Coding Interview Questions

Michael B. - "Write a function which Caesar ciphers all the strings so that the first character is "a". Use ascii code points and the modulo operator to do this. Use this function to create a hashmap between each string and the CC-a string. Then go through each key:value pair in the hashmap, and use the CC-a ciphered value as the key in a new defaultdict(list), adding the original string to the value field in the output."See full answer

GalacticInterviewer - "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"See full answer

Mahima M. - "class Solution: def lengthOfLIS(self, nums: List[int]) -> int: temp = [nums[0]] for num in nums: if temp[-1]< num: temp.append(num) else: index = bisect_left(temp,num) temp[index] = num return len(temp) "See full answer

Anonymous Unicorn - "input_logs = [ f"{senderid} {receiverid} {transaction_count}" "1 2 2", "3 2 42", "2 2 22", "1 1 12", "2 1 1", "2 5 4", "4 2 15" ] input_threshold = 20 exptected_output = [ list of user_ids that made more than 20 transactions sorted by number of transactions in descending order "3", # 42 transactions "2", # 27 transactions (22 + 1 + 4) #"4", # 15 transactions #"1" # 14 transactions (2 + 12 + 1) ] def gettopapi_users(logs, thres"See full answer

Rafał P. - "def changeString(org: str,target:str) -> bool: lOrg = len(org) lTarget = len(target) \# They have to be equal in lenght if lOrg != lTarget: return False counter1 = Counter(org) counter2 = Counter(target) \# Counter internally iterates through the input sequence, counts the number of times a given object occurs, and stores objects as keys and the counts as values. if counter1 != counter2: return False diff = sum(org[i] != target[i] for i in range(n)) return diff == 2 or (diff == 0 and any(v > 1 f"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

Anzhe M. - "It's a 2Sum question with duplicate array elements."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

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

Azeezat R. - "/* You are with your friends in a castle, where there are multiple rooms named after flowers. Some of the rooms contain treasures - we call them the treasure rooms. Each room contains a single instruction that tells you which room to go to next. * instructions1 and treasurerooms_1 * lily* --------- daisy sunflower | | | v v v jasmin --> tulip* violet* ----> rose* --> ^ | ^ ^ | "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

XponentShift32 - "solving to find a cycle in directed graph"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

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

Mridul J. - "We have a list of documents. We want to build an index that maps keywords to documents containing them. Then, given a query keyword, we can efficiently retrieve all matching documents. docs = [ "Python is great for data science", "C++ is a powerful language", "Python supports OOP and functional programming", "Weather today is sunny", "Weather forecast shows rain" ]"See full answer

Jaime A. - "I might be missing something but the solution, seems to be incorrect. ... , post_pairings AS ( SELECT ps.user_id, ps.postseqid AS failpostid, ps.postseqid + 1 AS nextpostid FROM post_seq AS ps WHERE ps.issuccessfulpost IS TRUE ) -- here ps.issuccessfulpost IS TRUE the condition should be FALSE -- in that way ps.postseqid is the actual failed post(failpostid) -- Additionally, at the end the join is assumming that the sequence id is going to match the post_id, wh"See full answer

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