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Java-based LeetCode algorithm problem solutions, regularly updated
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package g2301_2400.s2306_naming_a_company;

// #Hard #Array #String #Hash_Table #Bit_Manipulation #Enumeration
// #2022_06_15_Time_486_ms_(74.14%)_Space_83.3_MB_(58.98%)

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

/**
 * 2306 - Naming a Company\.
 *
 * Hard
 *
 * 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 idea_A and idea_B.
 * 2.  Swap the first letters of idea_A and idea_B with each other.
 * 3.  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.
 * 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**.
**/
public class Solution {
    private long count(Map> map, char a, char b) {
        if (!map.containsKey(a) || !map.containsKey(b)) {
            return 0;
        }
        long common = 0;
        Set first = map.get(a);
        Set 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> map = new HashMap<>();
        for (String idea : ideas) {
            char startChar = idea.charAt(0);
            Set 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;
    }
}