<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>

class HashRange:
    mod = 344555666677777
    base = randint(mod//10, mod*9//10)
    big = 2**63
    @staticmethod
    @cache
    def pow(w):
        if w == 0: return 1
        return HashRange.pow(w-1) * HashRange.base % HashRange.mod
    def __init__(self, s=[]):
        self.s = []
        self.pre = [0]
        for c in s: self.add(c)
    def add(self, c):
        self.s.append(c)
        self.pre.append((self.pre[-1]*HashRange.base + hash(c)+HashRange.big) % HashRange.mod)
    def query(self, l, r):
        return (self.pre[r] - self.pre[l]*HashRange.pow(r-l)) % HashRange.mod
    def pop(self):
        self.s.pop()
        self.pre.pop()

class Solution:
    def countPrefixSuffixPairs(self, words: List[str]) -> int:
        cnt = Counter()
        res = 0
        for S in words:
            rk = HashRange(S)
            N = len(S)
            for w in range(1, N+1):
                res += cnt[rk.query(0, w), rk.query(N-w, N)]
            cnt[rk.query(0, N), rk.query(0, N)] += 1
        return res
2336