Coding Interview Questions

Aryan D. - "int main() { int a1[7]={1,2,3,4,5,6,7}; int a2[7]={1,9,10,11,12,13,14}; vectorv; v.insert(v.begin(),begin(a1),end(a1)); v.insert(v.begin(),begin(a2),end(a2)); int a3[v.size()]; sort(v.begin(),v.end()); for(int i=0;i<v.size();i++) { a3[i]=v[i]; } } `"See full answer

Anonymous Tortoise - "SELECT i.item_category, o.order_date, SUM(o.orderquantity) AS totalunits_ordered FROM orders o JOIN items i ON o.itemid = i.itemid WHERE o.order_date >= DATE('now', '-6 days') GROUP BY i.item_category, o.order_date ORDER BY i.item_category ASC, o.order_date ASC;"See full answer

Nasit S. - "Depend on the array size and number of 0's theere."See full answer

Anonymous Armadillo - "I couldn't follow the solution offered here, but my solution seemed to pass 6/6 tests. Any feedback is welcome, thank you! def encrypt(word): en_word = "" for i in range(len(word)): if i == 0: en_word += chr(ord(word[0])+1) else: num = ord(word[i]) + ord(en_word[i-1]) while num > 122: num -= 26 en_word += chr(num) return en_word def decrypt(word): de_word = "" for i in range(len(word)): if i == 0: de_word += chr(ord(word[i]"See full answer

Anonymous Owl - "def getcheapestcost(rootNode): \# need to do DFS for each branch \# but this can be done recursively n = len(rootNode.children) if n == 0: return 0 else: min_cost = float('inf') for i in range(len(n)): tempcost = getcheapest_cost(rootNode.children[i]) if (tempcost < mincost): mincost = tempcost return min_cost + rootNode.cost \# A node class Node: \# Constructor to create a new node def init\(self, cost): self.cost = cost self.children = [] self.parent = None"See full answer

Aneesha K. - "Test case is wrong. It expects to sort in asc order of month_year. -- Write your query here SELECT strftime('%Y-%m', createdat) AS monthyear, COUNT(DISTINCT userid) AS numcustomers, COUNT(t.id) AS num_orders, SUM(price * quantity) AS order_amt FROM transactions t INNER JOIN products p ON t.product_id = p.id GROUP BY month_year ORDER BY month_year ; "See full answer

Gabriel P. - "Hi, my solution gives the exact numerical values as the proposed solution, but it doesn't pass the tests. Am I missing something, or is this a bug? def findrevenueby_city(transactions: pd.DataFrame, users: pd.DataFrame, exchange_rate: pd.DataFrame) -> pd.DataFrame: gets user city for each user id userids = users[['id', 'usercity']] and merge on transactions transactions = transactions.merge(user_ids, how='left"See full answer

Pk - "Without using the window function: SELECT w1.viewer_id FROM watch_time w1 JOIN watch_time w2 ON abs(w1.month - w2.month) = 1 AND w1.viewerid = w2.viewerid AND w1.year = w2.year GROUP BY w1.viewer_id HAVING sum(CASE WHEN w1.month > w2.month AND w1.watchhours > w2.watchhours THEN 1 ELSE 0 END) > 2 `"See full answer

Arshad P. - "Schema is wrong - id from product is mapped to id from transactions, id from product should point to product_id in transcations table"See full answer

Ashley M. - "sliding window"See full answer

Frederick A. - "This problem can be solved with two approaches Iterative approach Recursive approach Quite easy to think about the iterative approach, you can make use of a while loop in that case. But what if you want to make use of previously computed values? That case going for the recursive solution is quite useful. class Collatz: def init(self) -> None: self.cache = {} self.steps = 0 def steps_from(self, n) -> int: # base case if n == 1: "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

Prashanth A. - "We are asked to calculate Sum(over value) for time in (t - window_size, t) where key in (key criteria). To develop a function to set this up. Let w be the window size. I would have an observer of some kind note the key-value, and for the first w windows just add the value to a temporary variable in memory if the key meets the key criteria. Then it would delete the oldest value and add the new value if the new key meets the criteria. At each step after "w", we would take the sum / w and store"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

Jayveer S. - "WITH suspicious_transactions AS ( SELECT c.first_name, c.last_name, t.receipt_number, COUNT(t.receiptnumber) OVER (PARTITION BY c.customerid) AS noofoffences FROM customers c JOIN transactions t ON c.customerid = t.customerid WHERE t.receipt_number LIKE '%999%' OR t.receipt_number LIKE '%1234%' OR t.receipt_number LIKE '%XYZ%' ) SELECT first_name, last_name, receipt_number, noofoffences FROM suspicious_transactions WHERE noofoffences >= 2;"See full answer