I would like to ask about some algorithms related to checking if a customer can book a table at the store?
I will describe my problem with the following example:
Restaurant:
User M has a restaurant R. R is open from 08:00 to 17:00.
Restaurant R has 3 tables (T1, T2, T3), each table will have 6 seats.
R offers F1 food, which can be eaten within 2 hours.
Booking:
R has a customer C has booked a table T1 for 5 people with F1 food | B[0]
B[0] has a start time: 9AM
M is the manager of the store, so M wants to know if the YYYY-MM-DD date has been ordered by the customer or not?
My current algorithm is:
I will create an array with the elements as the number of minutes of the day, and their default value is 0
24 * 60 = 1440
=> I have: arr[1440] = [0, 0, 0, ...]
Next I will get all the bookings for the day YYYY-MM-DD. The result will be an array B[].
Then I will loop the array B[]
for b in B[]
I then keep looping for the start_time, to the end_time of b with step of 1 min.
for time = start_time, time <= end_time. time++
With each iteration I will reassign the value of the array arr with index as the corresponding number of minutes in the day to 1
(It is quite similar to Sieve of Eratosthenes)
Then what I need to do is iterate over the array arr 1 more time, if there is at least 1 value 0 in the array it means YYYY-MM-DDdate is still bookable.
But my algorithm will not be optimal if increase the number of tables that the store has, the number of days to check is many days (for example from 2022-01-01 -> 2022-02-01), ...
Thank you very much.
P/S: Regarding the technology background, I am currently using laravel 9
Related
I tried to solve the task which sounds like "Given the schedules of the days investors are available, determine how many meetings the owner can schedule". The owner is looking to meet new investors to get some funds for his company. The owner must respect the investor's schedule. Note that the owner can only have one meeting per day.
The schedule consists of 2 integer arrays, firstDay and lastDay. Each element in the array firstDay represents the first day an investor is available, and each element in lastDay represents the last day an investor is available, both inclusive.
Example:
firstDay = [1,2,3,3,3]
lastDay = [2,2,3,4,4]
There are 5 investors [i0, i1, i2, i3, i4]
The investor i0 is available from day 1 to day 2 inclusive [1,2]
The investor i1 is available in day 2 only [2,2]
The investor i2 is available in day3 only [3,3]
The investors i3 and i4 are available from day 3 to day 4 only [3,4]
The owner can only meet 4 investors out of 5: i0 in day 1, i1 in day 2, i2 in day 3 and i3 in day 4. The image below shows the scheduled meetings in green and blocked days are in gray.
A graphic shows the scheduled meetings
The task is to implement the function which takes 2 lists of integers as input parameters and returns integer result that represents the maximum number of meetings possible.
Constraints
array length - bigger or equal 1 and less or equal 100000
firstDay[i], lastDay[i] bigger or equal 1 and less or equal 100000 (i bigger than or equal 0 less than n)
firstDay[i] less or equal lastDay[i]
My implementation of this task is the following:
public static int countMeetings(List<int> firstDay, List<int> lastDay)
{
var count = 0;
count = firstDay.Concat(lastDay).Distinct().Count();
if (count > firstDay.Count)
{
count = firstDay.Count;
}
return count;
}
And this code successfully passes 8 of 12 provided tests. I'll be glad to see and discuss any working solutions to this issue. Thanks.
For the input
firstDay = [1,1,1]
lastDay = [5,5,5]
your code returns 2 however correct answer is 3
I have an array and I have to perform query and updates on it.
For queries, I have to find frequency of a particular number in a range from l to r and for update, I have to add x from some range l to r.
How to perform this?
I thought of sqrt{n} optimization but I don't know how to perform range updates with this time complexity.
Edit - Since some people are asking for an example, here is one
Suppose the array is of size n = 8
and it is
1 3 3 4 5 1 2 3
And there are 3 queries to help everybody explain about what I am trying to say
Here they are
q 1 5 3 - This means that you have to find the frequency of 3 in range 1 to 5 which is 2 as 3 appears on 2nd and 3rd position.
second is update query and it goes like this - u 2 4 6 -> This means that you have to add 6 in the array from range 2 to 4. So the new array will become
1 9 9 10 5 1 2 3
And the last query is again the same as first one which will now return 0 as there is no 3 in the array from position 1 to 5 now.
I believe things must be more clear now. :)
I developed this algorithm long time (20+ years) ago for Arithmetic coder.
Both Update and Retrieve are performed in O(log(N)).
I named this algorithm "Method of Intervals". Let I show you the example.
Imagine, we have 8 intervals, with numbers 0-7:
+--0--+--1--+--2-+--3--+--4--+--5--+--6--+--7--+
Lets we create additional set of intervals, each spawns pair of original ones:
+----01-----+----23----+----45-----+----67-----+
Thereafter, we'll create the extra one layer of intervals, spawn pairs of 2nd:
+---------0123---------+---------4567----------+
And at last, we create single interval, covers all 8:
+------------------01234567--------------------+
As you see, in this structure, to retrieve right border of the interval [5], you needed just add together length of intervals [0123] + [45]. to retrieve left border of the interval [5], you needed sum of length the intervals [0123] + [4] (left border for 5 is right border for 4).
