Coding Interview Questions

Zacharias E. - "SELECT pro.id, pro.title, pro.budget, COUNT(employeeid) AS numemployees, SUM(e.salary) as total_salaries FROM projects pro JOIN employeesprojects ep ON ep.projectid = pro.id JOIN employees e ON e.id = ep.employee_id GROUP BY project_id; `"See full answer

Dinar M. - "Even more faster and vectorized version, using np.linalg.norm - to avoid loop and np.argpartition to select lowest k. We dont need to sort whole array - we need to be sure that first k elements are lower than the rest. import numpy as np def knn(Xtrain, ytrain, X_new, k): distances = np.linalg.norm(Xtrain - Xnew, axis=1) k_indices = np.argpartition(distances, k)[:k] # O(N) selection instead of O(N log N) sort return int(np.sum(ytrain[kindices]) > k / 2.0) `"See full answer

Shatabdi P. - "def lowestearningemployees(employees: pd.DataFrame) -> pd.DataFrame: selectedcolumns = employees[['id','firstname','last_name','salary' ]] sorteddf = selectedcolumns.sort_values(by='salary', ascending=True) return sorted_df.head(3)"See full answer

Akshay D. - "SELECT name, type, CASE WHEN type = 'Electronic' THEN price-(0.10*price) WHEN type = 'Clothing' THEN price-(0.20*price) WHEN type = 'Grocery' THEN price-(0.05*price) WHEN type = 'Book' THEN price-(0.15*price) ELSE price END AS discounted_price FROM products; `"See full answer

Salome L. - "SELECT items.item_category, SUM(orders.orderquantity) AS totalunitsorderedlast7days FROM orders JOIN items ON orders.itemid = items.itemid WHERE orders.order_date BETWEEN DATE('now', '-6 days') AND DATE('now') GROUP BY items.item_category `"See full answer

Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

Komal S. - "a. Sort the array elements. b. take two pointers at index 0 and index Len-1; c. if the sum at the two pointers is target; break and return the pair d. if the sum is smaller, then move left pointer by 1 e. else move right pointer by 1; run the logic till the target is met or right pointer crosses the left pointer."See full answer

Akash C. - " from typing import List def find_first(array: List[int], num: int) -> int: left = 0 right = len(array) - 1 result = -1 # keep track of leftmost occurence found so far while left <= right: mid = (left + right) // 2 if array[mid] == num: result = mid #Found a potential result right = mid - 1 elif array[mid] < num: left = mid + 1 else: right = mid - 1 return result debug your code"See full answer

Anonymous Possum - "The unique id is not clear in this question"See full answer

Anonymous Wasp - "I was able to answer this question and the follow-up questions as well"See full answer

Tiago R. - "function findPrimes(n) { if (n < 2) return []; const primes = []; for (let i=2; i <= n; i++) { const half = Math.floor(i/2); let isPrime = true; for (let prime of primes) { if (i % prime === 0) { isPrime = false; break; } } if (isPrime) { primes.push(i); } } return primes; } `"See full answer

Rick E. - " from typing import Dict, List, Optional def max_profit(prices: Dict[str, int]) -> Optional[List[str]]: pass # your code goes here max = [None, 0] min = [None, float("inf")] for city, price in prices.items(): if price > max[1]: max[0], max[1] = city, price if price 0: return [min[0], max[0]] return None debug your code below prices = {'"See full answer

Bhavna S. - " with youngsuccrate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as yascrate from post where userid in (select userid from post_user where age between 0 and 18) group by post_month ), nonyoungsucc_rate as( select strftime('%m', postdate) AS postmonth, round(sum(issuccessfulpost)*1.0/count(issuccessfulpost),2)as nonyasc_rate from post where user_id in (select"See full answer

G B. - " select user_id, b.marketing_channel from user_sessions a Left join attribution b on b.sessionid = a.sessionid group by 1,2 HAVING sum(purchasevalue)>100 and min(adclick_timestamp) `"See full answer

Bradley E. - "I'm pretty sure Exponent's answer is wrong. In the snippet below, they use "pl.name = 'Telephones' to attempt to filter down to the Telephone transactions, but they do this within a LEFT JOIN which means all product_lines rows are returned. > LEFT JOIN product_lines pl > ON p.productlineid = pl.id > AND pl.name = 'Telephones' Below is my solution. Also, I didn't see anywhere that said the "amount" column was in cents instead of dollars, but I still divided by 100 to be consistent with Exp"See full answer

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

Sean L. - " import pandas as pd def findaveragedistance(gps_data: pd.DataFrame) -> pd.DataFrame: #0. IMPORTANT: get the unordered pairs gpsdata['city1']=gpsdata[['origin','destination']].min(axis=1) gpsdata['city2']=gpsdata[['origin','destination']].max(axis=1) #1. get the mean distance by cities avgdistance=gpsdata.groupby(['city1','city2'], as_index=False)['distance'].mean().round(2) avgdistance.rename(columns={'distance':"averagedistance"}, inplace=True) "See full answer

Alina G. - "Order the result in descending month is not applied in the solution"See full answer

Tiago R. - "function generateParentheses(n) { if (n < 1) { return []; } if (n === 1) { return ["()"]; } const combinations = new Set(); let previousCombinations = generateParentheses(n-1); for (let prev of previousCombinations) { for (let i=0; i < prev.length; i++) { combinations.add(prev.slice(0, i+1) + "()" + prev.slice(i+1)); } } return [...combinations]; } `"See full answer