Leetcode #2306: Naming a Company
In this guide, we solve Leetcode #2306 Naming a Company in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Hash Table, String, Enumeration
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def distinctNames(self, ideas: List[str]) -> int:
s = set(ideas)
f = [[0] * 26 for _ in range(26)]
for v in ideas:
i = ord(v[0]) - ord('a')
t = list(v)
for j in range(26):
t[0] = chr(ord('a') + j)
if ''.join(t) not in s:
f[i][j] += 1
ans = 0
for v in ideas:
i = ord(v[0]) - ord('a')
t = list(v)
for j in range(26):
t[0] = chr(ord('a') + j)
if ''.join(t) not in s:
ans += f[j][i]
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.