<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>

# 30. Substring with Concatenation of All Words


class TrieNode:
    def __init__(self) -> None:
        self.alphaMap = [None for _ in range(26)]
        self.isWord = 0
        self.currIndex = -1
        self.currIndexCount = 0

class Trie:
    def __init__(self) -> None:
        self.root = TrieNode()
    
    def addWord(self, word):
        currNode = self.root
        for i in word:
            if currNode.alphaMap[self.getIndex(i)] == None:
                newNode = TrieNode()
                currNode.alphaMap[self.getIndex(i)] = newNode
                currNode = newNode
            else:
                currNode = currNode.alphaMap[self.getIndex(i)]
        currNode.isWord+=1
    
    def containsWord(self, word):
        currNode = self.root
        exists = True
        for i in word:
            if currNode.alphaMap[self.getIndex(i)] == None:
                exists = False
                break
            else:
                currNode = currNode.alphaMap[self.getIndex(i)]
        if exists:
            exists = currNode.isWord > 0
        return exists   

    def containsWordComplex(self, word, index):
        currNode = self.root
        exists = True
        for i in word:
            if currNode.alphaMap[self.getIndex(i)] == None:
                exists = False
                break
            else:
                currNode = currNode.alphaMap[self.getIndex(i)]
        if exists:
            exists = currNode.isWord > 0
        return exists             

    def getIndex(self, alphabet):
        return ord(alphabet) - ord('a')
    
class Solution:
    def findSubstring(self, s: str, words):
        answer = []
        trie = Trie()
        for i in words:
            trie.addWord(i)
        wordsLength = len(words)
        wordL = len(words[0])

        #lenght
        l = len(s)
        if wordL == 1 and words.count(words[0]) == wordsLength and s[0] == words[0] and s.count(s[0]) == len(s):
            kk = l-(wordsLength*wordL) +1
            if kk>0:
                answer = [_ for _ in range(kk)]
                return answer
            else:
                return []  
        # loop to interate from i=0 len(s)
        i = 0
        while(i<=(l-(wordsLength*wordL))):
            currI = i
            isHope = False
            wordsCompleted = 0
            while(currI < l):
                localCurrNode = trie.root
                wordsCompleted
                while(currI <l and localCurrNode.alphaMap[trie.getIndex(s[currI])] != None):
                    localCurrNode = localCurrNode.alphaMap[trie.getIndex(s[currI])]
                    currI +=1
                if localCurrNode.isWord <= 0: # there's no valid word of s[i:currI]
                    isHope = False
                    break
                else:
                    if localCurrNode.currIndex != i:
                        localCurrNode.currIndex = i
                        localCurrNode.currIndexCount = 0
                    localCurrNode.currIndexCount +=1
                    if localCurrNode.currIndexCount > localCurrNode.isWord:
                        isHope = False
                        break
                wordsCompleted+=1
                if wordsCompleted >= wordsLength:
                    isHope= True
                    break
            if isHope:
                answer.append(i)
            i+=1
        return answer
        
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