Coding Interview Questions

Sarvesh G. - "def calc(expr): ans = eval(expr) return ans your code goes debug your code below print(calc("1 + 1")) `"See full answer

Tiago R. - "function addChildren(root, val, inorder) { const rootInOrderIndex = inorder.indexOf(root.value); const childrenInOrderIndex = inorder.indexOf(val); if (childrenInOrderIndex < rootInOrderIndex) { if (!root.left) { root.left = new TreeNode(val); } else { addChildren(root.left, val, inorder); } } else { if (!root.right) { root.right = new TreeNode(val); } else { addChildren(root.right,"See full answer

Marcos G. - "with my_table as (select * , rownumber() over(order by customerid) as row_index from customers) select customer_id, customer_name from my_table where row_index % 3 = 0"See full answer

Yash N. - "import time class Task: def init\(self, description, interval=None): self.description = description self.interval = interval self.next_run = time.time() class SimpleTaskScheduler: def init\(self): self.tasks = [] def add_task(self, description, interval=None): self.tasks.append(Task(description, interval)) def run(self, duration=60): end_time = time.time() + duration while time.time() < end_time: curr"See full answer

Anonymous Owl - "def validateIP(ip): """ @param ip: str @return: bool """ \# ip needs to be in X.X.X.X \# X is from 0 to 255 \# split the ip at "." split = ip.split('.') if (len(split) != 4): return False for number in split: if (int(number) 255): return False return True"See full answer

Jaime A. - "I might be missing something but the solution, seems to be incorrect. ... , post_pairings AS ( SELECT ps.user_id, ps.postseqid AS failpostid, ps.postseqid + 1 AS nextpostid FROM post_seq AS ps WHERE ps.issuccessfulpost IS TRUE ) -- here ps.issuccessfulpost IS TRUE the condition should be FALSE -- in that way ps.postseqid is the actual failed post(failpostid) -- Additionally, at the end the join is assumming that the sequence id is going to match the post_id, wh"See full answer

Richard B. - "Select interface, Count(case when issuccessfulpost then 1 end) as post_success, Count() as postattempt, ROUND((COUNT(CASE WHEN issuccessfulpost THEN 1 END) * 100 / COUNT()), 2) AS postsuccess_rate from post where interface like 'Iphone%' group by 1 order by postsuccessrate desc `"See full answer

Tiago R. - "function swap(arr, i, j) { const temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } function sortKMessedArray(arr, k) { for (let i=0; i < arr.length; i++) { for (let j=1; j <= k; j++) { if (arr[i+j] < arr[i]) { swap(arr, i, i+j); } } } return arr; } `"See full answer

Lucas G. - "select sub.name subreddit_name, count(distinct us.userid) totalusers from user_subreddit as us left join subreddit as sub on us.subredditid = sub.subredditid group by us.subreddit_id having count(distinct us.user_id) > 3"See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Tiago R. - "function areSentencesSimilar(sentence1, sentence2, similarPairs) { if (sentence1.length !== sentence2.length) return false; for (let i=0; i (w1 === word1 && !visited.has(w2)) || (w2 === word1 && !visited.has(w1))); if (!edge) { "See full answer

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"See full answer

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

Rishi G. - "Problem Statement: The Fibonacci sequence is defined as F(n) = F(n-1) + F(n-2) with F(0) = 1 and F(1) = 1. The solution is given in the problem statement itself. If the value of n = 0, return 1. If the value of n = 1, return 1. Otherwise, return the sum of data at (n - 1) and (n - 2). Explanation: The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, typically starting with 0 and 1. Java Solution: public static int fib(int n"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

G B. - "-- Write your query here Select teamid,playerid,top_Score from( with high_score as( select player_id, max(gamescore) topscore from scores group by player_id ) select team_id, a.player_id, top_score, rank() over(partition by teamid order by topscore desc) rnk from players a left join highscore b on b.playerid = a.player_id) where rnk <=2"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

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