How do you solve Naming a Company on LeetCode?

Naming a Company (LeetCode #2306) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Grouping ideas by suffixes allows for efficient counting of valid pairs.

What is the brute force solution for Naming a Company?

The brute force approach for Naming a Company is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible pair of names from the list. For each pair, we swap the first letters and check if the new names are valid by ensuring they don't exist in the original list. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Naming a Company?

Think about how to reduce the number of checks needed by grouping similar items. Always consider edge cases, such as names that might be too similar. Explain your thought process clearly to the interviewer as you work through the problem.

What are common mistakes when solving Naming a Company?

Not considering pairs of names that share the same suffix. Failing to check if new names exist in the original list correctly.

#2306

Naming a Company

Hard

ArrayHash TableStringBit ManipulationEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach groups ideas by their suffixes (everything except the first letter) and counts how many ideas share the same suffix. This allows us to quickly determine valid pairs without checking every combination.

⚙️

Algorithm

5 steps

1Step 1: Create a map to count the number of ideas for each suffix.
2Step 2: Loop through each idea and populate the suffix count map.
3Step 3: For each unique first letter, calculate valid pairs using the counts from the suffix map.
4Step 4: For each pair of different first letters, multiply the counts of their suffixes to get valid combinations.
5Step 5: Return the total valid combinations.

solution.py12 lines

1def distinctNames(ideas):
2    from collections import defaultdict
3    suffix_count = defaultdict(int)
4    for idea in ideas:
5        suffix_count[idea[1:]] += 1
6    valid_names_count = 0
7    first_letters = set(idea[0] for idea in ideas)
8    for letter_a in first_letters:
9        for letter_b in first_letters:
10            if letter_a != letter_b:
11                valid_names_count += (len(ideas) - suffix_count[letter_a]) * (len(ideas) - suffix_count[letter_b])
12    return valid_names_count

ℹ

Complexity note: The time complexity is O(n) because we only loop through the list a couple of times. The space complexity is O(n) due to the suffix count map storing counts for each unique suffix.

1Grouping ideas by suffixes allows for efficient counting of valid pairs.
2Swapping first letters only matters if the resulting names are not in the original list.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.