Count Prefix and Suffix Pairs II
<p>You are given a <strong>0-indexed</strong> string array <code>words</code>.</p>

<p>Let&#39;s define a <strong>boolean</strong> function <code>isPrefixAndSuffix</code> that takes two strings, <code>str1</code> and <code>str2</code>:</p>

<ul>
	<li><code>isPrefixAndSuffix(str1, str2)</code> returns <code>true</code> if <code>str1</code> is <strong>both</strong> a <span data-keyword="string-prefix">prefix</span> and a <span data-keyword="string-suffix">suffix</span> of <code>str2</code>, and <code>false</code> otherwise.</li>
</ul>

<p>For example, <code>isPrefixAndSuffix(&quot;aba&quot;, &quot;ababa&quot;)</code> is <code>true</code> because <code>&quot;aba&quot;</code> is a prefix of <code>&quot;ababa&quot;</code> and also a suffix, but <code>isPrefixAndSuffix(&quot;abc&quot;, &quot;abcd&quot;)</code> is <code>false</code>.</p>

<p>Return <em>an integer denoting the <strong>number</strong> of index pairs </em><code>(i<em>, </em>j)</code><em> such that </em><code>i &lt; j</code><em>, and </em><code>isPrefixAndSuffix(words[i], words[j])</code><em> is </em><code>true</code><em>.</em></p>

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

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;aba&quot;,&quot;ababa&quot;,&quot;aa&quot;]
<strong>Output:</strong> 4
<strong>Explanation:</strong> In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix(&quot;a&quot;, &quot;aba&quot;) is true.
i = 0 and j = 2 because isPrefixAndSuffix(&quot;a&quot;, &quot;ababa&quot;) is true.
i = 0 and j = 3 because isPrefixAndSuffix(&quot;a&quot;, &quot;aa&quot;) is true.
i = 1 and j = 2 because isPrefixAndSuffix(&quot;aba&quot;, &quot;ababa&quot;) is true.
Therefore, the answer is 4.</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;pa&quot;,&quot;papa&quot;,&quot;ma&quot;,&quot;mama&quot;]
<strong>Output:</strong> 2
<strong>Explanation:</strong> In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix(&quot;pa&quot;, &quot;papa&quot;) is true.
i = 2 and j = 3 because isPrefixAndSuffix(&quot;ma&quot;, &quot;mama&quot;) is true.
Therefore, the answer is 2.  </pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;abab&quot;,&quot;ab&quot;]
<strong>Output:</strong> 0
<strong>Explanation: </strong>In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix(&quot;abab&quot;, &quot;ab&quot;) is false.
Therefore, the answer is 0.</pre>

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10<sup>5</sup></code></li>
	<li><code>words[i]</code> consists only of lowercase English letters.</li>
	<li>The sum of the lengths of all <code>words[i]</code> does not exceed <code>5 * 10<sup>5</sup></code>.</li>
</ul>
Trie; Array; String; String Matching; Hash Function; Rolling Hash
class Node:
    __slots__ = ["children", "cnt"]

    def __init__(self):
        self.children = {}
        self.cnt = 0


class Solution:
    def countPrefixSuffixPairs(self, words: List[str]) -> int:
        ans = 0
        trie = Node()
        for s in words:
            node = trie
            for p in zip(s, reversed(s)):
                if p not in node.children:
                    node.children[p] = Node()
                node = node.children[p]
                ans += node.cnt
            node.cnt += 1
        return ans
