Wayfire的最新coding面试题来了!让我们深入了解这些面试题,看看他们究竟在考察什么吧。今天的问题是常规的算法题 + 一个sql题,本文特别适合那些需要深入准备Wayfire面试的候选人。
Q1: String Compression Challenge
Question Description: A student decides to perform some operations on big words to compress them, so they become easy to remember. An operation consists of choosing a group of K consecutive equal characters and removing them. The student keeps performing this operation as long as it is possible. Determine the final word after the operation is performed.
Example:
- Input:
word = "abbcccbb"
,k = 3
- Remove
k = 3
characters 'c', now word = "abbbb". - Remove 3 characters 'b', so the final word is "a".
- Remove
- Output:
"a"
It can be easily proven that the final word will be unique. Also, it is guaranteed that the final word consists of at least one character.
Function Description: Complete the function compressWord
with the following parameters:
string word
: a string of lowercase English lettersint k
: the number of consecutive equal characters
Returns:
string
: the final word after all operations.
Constraints:
- 1 ≤ |word| ≤ 10^5
Solution Approach:
To tackle this problem, we can use a stack-based approach to efficiently manage the compression operations. Here’s a concise plan:
- Stack Usage: Use a stack to keep track of characters and their counts.
- Character Counting: Traverse the word and count consecutive characters.
- Removal Condition: If a character’s count reaches
k
, remove those characters from the stack. - Reconstruction: Reconstruct the final word from the remaining characters in the stack.
Q2: Cities With More Customers Than Average
Question Description: Write a query which will return all cities with more customers than the average number of customers of all cities. For each such city, return the country name, the city name and the number of customers. Order the result by country name ascending.
Table Definitions and Data Sample:
Table: country
id | country_name |
---|---|
1 | Austria |
2 | Germany |
3 | United Kingdom |
Table: city
id | city_name | postal_code | country_id |
---|---|---|---|
1 | Wien | 1010 | 1 |
2 | Berlin | 10115 | 2 |
3 | Hamburg | 20095 | 2 |
4 | London | EC4V 4AD | 3 |
Table: customer
id | customer_name | city_id |
---|---|---|
1 | John | 1 |
2 | Maria | 2 |
3 | Alex | 3 |
4 | Sara | 4 |
Solution Approach:
To solve this problem, we need to perform the following steps:
- Calculate the average number of customers per city.
- Select cities with a customer count above this average.
- Join the tables to get the country names and sort the result.
Interview Follow-Up Questions:
1. Difference between JOIN
and LEFT JOIN
:
- Question: Why use
LEFT JOIN
instead ofJOIN
? - Explanation: A
JOIN
(orINNER JOIN
) returns rows when there is a match in both tables. ALEFT JOIN
returns all rows from the left table and the matched rows from the right table; if no match is found, NULLs are returned for columns from the right table. - Reason: Using
LEFT JOIN
ensures all cities are included, even if they don't have customers. This approach is essential to avoid missing data, especially if some cities do not have customers or some customers do not have corresponding city entries.
2. Handling Missing Data:
- Question: What if the city doesn't have customers or what if the customer doesn't have the city listed?
- Explanation: This scenario highlights the importance of using
LEFT JOIN
to ensure that all cities are listed, regardless of whether they have customers or not. If a city has no customers, it will still appear in the results with a NULL count for customers, ensuring comprehensive data representation.
通过解析这些问题和解答,求职者不仅能展示他们的SQL能力,还能展示他们在数据处理和逻辑推理方面的能力。这些能力对于Wayfire的面试至关重要。祝大家面试顺利!
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