Booking.com Interview Question for Software Engineer / Developers


Country: Netherlands
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
4
of 6 vote

The question is similar to minimum number of plateforms needed for the bus/rail station.

My solution for this very specific question is -

A[] = {1,2,10,5,5} // checkin array
D[] = {4,5,12,9,12} // checkout array

sort(A) // {1,2,5,5,10}
sort(D) // {4,5,9,12,12}

count = 1  // first person checked in
int i = 1; // to traverse A
int j = 0; // to traverse D
int max= 0;
int day = A[i];

while(i < n && j < n) {

	if(A[i] <= D[j]) {
		count++;
		i++;
		if(count > max) {
			max = count ;
			day = A[i];
		}
	} else {
		count--;
		j++;
	}

}

return day;

The time complexity will be O(nlogn).

Explanation is below -

C - checkin
D - checkout
count - number of people at same time in hotel

Day ---- C/D ------ count
1 C 1
2 C 2 // Increase count on checkin
4 D 1 // Decrease count on checkout
5 C 2
5 C 3 // maximum number of people in hotel -- return the day
9 D 2
12 D 1
12 D 0

- azambhrgn February 01, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

i++ should be moved after you assign day

- Anju July 04, 2019 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

public static int getFullestDay(List<Schedule> schedules) {
	if (schedules == null || schedules.size() == 0) {
		return -1;
	}

	final HashMap<Integer, Integer> dayAndCapacity = new HashMap<>();

	for (Pair<Integer, Integer> schedule : schedules) {
		for (int day = schedule.check-in; day <= schedule.check-out; day ++) {
			dayAndCapacity.put(day, dayAndCapacity.get(day) + 1);
		}
	}

	int fullestDay = 0;
	int highestCapacity = 0;
	for (Entry<Integer, Integer> entry : dayAndCapacity.entrySet()) {
		if (entry.getValue() > highestCapacity) {
			highestCapacity = entry.getValue();
			fullestDay = entry.getKey();
		}
	}

	return highestDay;
}

public class Schedule {
	public int check-in;
	public int check-out;
}

- feistc February 05, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

Incrementing each day in a range (initialize it with 0 when first occurs) and then find the max by iterating:

import java.util.*;

public class MaxNumOfTravelers {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        Map<Integer, Integer> numberOfTravelers = new HashMap();
        for(int n = 0; n < N; n++){
            int start = scanner.nextInt();
            int finish = scanner.nextInt();
            for(int i = start; i <= finish; i++){
                numberOfTravelers.putIfAbsent(i,0);
                numberOfTravelers.put(i, numberOfTravelers.get(i)+1);
            }
        }

        int maxTravelers = Integer.MIN_VALUE;
        int maxDay = -1;
        for(Map.Entry<Integer, Integer> entry : numberOfTravelers.entrySet()){
            if(entry.getValue() > maxTravelers){
                maxTravelers = entry.getValue();
                maxDay = entry.getKey();
            }
        }

        System.out.println("Max day - " + maxDay);
    }
}

- Anatolii.Stepaniuk June 28, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

input = [(1,4),(2,5),(10,12),(5,9),(5,12)]
days = {}
for i in input:
	for k in range(i[0],i[1]+1):
		if k in days:
			days[k] += 1
		else:
			days[k] = 1

print(sorted(days.items(),key=lambda x: x[1],reverse=True)[0][0])

- saurabh July 17, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

The question is not very clear because it depends upon the initial no. of people in hotel.

- Harsh Bhardwaj February 05, 2017 | Flag Reply
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0
of 0 vote

The problem stated does not look like a real one. In real world if one person leaves on day 5 and to persons arrive on day 5, there are only 2 guests in the hotel on that day. Because visitors usually leave before new visitors arrive.
According to this amendment my solution would be

import heapq


def high_day(guests):
    schedule = []
    for arrive, leave in guests:
        heapq.heappush(schedule, (arrive, 1))
        heapq.heappush(schedule, (leave, -1))

    current_guests = max_day = max_day_guests = 0
    while (schedule):
        day, inout = heapq.heappop(schedule)
        current_guests += inout
        if current_guests > max_day_guests:
            max_day = day
            max_day_guests = current_guests
    print max_day


if __name__ == "__main__":
    guests = [
        (1, 4),
        (2, 5),
        (10, 12),
        (5, 9),
        (5, 12),
    ]
    high_day(guests)

Complexity is O(n * log n). This is the time it takes to build a heap from list of guests and also the time it takes to iterate over heap. And O(2 * n * log n) is equivalent to O(n * log n)

- s.a.kovalev February 24, 2017 | Flag Reply
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0
of 0 vote

List<int> periods = new List<int>();
string endChar = "";
while (endChar != "===")
{
string s = Console.ReadLine();
if (s == "===")
{
endChar = "===";
break;
}
int[] r = Array.ConvertAll(s.Split(' '), int.Parse);
for (int i = r[0]; i <= r[1]; i ++)
{
periods.Add(i);
}
}
var most = (from i in periods group i by i into grp orderby grp.Count() descending select grp.Key).First();
Console.WriteLine(most);
Console.ReadLine();

- Omar Mer April 30, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{{ int[] A = { 1, 2, 10, 5, 5 }; // checkin array
int[] D = { 4, 5, 12, 9, 12 }; // checkout array

A = A.OrderBy(a => a).ToArray<int>();// {1,2,5,5,10}

D = D.OrderBy(d => d).ToArray<int>(); // {4,5,9,12,12}

int[] result = new int[Math.Max(A[A.Length - 1], D[D.Length - 1])];

for (int i = 0; i < A.Length; i++)
{
int min = A[i];
int max = D[i];

for (int j = min - 1; j < max; j++)
{
result[j] += 1;
}
}

int maxi = result.Max();
int day = result.ToList().IndexOf(maxi) + 1;

Console.WriteLine(day);
Console.ReadKey();
}}

- bshyam September 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

2 Solutions

1) Use a HashMap. Map the Day with Count. Finally, Days with max-count is the answer.

2) Append all days in Array-list. Sort it. Days with max repetition is the answer.

- Aravind Prasad October 29, 2018 | Flag Reply


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