            if(j==word_arr_len)//means this word is matching requirements
            {
                ans.push_back(i);
                checker[x]=true;//to avoid checking this string in further iterations
            }
            else checker[x]=false;
            //above line means we have matched all words in the words array
        }

        return ans;
    }
};
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