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:
ideas, call them ideaA and ideaB.ideaA and ideaB with each other.ideas, then the name ideaA ideaB (the concatenation of ideaA and ideaB, separated by a space) is a valid company 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 * 1041 <= ideas[i].length <= 10ideas[i] consists of lowercase English letters.ideas are unique.In this approach, we will iterate over all possible pairs of ideas and attempt to swap their first letters to form new names. We will then check if these new names are unique by not being present in the original list of ideas. This approach straightforwardly uses nested loops to examine each pair, which results in an O(n^2) time complexity for the pair-wise name generation.
This implementation iterates over each possible pair of ideas and attempts to swap their first letters. If the new names generated are not found among the original ideas, it counts such pairs as valid. It restores the original names after checking for existence.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n^2 * m), where n is the number of ideas, and m is the average length of the ideas, due to the nested loop and string comparison.
Space Complexity: O(1) as we are not using any additional space proportional to input size beyond function parameters.
This improved approach seeks to avoid redundant swaps by organizing names into groups based on their prefix letters. If two sets have no overlap with original names when swapping the first characters of the names from those sets, we count the combinations. By limiting unnecessary swaps, this method reduces repetitive calculations.
This C solution organizes names by their starting letters and checks pairwise swapping between distinct prefix sets. By checking overlapping only within sets of the same initial character, it reduces redundant checks and improves efficiency over brute force.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n^2 * m) but more efficient than brute force due to reduced checking within grouped sets.
Space Complexity: O(n) since we store ideas organized by prefix.
| Approach | Complexity |
|---|---|
| Brute Force Approach | Time Complexity: O(n^2 * m), where n is the number of ideas, and m is the average length of the ideas, due to the nested loop and string comparison. |
| Optimized Approach Using Prefix Sets | Time Complexity: O(n^2 * m) but more efficient than brute force due to reduced checking within grouped sets. |
HOW TO NAME YOUR COMPANY (get the best Trademark!) • Jim Hart • 191,147 views views
Watch 9 more video solutions →Practice Naming a Company with our built-in code editor and test cases.
Practice on FleetCode