<p>You are given an array of strings <code>products</code> and a string <code>searchWord</code>.</p>

<p>Design a system that suggests at most three product names from <code>products</code> after each character of <code>searchWord</code> is typed. Suggested products should have common prefix with <code>searchWord</code>. If there are more than three products with a common prefix return the three lexicographically minimums products.</p>

<p>Return <em>a list of lists of the suggested products after each character of </em><code>searchWord</code><em> is typed</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> products = [&quot;mobile&quot;,&quot;mouse&quot;,&quot;moneypot&quot;,&quot;monitor&quot;,&quot;mousepad&quot;], searchWord = &quot;mouse&quot;
<strong>Output:</strong> [[&quot;mobile&quot;,&quot;moneypot&quot;,&quot;monitor&quot;],[&quot;mobile&quot;,&quot;moneypot&quot;,&quot;monitor&quot;],[&quot;mouse&quot;,&quot;mousepad&quot;],[&quot;mouse&quot;,&quot;mousepad&quot;],[&quot;mouse&quot;,&quot;mousepad&quot;]]
<strong>Explanation:</strong> products sorted lexicographically = [&quot;mobile&quot;,&quot;moneypot&quot;,&quot;monitor&quot;,&quot;mouse&quot;,&quot;mousepad&quot;].
After typing m and mo all products match and we show user [&quot;mobile&quot;,&quot;moneypot&quot;,&quot;monitor&quot;].
After typing mou, mous and mouse the system suggests [&quot;mouse&quot;,&quot;mousepad&quot;].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> products = [&quot;havana&quot;], searchWord = &quot;havana&quot;
<strong>Output:</strong> [[&quot;havana&quot;],[&quot;havana&quot;],[&quot;havana&quot;],[&quot;havana&quot;],[&quot;havana&quot;],[&quot;havana&quot;]]
<strong>Explanation:</strong> The only word &quot;havana&quot; will be always suggested while typing the search word.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= products.length &lt;= 1000</code></li>
	<li><code>1 &lt;= products[i].length &lt;= 3000</code></li>
	<li><code>1 &lt;= sum(products[i].length) &lt;= 2 * 10<sup>4</sup></code></li>
	<li>All the strings of <code>products</code> are <strong>unique</strong>.</li>
	<li><code>products[i]</code> consists of lowercase English letters.</li>
	<li><code>1 &lt;= searchWord.length &lt;= 1000</code></li>
	<li><code>searchWord</code> consists of lowercase English letters.</li>
</ul>

import sortedcontainers
class Solution:
    def suggestedProducts(self, products: List[str], searchWord: str) -> List[List[str]]:
        # i suppose you could sort products lexicographically, and then
        # do a binary search for the beginning of words that match the 
        # search string up to the current index of search word,
        # a binary search to the last match, add the first 
        # (up to) 3 to the first list, and then do a binary search for the
        # first two letters and so on

        # lets just do the trie approach first, since its a more efficient
        # approach in terms of time complexity
        class Node:
            def __init__(self, children=None):
                self.val = sortedcontainers.SortedList()
                self.children = [None for _ in range(26)]
            
            def addWordToVal(self, newVal):
                self.val.add(newVal)
                if len(self.val) > 3:
                    self.val.pop()
            
        trie = Node()
        firstLetter = searchWord[0]

        for p in products:
            currNode = trie
            for i, c in enumerate(p):
                if i >= len(searchWord) or c != searchWord[i]:
                    break
                index = ord(c) - ord('a')
                if currNode.children[index] is None:
                    currNode.children[index] = Node()
                currNode = currNode.children[index]
                currNode.addWordToVal(p)
    
        currNode = trie
        res = []
        for s in searchWord:
            index = ord(s) - ord('a')
            if currNode.children[index] is None:
                break
            currNode = currNode.children[index]
            res.append(currNode.val)
        
        while len(res) < len(searchWord):
            res.append([])
        
        return res
            

        
                


                


            
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