Coding Interview Questions

Laura U. - "productssold = set(transactions['productid']) unsoldproducts = products.loc[~products['id'].isin(productssold)] return unsold_products[["id", "name", "stock"]] `"See full answer

Derrick M. - "SELECT o.order_amount FROM orders o JOIN departments d ON d.departmentid = o.departmentid WHERE d.department_name = 'Fashion' ORDER BY order_amount DESC LIMIT 1 OFFSET 1; `"See full answer

XponentShift32 - "solving to find a cycle in directed graph"See full answer

Michael B. - "Problem: Given an input string txt consisting of alphanumeric characters and the parentheses characters '(' & ')', write a function which removes the minimum number of characters to return a version of the string with properly balanced parenthesis. Answer: You can do this with a counter. Psuedo-Python Start with counter = 0 output = [] Iterate through the string, every time you encounter a '(', increment the counter. Add the character to the output. If you encounter a ')', decrement the coun"See full answer

Mridul J. - "We have a list of documents. We want to build an index that maps keywords to documents containing them. Then, given a query keyword, we can efficiently retrieve all matching documents. docs = [ "Python is great for data science", "C++ is a powerful language", "Python supports OOP and functional programming", "Weather today is sunny", "Weather forecast shows rain" ]"See full answer

Mahima M. - "class Solution: def lengthOfLIS(self, nums: List[int]) -> int: temp = [nums[0]] for num in nums: if temp[-1]< num: temp.append(num) else: index = bisect_left(temp,num) temp[index] = num return len(temp) "See full answer

Azeezat R. - "/* You are with your friends in a castle, where there are multiple rooms named after flowers. Some of the rooms contain treasures - we call them the treasure rooms. Each room contains a single instruction that tells you which room to go to next. * instructions1 and treasurerooms_1 * lily* --------- daisy sunflower | | | v v v jasmin --> tulip* violet* ----> rose* --> ^ | ^ ^ | "See full answer

Maliki U. - "Here is my implementation: select marketing_channel, AVG(purchasevalue) as avgpurchase_value from attribution group by marketing_channel order by avgpurchasevalue DESC ; There is no need to copy and past the line of code for calculating the average into order by, just Alias is enough because going by the order of execution in sql, Always, order by is executed after executing select clause."See full answer

Steve M. - "def split_count(s): return 2**(len(s)-1) `"See full answer

Anonymous Condor - "Yes, I need to compare the first half of the first string with the reverse order of the second half of the second string. Repeat this process to the first half of the second string and the second half of the first string."See full answer

Jayveer S. - "with jay as (select teamid, playername,maxscore,denserank()over(partition by teamid order by maxscore desc ) as ranks from (select p.teamid,p.playername,max(s.gamescore) as maxscore from players as p join scores as s on p.playerid=s.playerid group by 1,2) as inner_query) select teamid,playername,max_score from jay where ranks<=2"See full answer

Alina G. - "SELECT upsellcampaignid, COUNT(DISTINCT trans.userid) AS eligibleusers FROM campaign JOIN "transaction" AS trans ON transactiondate BETWEEN datestart AND date_end JOIN user ON trans.userid = user.userid WHERE iseligibleforupsellcampaign = 1 GROUP BY upsellcampaignid `"See full answer

Mohit C. - "-- LTV = Sum of all purchases made by that user -- order the results by desc on LTV select u.user_id, sum(a.purchase_value) as LTV from user_sessions u join attribution a on u.sessionid = a.sessionid group by u.user_id order by sum(a.purchase_value) desc"See full answer

Reno S. - "import Foundation func spiralCopy(inputMatrix: [[Int]]) -> [Int] { let arr = inputMatrix var top = 0, down = arr.count - 1 var left = 0, right = arr[0].count - 1 if top == down && left == right { return arr[top] } var ans: [Int] = [] while top <= down && left <= right { for i in left..<right { ans.append(arrtop) } for i in top..<down { ans.append(arri) } fo"See full answer

Anonymous Roadrunner - "-- Write your query here select avg(julianday(dateend) - julianday(datestart)) as average_duration from campaign; `"See full answer

Gaurav K. - "I think sliding window will work here and it is the most optimized approach to solve this question."See full answer