2306. Naming a Company 

Description

You are given an array of strings ideas 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:

Choose 2 distinct names from ideas, call them idea_A and idea_B.
Swap the first letters of idea_A and idea_B with each other.
If both of the new names are not found in the original ideas, then the name idea_A idea_B (the concatenation of idea_A and idea_B, separated by a space) is a valid company name.
Otherwise, it is not a valid name.

Return the number of distinct valid names for the company.

Example 1:

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

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

Example 2:

Input: ideas = ["lack","back"]
Output: 0
Explanation: There are no valid selections. Therefore, 0 is returned.

Constraints:

2 <= ideas.length <= 5 * 10⁴
1 <= ideas[i].length <= 10
ideas[i] consists of lowercase English letters.
All the strings in ideas are unique.

Solutions

Solution 1: Enumeration and Counting

We define $f[i][j]$ to represent the number of strings in $\textit{ideas}$ that start with the $i$-th letter and, when replaced with the $j$-th letter, do not exist in $\textit{ideas}$. Initially, $f[i][j] = 0$. Additionally, we use a hash table $s$ to record the strings in $\textit{ideas}$, allowing us to quickly determine whether a string is in $\textit{ideas}$.

Next, we traverse the strings in $\textit{ideas}$. For the current string $v$, we enumerate the first letter $j$ after replacement. If the string obtained by replacing $v$ is not in $\textit{ideas}$, we update $f[i][j] = f[i][j] + 1$.

Finally, we traverse the strings in $\textit{ideas}$ again. For the current string $v$, we enumerate the first letter $j$ after replacement. If the string obtained by replacing $v$ is not in $\textit{ideas}$, we update the answer $\textit{ans} = \textit{ans} + f[j][i]$.

The final answer is $\textit{ans}$.

The time complexity is $O(n \times m \times |\Sigma|)$, and the space complexity is $O(|\Sigma|^2)$. Here, $n$ and $m$ are the number of strings in $\textit{ideas}$ and the maximum length of the strings, respectively, and $|\Sigma|$ is the character set of the strings, with $|\Sigma| \leq 26$ in this problem.

Python3

class Solution:
    def distinctNames(self, ideas: List[str]) -> int:
        s = set(ideas)
        f = [[0] * 26 for _ in range(26)]
        for v in ideas:
            i = ord(v[0]) - ord('a')
            t = list(v)
            for j in range(26):
                t[0] = chr(ord('a') + j)
                if ''.join(t) not in s:
                    f[i][j] += 1
        ans = 0
        for v in ideas:
            i = ord(v[0]) - ord('a')
            t = list(v)
            for j in range(26):
                t[0] = chr(ord('a') + j)
                if ''.join(t) not in s:
                    ans += f[j][i]
        return ans

Java

class Solution {
    public long distinctNames(String[] ideas) {
        Set<String> s = new HashSet<>();
        for (String v : ideas) {
            s.add(v);
        }
        int[][] f = new int[26][26];
        for (String v : ideas) {
            char[] t = v.toCharArray();
            int i = t[0] - 'a';
            for (int j = 0; j < 26; ++j) {
                t[0] = (char) (j + 'a');
                if (!s.contains(String.valueOf(t))) {
                    ++f[i][j];
                }
            }
        }
        long ans = 0;
        for (String v : ideas) {
            char[] t = v.toCharArray();
            int i = t[0] - 'a';
            for (int j = 0; j < 26; ++j) {
                t[0] = (char) (j + 'a');
                if (!s.contains(String.valueOf(t))) {
                    ans += f[j][i];
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long distinctNames(vector<string>& ideas) {
        unordered_set<string> s(ideas.begin(), ideas.end());
        int f[26][26]{};
        for (auto v : ideas) {
            int i = v[0] - 'a';
            for (int j = 0; j < 26; ++j) {
                v[0] = j + 'a';
                if (!s.count(v)) {
                    ++f[i][j];
                }
            }
        }
        long long ans = 0;
        for (auto& v : ideas) {
            int i = v[0] - 'a';
            for (int j = 0; j < 26; ++j) {
                v[0] = j + 'a';
                if (!s.count(v)) {
                    ans += f[j][i];
                }
            }
        }
        return ans;
    }
};

Go

func distinctNames(ideas []string) (ans int64) {
	s := map[string]bool{}
	for _, v := range ideas {
		s[v] = true
	}
	f := [26][26]int{}
	for _, v := range ideas {
		i := int(v[0] - 'a')
		t := []byte(v)
		for j := 0; j < 26; j++ {
			t[0] = 'a' + byte(j)
			if !s[string(t)] {
				f[i][j]++
			}
		}
	}

	for _, v := range ideas {
		i := int(v[0] - 'a')
		t := []byte(v)
		for j := 0; j < 26; j++ {
			t[0] = 'a' + byte(j)
			if !s[string(t)] {
				ans += int64(f[j][i])
			}
		}
	}
	return
}