<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:
        initial_groups = [set() for _ in range(26)]
        for i in ideas:
            idx = ord(i[0]) - ord('a')
            initial_groups[idx].add(i[1:])
        
        print(initial_groups)
        res = 0
        for i in range(25):
            for j in range(i + 1, 26):
                a = initial_groups[i]
                b = initial_groups[j]
                intersection = len(a & b)
                a_vals = len(a) - intersection
                b_vals = len(b) - intersection
                res += 2 * a_vals * b_vals

        return res

        # first_letters_ctr = {}
        # word_end_dict = defaultdict(list)
        # for i in ideas:
        #     if i[0] in first_letters_ctr:
        #         first_letters_ctr[i[0]] += 1
        #     else:
        #         first_letters_ctr[i[0]] = 1
        #     end = i[1:]
        #     word_end_dict[end].append(i[0])

        # invalid_pair_count = 0
        # # Calculating first letter same 
        # for key, value in first_letters_ctr.items():
        #     if value > 1:
        #         temp = first_letters_ctr[key]
        #         temp = temp * (temp-1)
        #         invalid_pair_count += temp
        
        # # # Calculating other letters same
        # for key, value in word_end_dict.items():
        #     if len(value) == 1:
        #         continue
        #     temp = []
        #     for letter in value:
        #         temp.append(first_letters_ctr[letter])
            
        #     invalid_vals = 1
        #     for i in temp: 
        #         invalid_vals *= i
        #     # print(temp, invalid_vals)
        #     invalid_pair_count += (2 * invalid_vals)

        # print(first_letters_ctr, word_end_dict, invalid_pair_count)
        # all_combos = len(ideas) * (len(ideas) - 1)
        # return all_combos - invalid_pair_count

32600