Introduction

In this blog post, we will explore how to solve the LeetCode problem "2306" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

• Efficient Grouping: Group ideas by their initial character. This allows us to quickly compare sets of suffixes for different starting letters.
  • Counting Valid Pairs: For each pair of initial characters, determine how many suffixes are unique to each group after the swap. The product of these counts gives the number of valid names that can be formed with those initial letters in one order.
  • Avoiding Duplicates: Multiply the count by 2 to account for both possible orders of the names (ideaA ideaB and ideaB ideaA).

• Runtime Complexity: O(N), where N is the total number of characters across all strings in ideas. This is because we iterate through the strings once to create the groups and then iterate through possible pairs of first letters (which is at most 26*26).
• Storage Complexity: O(N), where N is the total number of characters across all strings in ideas, since in the worst case the set suffixes in the group will store all suffixes of the strings.

Code

    def distinctNames(ideas):
    groups = [set() for _ in range(26)]
    for idea in ideas:
        groups[ord(idea[0]) - ord('a')].add(idea[1:])

    ans = 0
    for i in range(26):
        for j in range(i + 1, 26):
            common = len(groups[i] & groups[j])
            ans += (len(groups[i]) - common) * (len(groups[j]) - common)

    return ans * 2