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

Victor N. - "with top_players as( Select team_id, player_name, max(gamescore) as maxscore ,denserank() over(partition by teamid order by max(game_score) desc) rk from players p join scores s on p.playerid=s.playerid group by player_name, team_id ) Select team_id, player_name, max_score from top_players where rk<=2 order by teamid, maxscore desc `"See full answer

Nitin P. - "public static void sortBinaryArray(int[] array) { int len = array.length; int[] res = new int[len]; int r=len-1; for (int value : array) { if(value==1){ res[r]= 1; r--; } } System.out.println(Arrays.toString(res)); } `"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

Shakshi R. - "def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]: if not preorder or not inorder: return None root = TreeNode(preorder[0]) mid = inorder.index(preorder[0]) root.left = self.buildTree(preorder[1:mid+1], inorder[:mid]) root.right = self.buildTree(preorder[mid+1:], inorder[mid+1:]) return root"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

Alexey T. - "-- The text of the task is a bit confusing. If the status is repeated several -- times, then in the end you should show as start_date the date of the first -- occurrence, and in end_date the date of the last occurrence of this status, -- and not the date of the beginning of the next status with t1 as (select order_id, status, orderdate as startdate, lead(orderdate) over (partition by orderid order by orderdate) as enddate, ifnull(lag(status) over (partition by order_id order by or"See full answer

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"See full answer

Jacky T. - "SELECT order_amount FROM ( SELECT *, rank() OVER(ORDER BY order_amount desc) as ranking FROM departments d LEFT JOIN orders o ON d.departmentid = o.departmentid LEFT JOIN customers c ON o.customerid = c.customerid WHERE department_name = 'Fashion' ) where ranking = 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

Shashwat K. - "Binary Search on the array and after than compare the numbers at low and the high pointers whichever is closest is the answer. Because after the binary search low will be pointing to a number which is immediate greater than x and high will be pointing to a number which is immediate lesser than x. int low = 0; int high = n-1; while(low <= high){ int mid = (low + high) / 2; if(x == arr[mid]) return mid; //if x is already present then it will be the closest else if(x < arr[mid]) high"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

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

Rohit B. - "class Solution: def missingNumber(self, nums: list[int]) -> int: Sorting approach n = len(nums) s = n*(n+1)//2 r = s - sum(nums) return self.r l = [3,0,1] print(missingNumber(l))"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

B. T. - "Example: bucket A: 3 liters capacity bucket B: 5 liters capacity goal: 4 liters You are asked to print the logical sequence to get to the 4 liters of water in one bucket. Follow up: How would you solve the problem if you have more than 2 buckets of water?"See full answer

Aneesha K. - "-- filter for december and november data -- the total order amount per depatment per month -- department, month, order_amount with monthly_orders AS ( SELECT department_id, strftime('%m', order_date) AS month, SUM(orderamount) AS orderamount FROM orders WHERE orderdate >= '2022-11-01' AND orderdate < '2023-01-01' group by department_id, month ), -- -- add difference from this month to last ( use lag ) monthly_comp"See full answer

Anonymous Wasp - "I was able to provide the optimal approach and coded it up"See full answer

Jessica C. - "select customer_id, order_date, orderid as earliestorder_id from ( select customer_id, order_date, order_id, rownumber() over (partition by customerid, orderdate order by orderdate) as orderrankper_customer from orders ) sub_table where orderrankper_customer=1 order by orderdate, customerid; Standard solution assumed that the orderid indicates which order comes in first. However this is not always the case, and sometime orderid can be random number withou"See full answer