{x}
blog image

Naming a Company

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.

Solution Explanation: Naming a Company

This problem involves finding the number of distinct valid company names that can be formed by swapping the first letters of two distinct names from a given list. The validity of a company name depends on whether the two modified names (after the letter swap) are not present in the original list.

Approach: Enumeration and Counting

The most straightforward approach is to enumerate all possible pairs of distinct names from the input list (ideas), swap their first letters, and check if the resulting names are valid. This can be optimized using a hash table (or set) for efficient lookup of existing names.

Algorithm:

  1. Create a set: First, create a set (s) containing all the names from the ideas list. This allows for O(1) lookup time to check if a name exists.

  2. Iterate through pairs: Iterate through all possible pairs of distinct names from ideas. This requires nested loops.

  3. Swap first letters: For each pair ( ideaA, ideaB), swap their first letters to create two new names (newIdeaA, newIdeaB).

  4. Check for validity: Check if both newIdeaA and newIdeaB are not present in the set s. If they are not, the combined name (ideaA + " " + ideaB) is a valid company name.

  5. Count valid names: Increment a counter for each valid company name found.

  6. Return the count: Return the final count of distinct valid company names.

Time and Space Complexity Analysis:

  • Time Complexity: The dominant factor in the time complexity is the nested loop iterating through all pairs of names in the ideas list. This results in O(n^2), where n is the number of names in the ideas list. The operations within the loop (swapping letters and set lookups) take constant time, O(1).

  • Space Complexity: The space complexity is dominated by the set s, which stores all the names from ideas. This requires O(n) space, where n is the number of names.

Code Implementation (Python):

def distinctNames(ideas):
    s = set(ideas)  # Create a set for efficient lookup
    count = 0
    for i in range(len(ideas)):
        for j in range(i + 1, len(ideas)):  # Avoid duplicate pairs
            ideaA = list(ideas[i])
            ideaB = list(ideas[j])
 
            # Swap first letters
            ideaA[0], ideaB[0] = ideaB[0], ideaA[0]
            newIdeaA = "".join(ideaA)
            newIdeaB = "".join(ideaB)
 
            # Check validity
            if newIdeaA not in s and newIdeaB not in s:
                count += 1
 
    return count

The code in other languages (Java, C++, Go) follows a similar approach, using appropriate data structures (sets or hash sets) for efficient name lookups. The overall time and space complexity remain the same.