Coding Interview Questions

E L. - "WITH CTE AS ( SELECT *, ROWNUMBER()OVER(PARTITION BY utxoid ORDER BY transactionid) AS trxrk FROM transactions JOIN transaction_inputs USING (transaction_id) JOIN utxo USING (utxo_id) ) SELECT transaction_id AS InvalidTransactionId FROM CTE WHERE sender!=address OR trx_rk > 1 `"See full answer

Alvin P. - "SELECT e1.empid AS manageremployee_id, e1.empname AS managername, COUNT(e2.empid) AS numberofdirectreports FROM employees AS e1 INNER JOIN employees AS e2 ON e2.managerid = e1.empid GROUP BY e1.emp_id HAVING COUNT(e2.emp_id) >= 2 ORDER BY numberofdirectreports DESC, managername ASC `"See full answer

Anonymous Roadrunner - "-- Write your query here select count(distinct o.customer_id) as customers, d.department_name from orders o join departments d using (department_id) where extract(year from o.order_date) = 2022 and d.department_name in ('Electronics', 'Fashion') group by 2; `"See full answer

Gabriella F. - "Table user is empy....... Problem with this problem "See full answer

Matthew K. - " function diffBetweenTwoStrings(source, target) { /** @param source: string @param target: string @return: string[] */ let dp = new Array(source.length+1).fill().map(() => Array(target.length+1).fill(0)) for (let i = source.length; i>= 0; i--) { for (let j = target.length; j>= 0; j--) { if (i === source.length) { dpi = target.length - j } else if (j === target.length) { dpi = sou"See full answer

Caleb S. - " import pandas as pd def findimprovingstudents(transcript: pd.DataFrame) -> pd.DataFrame: summary = transcript.pivottable(index='studentid', values = 'yearlygpa', aggfunc = 'sum',columns = 'year').resetindex() summary['average_gpa'] = round((summary[2023] + summary[2022] + summary[2021])/3,2) return summary(summary[2023] > summary[2022]) & (summary[2022] > summary[2021])] #yn > yn-1, yn-1 > yn-2, yn-3 debug your co"See full answer

Chao peter Y. - "The question is incomplete --- the code only passes if you return the data frame sorted by BOTH department name AND rank. While in the problem description, it mentions to only rank by department name: "The results should be ordered by department name." Not a big difference I know, but students shouldn't need to look into the solution to get the necessary knowledge to answer the question."See full answer

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"See full answer

Tiago R. - "function isPalindrome(s, start, end) { while (s[start] === s[end] && end >= start) { start++; end--; } return end <= start; } function longestPalindromicSubstring(s) { let longestPalindrome = ''; for (let i=0; i < s.length; i++) { let j = s.length-1; while (s[i] !== s[j] && i <= j) { j--; } if (s[i] === s[j]) { if (isPalindrome(s, i, j)) { const validPalindrome = s.substring(i, j+1"See full answer

Anmol R. - "The example given is wrong. The 2nd test case should have answer 3, as we can get to 6 by using 3 coins of denomination 2."See full answer

Divya R. - "public Double calculateRatio(String source, String destination) { Double ratio=1.0; while(graph.containsKey(source) && !visited.contains(source)) { visited.add(source); Map valueMap=graph.get(source); if(valueMap.containsKey(destination)) { return ratio*=valueMap.get(destination); } Map.Entry firstEntry=valueMap.entrySet().iterator().next(); source=firstEntry.getKey(); ratio*=firstEntry.getValue(); System.out.println("Entered"); } return null; }"See full answer

Sai R. - "def countuniqueoutfits(totalpants: int, uniquepants: int, totalshirts: int, uniqueshirts: int, totalhats: int, uniquehats: int) -> int: """ Number of unique outfits can simply be defined by (uniquepantschoose1uniqueshirtschoose1uniquehatschoose_1) (uniquepantschoose1*uniqueshirtschoose1) # Not wearing a hat nchoosek is n """ res = (uniquepants*uniqueshirtsuniquehats) + (uniquepantsunique_shirts) return res print(countuniqueoutfits(2, 1, 1, 1, 3, 2))"See full answer

Sean L. - " debug your code below departments = pd.DataFrame({ 'id': [1, 2, 3, 4, 5], 'name': ['Reporting', 'Engineering', 'Marketing', 'Biz Dev', 'Silly Walks'] }) employees = pd.DataFrame({ 'id': [1, 2, 3, 4, 5, 6], 'first_name': ['John', 'Ava', 'Cailin', 'Mike', 'Ian', 'John'], 'last_name': ['Smith', 'Muffinson', 'Ninson', 'Peterson', 'Peterson', 'Mills'], 'salary': [20000, 10000, 30000, 20000, 80000, 50000], 'department_id': [1, 5, 2, 2, 2, 3] }) projects = p"See full answer

Sean L. - " import pandas as pd from datetime import datetime def findfastestlike(log: pd.DataFrame) -> pd.DataFrame: log=log.sortvalues(['userid','timestamp']) #get the prev event, time by user log['prevevent'] = log.groupby('userid')['event'].shift(1) log['prevtimestamp'] = log.groupby('userid')['timestamp'].shift(1) True only on rows where the previous event was a login and the current event is a like log['loginlike'] = (log['prevevent'] == 'log"See full answer