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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:

  1. Choose 2 distinct names from ideas, call them ideaA and ideaB.
  2. Swap the first letters of ideaA and ideaB with each other.
  3. If both of the new names are not found in the original ideas, then the name ideaA ideaB (the concatenation of ideaA and ideaB, separated by a space) is a valid company name.
  4. 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 Counting

We define $f[i][j]$ to represent the number of strings in $ideas$ that start with the $i$th letter and are not in $ideas$ after being replaced with the $j$th letter. Initially, $f[i][j] = 0$. Additionally, we use a hash table $s$ to record the strings in $ideas$, which allows us to quickly determine whether a string is in $ideas$.

Next, we traverse the strings in $ideas$. For the current string $v$, we enumerate the first letter $j$ after replacement. If the string after $v$ is replaced is not in $ideas$, then we update $f[i][j] = f[i][j] + 1$.

Finally, we traverse the strings in $ideas$ again. For the current string $v$, we enumerate the first letter $j$ after replacement. If the string after $v$ is replaced is not in $ideas$, then we update the answer $ans = ans + f[j][i]$.

The final answer is $ans$.

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 $ideas$ and the maximum length of the strings, respectively, and $|\Sigma|$ is the character set that appears in the string. In this problem, $|\Sigma| \leq 26$.

  • 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;
    }
}

  • 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;
    }
};

  • 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


  • 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
}

  • function distinctNames(ideas: string[]): number {
    const s = new Set(ideas);
    const f: number[][] = Array(26)
        .fill(0)
        .map(() => Array(26).fill(0));
    for (const v of s) {
        const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
        const t = [...v];
        for (let j = 0; j < 26; ++j) {
            t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
            if (!s.has(t.join(''))) {
                f[i][j]++;
            }
        }
    }
    let ans = 0;
    for (const v of s) {
        const i = v.charCodeAt(0) - 'a'.charCodeAt(0);
        const t = [...v];
        for (let j = 0; j < 26; ++j) {
            t[0] = String.fromCharCode('a'.charCodeAt(0) + j);
            if (!s.has(t.join(''))) {
                ans += f[j][i];
            }
        }
    }
    return ans;
}

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