Coding Interview Questions

JOBHUNTER - "def findAlibaba(countOfRooms, strategy): #countofrooms: num rooms #listRooms rooms to look for alibabba possiblePlaces = [] #initialize rooms for i in range(countOfRooms): possiblePlaces.append(True) for i in range(len(strategy)): roomToCheck = strategy[i] #Room is marked as unavailable possiblePlaces[roomToCheck] = False #Next day calculatins nextDayPlaces = [] for j in range(countOfRooms): nextDayPla"See full answer

Derrick M. - "SELECT COUNT(DISTINCT o.customerid) AS customers, d.departmentname FROM orders o INNER JOIN departments d ON d.departmentid = o.departmentid WHERE d.departmentname IN ('Electronics','Fashion') AND o.orderdate BETWEEN '2022-01-01' AND '2022-12-31' GROUP BY d.department_name; `"See full answer

Bryan L. - "with var1 as (select *, rank() over(order by score desc) as srank from players) select player_name, score, srank as ranking from var1 where srank in (4, 6, 11) `"See full answer

Divya K. - "You will need to start from Browser and go all the way up to Analytic systems and methods. Everything needs to be covered"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

Nils G. - "Bitshift the number to the right and keep track of the 1's you encounter. If you bitshift it completely and only encounter one 1, it is a power of two."See full answer

Kwaku K. - "Use a BFS"See full answer

Brajesh J. - "dp"See full answer

Alireza K. - "import random def coin_flip(): x=4*[0]+[1] res=[] for i in range(20): res.append(random.choice(x)) return res res=[0,0] # [head,tail] for j in range(1000): temp=coin_flip() res[0]+=sum(temp) #head res[1]+=(20-sum(temp)) #tail"See full answer

B. T. - "Example: bucket A: 3 liters capacity bucket B: 5 liters capacity goal: 4 liters You are asked to print the logical sequence to get to the 4 liters of water in one bucket. Follow up: How would you solve the problem if you have more than 2 buckets of water?"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

Anonymous Wolf - "Construct a min-heap either inplace, or by making a copy of the array and then applying heapify on that copy. This is done in O(n) time. Maintain two zero-initialised variables - sum and count. Keep popping off the heap while sum < k, and update count. In the worst case you will have to do n pops, and each pop is O(log n), so the algorithm would take O(n log n) in total. Space complexity depends on whether you're allowed to modify inplace or not, so either O(1) or O(n) respectively."See full answer

Anonymous Porcupine - "Coded the solution using this approach that is frequency table and counting. it is Leetcode 791"See full answer