Data Scientist Coding Interview Questions

Ravi K. - "select e.firstname as firstname, m.salary as manager_salary from employees e join employees m on e.manager_id = m.id where e.salary > m.salary; `"See full answer

Yeshwanth D. - "Was this for an entry level engineer role?"See full answer

KAI - "Answer: select fromcaller, count(DISTINCT tocallee) as num_calls from calls group by fromcaller having count(DISTINCT tocallee) >= 3 Setup: CREATE TABLE calls ( from_caller VARCHAR(20), to_callee VARCHAR(20) ); INSERT INTO calls (fromcaller, tocallee) VALUES ('Alice', 'Bob'), ('Charlie', 'Dave'), ('Alice', 'Frank'), ('Charlie', 'Heidi'), ('Charlie', 'Judy'); "See full answer

Salome L. - "SELECT MIN(id) AS id, TRIM(LOWER(email)) AS cleaned_email FROM users GROUP BY cleaned_email ORDER BY id `"See full answer

Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

Richard W. - "int getMaxWater(vector& nums) { int n = nums.size(); int mx = INT_MIN; int i=0, j=n-1; while(i<j) { int water = (j - i) * min(nums[i], nums[j]); mx = max(mx, water); if(nums[i] < nums[j]){ i++; } else { j--; } } return mx; } `"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

David I. - "WITH filtered_posts AS ( SELECT p.user_id, p.issuccessfulpost FROM post p WHERE p.postdate >= '2023-11-01' AND p.postdate < '2023-12-01' ), post_summary AS ( SELECT pu.user_type, COUNT(*) AS post_attempt, SUM(CASE WHEN fp.issuccessfulpost = 1 THEN 1 ELSE 0 END) AS post_success FROM filtered_posts fp JOIN postuser pu ON fp.userid = pu.user_id GROUP BY pu.user_type ) SELECT user_type, post_success, post_attempt, CAST(postsuccess AS FLOAT) / postattempt AS postsuccessrate FROM po"See full answer

Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Anonymous Prawn - "Would be better to adjust resolution in the video player directly."See full answer

Temesgen B. - "Very Good Explanation Thanks For This Rely Good Explanation"See full answer

Bradley E. - "Here's a simpler solution: select u.username , count(p.postid) as countposts from posts as p join users as u on p.userid = u.userid where p.likes >= 100 group by 1 order by 2 desc, 1 asc limit 3 `"See full answer

Vysali K. - "Limit and rank() only works if there are no 2 employees with same salary ( which is okay for this use case) For the query to pass all the test results, we need to use dense_rank with ranked_employees as ( select id, firstname, lastname, salary, denserank() over(order by salary desc) as salaryrank from employees ) select id, firstname, lastname, salary from ranked_employees where salary_rank <= 3 `"See full answer

Bala G. - "WITH ActiveUsersYesterday AS ( SELECT DISTINCT user_id FROM user_activity WHERE activity_date = CAST(GETDATE() - 1 AS DATE) ), VideoCallUsersYesterday AS ( SELECT DISTINCT user_id FROM video_calls WHERE call_date = CAST(GETDATE() - 1 AS DATE) ) SELECT (CAST(COUNT(DISTINCT v.userid) AS FLOAT) / NULLIF(COUNT(DISTINCT a.userid), 0)) * 100 AS percentagevideocall_users FROM ActiveUsersYesterday a LEFT JOIN VideoCallUsersYesterday v ON a.userid = v.userid;"See full answer

Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

Erjan G. - "def reverseString(s): chars = list(s) l, r = 0, len(s) - 1 while l < r: chars[l], chars[r] = chars[r], chars[l] l += 1 r -= 1 reversed_str = "".join(chars) return reversed_str `"See full answer

Rishabh L. - "with empbysalary as ( select id, firstname, lastname, salary, department_id, rank() over (partition by department_id order by salary desc) as rnk from employees ) select d.name as department_name, e.id as employee_id, e.firstname, e.lastname, e.salary from empbysalary e join departments d on e.department_id=d.id where e.rnk=1 order by 1; `"See full answer

Tiago R. - "function isValid(s) { const stack = []; for (let i=0; i < s.length; i++) { const char = s.charAt(i); if (['(', '{', '['].includes(char)) { stack.push(char); } else { const top = stack.pop(); if ((char === ')' && top !== '(') || (char === '}' && top !== '{') || (char === ']' && top !== '[')) { return false; } } } return stack.length === 0"See full answer

Chase C. - "Wording is a bit confusing on this one. This one can be interpreted as asking for count of unmatched bookings per user. By saying average here, what they really mean "what is the proportion of unmatched bookings to total bookings for each user." Important clarifying question to ask here if this was asked in a real interview!"See full answer