2306. Naming a Company

Problem

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 ideaA and ideaB.

• Swap the first letters of ideaA and ideaB with each other.

• 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.

• 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 * 10^4

• 1 <= ideas[i].length <= 10

• ideas[i] consists of lowercase English letters.

• All the strings in ideas are unique.

Solution

class Solution {
    private long count(Map<Character, Set<String>> map, char a, char b) {
        if (!map.containsKey(a) || !map.containsKey(b)) {
            return 0;
        }
        long common = 0;
        Set<String> first = map.get(a);
        Set<String> second = map.get(b);
        for (String c : first) {
            if (second.contains(c)) {
                common++;
            }
        }
        long uniqueA = first.size() - common;
        long uniqueB = second.size() - common;
        return uniqueA * uniqueB * 2L;
    }

    public long distinctNames(String[] ideas) {
        long ans = 0;
        Map<Character, Set<String>> map = new HashMap<>();
        for (String idea : ideas) {
            char startChar = idea.charAt(0);
            Set<String> values = map.getOrDefault(startChar, new HashSet<>());
            values.add(idea.substring(1));
            map.put(startChar, values);
        }
        for (int i = 0; i <= 26; i++) {
            for (int j = i + 1; j <= 26; j++) {
                long unique = count(map, (char) (i + 'a'), (char) (j + 'a'));
                ans += unique;
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).