Coding Interview Questions

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

Manoj R. - "def traprainwater(height: List[int]) -> int: n = len(height) totalwaterlevel = 0 for i in range(n): j = i+1 while j = n: break rows = j - i -1 intrwaterlevel = min(height[j], height[i]) * rows for k in range(i+1, j): intrwaterlevel -= height[k] totalwaterlevel += intrwaterlevel i = j return totalwaterlevel"See full answer

Gowtami K. - "select a.playername as player1, b.playername as player2, abs(a.level-b.level) as level_disparity from players a cross join players b on a.playername < b.playername where abs(a.level-b.level) <=5 order by level_disparity `"See full answer

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

Vince S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["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

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

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

Gabriele G. - "Less efficient version, yet effective for the interview: def is_palindrome(s: str) -> bool: dim = len(s) if dim str: dim = len(s) if dim < 2: return s left = 0 longest = "" while left < dim: righ"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

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

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

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

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