<p>You are given an array of strings <code>ideas</code> that represents a list of names to be used in the process of naming a company. The process of naming a company is as follows:</p>

<ol>
	<li>Choose 2 <strong>distinct</strong> names from <code>ideas</code>, call them <code>idea<sub>A</sub></code> and <code>idea<sub>B</sub></code>.</li>
	<li>Swap the first letters of <code>idea<sub>A</sub></code> and <code>idea<sub>B</sub></code> with each other.</li>
	<li>If <strong>both</strong> of the new names are not found in the original <code>ideas</code>, then the name <code>idea<sub>A</sub> idea<sub>B</sub></code> (the <strong>concatenation</strong> of <code>idea<sub>A</sub></code> and <code>idea<sub>B</sub></code>, separated by a space) is a valid company name.</li>
	<li>Otherwise, it is not a valid name.</li>
</ol>

<p>Return <em>the number of <strong>distinct</strong> valid names for the company</em>.</p>

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

<pre>
<strong>Input:</strong> ideas = [&quot;coffee&quot;,&quot;donuts&quot;,&quot;time&quot;,&quot;toffee&quot;]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The following selections are valid:
- (&quot;coffee&quot;, &quot;donuts&quot;): The company name created is &quot;doffee conuts&quot;.
- (&quot;donuts&quot;, &quot;coffee&quot;): The company name created is &quot;conuts doffee&quot;.
- (&quot;donuts&quot;, &quot;time&quot;): The company name created is &quot;tonuts dime&quot;.
- (&quot;donuts&quot;, &quot;toffee&quot;): The company name created is &quot;tonuts doffee&quot;.
- (&quot;time&quot;, &quot;donuts&quot;): The company name created is &quot;dime tonuts&quot;.
- (&quot;toffee&quot;, &quot;donuts&quot;): The company name created is &quot;doffee tonuts&quot;.
Therefore, there are a total of 6 distinct company names.

The following are some examples of invalid selections:
- (&quot;coffee&quot;, &quot;time&quot;): The name &quot;toffee&quot; formed after swapping already exists in the original array.
- (&quot;time&quot;, &quot;toffee&quot;): Both names are still the same after swapping and exist in the original array.
- (&quot;coffee&quot;, &quot;toffee&quot;): Both names formed after swapping already exist in the original array.
</pre>

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

<pre>
<strong>Input:</strong> ideas = [&quot;lack&quot;,&quot;back&quot;]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no valid selections. Therefore, 0 is returned.
</pre>

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

<ul>
	<li><code>2 &lt;= ideas.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= ideas[i].length &lt;= 10</code></li>
	<li><code>ideas[i]</code> consists of lowercase English letters.</li>
	<li>All the strings in <code>ideas</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def distinctNames(self, ideas: List[str]) -> int:
        res = 0
        ideas_set = set(ideas)

        c_to_suffix = {}
        for w in ideas_set:
            if w[0] not in c_to_suffix:
                c_to_suffix[w[0]] = set()
            c_to_suffix[w[0]].add(w[1:])

        # suffix_to_c = collections.defaultdict(set)
        # for w in ideas_set:
        #     suffix_to_c[w[1:]].add(w[0])

        for x in c_to_suffix:
            for y in c_to_suffix:
                if x == y:
                    continue
                intersection = 0
                for suffix in c_to_suffix[x]:
                    if suffix in c_to_suffix[y]:
                        intersection += 1
                not_shared = (len(c_to_suffix[x]) - intersection) * (len(c_to_suffix[y]) - intersection)
                res += not_shared

        return res

































        # # group suffixes by first letter
        # group = collections.defaultdict(set)
        # for s in ideas:
        #     group[s[0]].add(s[1:])

        # # observe that there is no point in switching words that start with the
        # # same letter
        # # also observe that after we switch the first letter of two words a,b (that
        # # have different starting letter) to get the words c,d we have that c is in
        # # ideas if and only if a[1] = c[1] lies in both group[a[1]] and group[b[1]].
        # res = 0
        # keys = list(group.keys())
        # for i in range(len(keys)):
        #     for j in range(i + 1, len(keys)):
        #         key1, key2 = keys[i], keys[j]
        #         intersection = 0
        #         for suffix in group[key1]:
        #             if suffix in group[key2]:
        #                 intersection += 1
        #         total = (len(group[key1]) - intersection) * (len(group[key2]) - intersection)
        #         res += 2 * total
        # return res
30100