2306. Naming a Company - Explanation

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

        for i in range(n):
            for j in range(i + 1, n):
                A, B = ideas[j][0] + ideas[i][1:], ideas[i][0] + ideas[j][1:]
                if A not in ideasSet and B not in ideasSet:
                    res.add(A + ' ' + B)
                    res.add(B + ' ' + A)

        return len(res)

class Solution:
    def distinctNames(self, ideas: List[str]) -> int:
        wordMap = collections.defaultdict(set)
        for w in ideas:
            wordMap[w[0]].add(w[1:])

        res = 0
        for char1 in wordMap:
            for char2 in wordMap:
                if char1 == char2:
                    continue

                intersect = sum(1 for w in wordMap[char1] if w in wordMap[char2])
                distinct1 = len(wordMap[char1]) - intersect
                distinct2 = len(wordMap[char2]) - intersect
                res += distinct1 * distinct2

        return res

class Solution:
    def distinctNames(self, ideas: List[str]) -> int:
        suffixes = [set() for _ in range(26)]
        for w in ideas:
            suffixes[ord(w[0]) - ord('a')].add(w[1:])

        res = 0
        for i in range(26):
            for j in range(i + 1, 26):
                intersect = len(suffixes[i] & suffixes[j])
                res += 2 * (len(suffixes[i]) - intersect) * (len(suffixes[j]) - intersect)

        return res

class Solution:
    def distinctNames(self, ideas: List[str]) -> int:
        mp = defaultdict(lambda: [False] * 26)
        count = [[0] * 26 for _ in range(26)]
        res = 0

        for s in ideas:
            first_char = ord(s[0]) - ord('a')
            suffix = s[1:]
            mp[suffix][first_char] = True

        for suffix, arr in mp.items():
            for i in range(26):
                if arr[i]:
                    for j in range(26):
                        if not arr[j]:
                            count[i][j] += 1
                            res += count[j][i]

        return 2 * res