<p>You are given an array <code>words</code> of size <code>n</code> consisting of <strong>non-empty</strong> strings.</p>

<p>We define the <strong>score</strong> of a string <code>word</code> as the <strong>number</strong> of strings <code>words[i]</code> such that <code>word</code> is a <strong>prefix</strong> of <code>words[i]</code>.</p>

<ul>
	<li>For example, if <code>words = [&quot;a&quot;, &quot;ab&quot;, &quot;abc&quot;, &quot;cab&quot;]</code>, then the score of <code>&quot;ab&quot;</code> is <code>2</code>, since <code>&quot;ab&quot;</code> is a prefix of both <code>&quot;ab&quot;</code> and <code>&quot;abc&quot;</code>.</li>
</ul>

<p>Return <em>an array </em><code>answer</code><em> of size </em><code>n</code><em> where </em><code>answer[i]</code><em> is the <strong>sum</strong> of scores of every <strong>non-empty</strong> prefix of </em><code>words[i]</code>.</p>

<p><strong>Note</strong> that a string is considered as a prefix of itself.</p>

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

<pre>
<strong>Input:</strong> words = [&quot;abc&quot;,&quot;ab&quot;,&quot;bc&quot;,&quot;b&quot;]
<strong>Output:</strong> [5,4,3,2]
<strong>Explanation:</strong> The answer for each string is the following:
- &quot;abc&quot; has 3 prefixes: &quot;a&quot;, &quot;ab&quot;, and &quot;abc&quot;.
- There are 2 strings with the prefix &quot;a&quot;, 2 strings with the prefix &quot;ab&quot;, and 1 string with the prefix &quot;abc&quot;.
The total is answer[0] = 2 + 2 + 1 = 5.
- &quot;ab&quot; has 2 prefixes: &quot;a&quot; and &quot;ab&quot;.
- There are 2 strings with the prefix &quot;a&quot;, and 2 strings with the prefix &quot;ab&quot;.
The total is answer[1] = 2 + 2 = 4.
- &quot;bc&quot; has 2 prefixes: &quot;b&quot; and &quot;bc&quot;.
- There are 2 strings with the prefix &quot;b&quot;, and 1 string with the prefix &quot;bc&quot;.
The total is answer[2] = 2 + 1 = 3.
- &quot;b&quot; has 1 prefix: &quot;b&quot;.
- There are 2 strings with the prefix &quot;b&quot;.
The total is answer[3] = 2.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;abcd&quot;]
<strong>Output:</strong> [4]
<strong>Explanation:</strong>
&quot;abcd&quot; has 4 prefixes: &quot;a&quot;, &quot;ab&quot;, &quot;abc&quot;, and &quot;abcd&quot;.
Each prefix has a score of one, so the total is answer[0] = 1 + 1 + 1 + 1 = 4.
</pre>

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 1000</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

class Node:
    def __init__(self, isWord = False, adj = None, value = 0):
        self.isWord= False
        self.adj = adj
        self.value = value

class Trie:
    def __init__(self):
        self.node = Node()
    def addWord(self, word):
        def dfs(word, node):
            node.value += 1
            if len(word) == 0:
                node.isWord = True
                return
            if node.adj == None:
                node.adj = dict()
            if word[0] in node.adj:
                dfs(word[1:], node.adj[word[0]])
            else:
                newNode = Node()
                node.adj[word[0]] = newNode
                dfs(word[1:], newNode)
        dfs(word, self.node)
        # make this correct
        self.node.value = 0
    def searchMod(self, word):
        ans = [0]
        def dfs(word, node):
            ans[0] += node.value
            if len(word) == 0:
                return node.isWord
            if node.adj == None:
                return False
            if word[0] in node.adj:
                dfs(word[1:], node.adj[word[0]])
            return False
        dfs(word, self.node)
        return ans[0]

class Solution:
    def sumPrefixScores(self, words: List[str]) -> List[int]:
        n = len(words)
        ans = []
        trie = Trie()
        for w in words:
            trie.addWord(w)
        for w in words:
            ans.append(trie.searchMod(w))
        return ans
3682