Of course, left border of the interval [0] is always = 0.
When you'll watch this proposed structure carefully, you will see, the odd elements in the each layers aren't needed. I say, you do not needed elements 1, 3, 5, 7, 23, 67, 4567, since these elements aren't used, during Retrieval or Update.
Lets we remove the odd elements and make following remuneration:
+--1--+--x--+--3-+--x--+--5--+--x--+--7--+--x--+
+-----2-----+-----x----+-----6-----+-----x-----+
+-----------4----------+-----------x-----------+
+----------------------8-----------------------+
As you see, with this remuneration, used the numbers [1-8]. Lets they will be array indexes. So, you see, there is used memory O(N).
To retrieve right border of the interval [7], you needed add length of the values with indexes 4,6,7. To update length of the interval [7], you needed add difference to all 3 of these values. As result, both Retrieval and Update are performed for Log(N) time.
Now is needed algorithm, how by the original interval number compute set of indexes in this data structure. For instance - how to convert:
1 -> 1
2 -> 2
3 -> 3,2
...
7 -> 7,6,4
This is easy, if we will see binary representation for these numbers:
1 -> 1
10 -> 10
11 -> 11,10
111 -> 111,110,100
As you see, in the each chain - next value is previous value, where rightmost "1" changed to "0". Using simple bit operation "x & (x - 1)", we can wtite a simple loop to iterate array indexes, related to the interval number:
int interval = 7;
do {
int index = interval;
do_something(index);
} while(interval &= interval - 1);
I have some problems with defining a algorithm that will calculate a ranking number for a dentist.
Assume, we have three different dentists:
dentist number 1: Got 125 patients and out of the 125 patients the
dentist have booked a time with 75 of them. 60% of them got a time.
dentist number 2: Got 5 patients and out of the 5 patients the
dentist have booked a time with 4 of them. 80% of them got a time.
dentist number 3: Got 25 patients and out of the 14 patients the
dentist have booked a time with 14 of them. 56% got a time.
If we use the formula:
patients booked time with / totalpatients * 100
it will not be the right way to calculate the ranking, as we will get an output of the higher percentage is, the better the dentist is, but it's wrong. By doing it in that way, the dentists would have a ranking:
dentist number 2 would have a ranking of 1. (80% got a time).
dentist number 1 would have a ranking of 2 (60% got a time).
dentist number 3 would have a ranking of 3. (56% got a time).
But, it should be in this way:
dentist number 1 = ranking 1
dentist number 2 = ranking 2
dentist number 3 = ranking 3
I don't know to make a algorithm that also takes the amount of patients as a factor to the ranking-calculation.
It is quite arbitrary how you define what makes a better dentist in terms of number of patients and the percentage of those that have an appointment with them.
Let's call the number of patients P, the number of those that have an appointment A, and the function determining how "good" a dentist is f. So f would be a function of P and A: f(P, A).
One component of f could indeed be what you already calculated: A/P.
Another component would have to be P, but I would think that the effect on f(P, A) of increasing P with 1 would be much higher for a low P, than for a high P, so this component should not be a linear function. It would also be practical if this component would have a value between 0 and 1, just like the other component.
Taking all this together, I suggest this definition of f, which will give a number between 0 and 1:
f(P,A) = 1/3 * P/(10 + P) + 2/3 * A/P
For the different dentists, this results in:
1: 1/3 * 125/135 + 2/3 * 75/125 = 0.7086419753...
2: 1/3 * 5/15 + 2/3 * 4/5 = 0.6444444444...
3: 1/3 * 25/35 + 2/3 * 14/25 = 0.6114285714...
You could play a bit with the constant factors in the formula, like increasing the term 10. Or you could change the factors 1/3 and 2/3 making sure that their sum is 1.
This is just one way to do it. There are an infinity of other ways...
We have one distribution center ( ware house ) and we are getting orders in real time whose time/distance from ware house and other order locations is known.
time matrix=
W O1 O2 O3
W 0 5 20 2
O1 5 0 21 7
O2 20 21 0 11
O3 2 7 11 0
order time of O1= 10:00 AM
order time of O2= 10:20 AM
order time of O3= 10:25 AM
I want to club as many as order possible such that delivery time of any order does not exceed by 2 hours of its order time. Thus the question is to reduce the number of trips(Trip is when delivery agent goes for delivery).
I am trying to come up with algorithm for this. there are two competing factors when
We can combine all the orders in the sequence as they are coming till it satisfies the constraint of delivery of the order within 2 hours of its ordering time.
We can modify above approach to find the bottleneck order(due to which we can not club more order now in approach 1). and pull it out from trip1 and make it a part of trip 2(new order) and wait for other orders to club it with trip1 or trip2 depending.
All the orders are coming in realtime. What will be the best approach to conquer this situation. Let me know if you need more clarity on this.
Very safe and easy algorithm which is guaranteed to not exceed the maximal waiting time for an order:
Let TSP() be a function which returns the estimate of time spent to visit given places. The estimate is pessimistic, i.e. the actual ride time can be shorter or equals to estimate, but not longer. For the good start you can implement TSP() very easily in a greedy way: from each place go to the nearest place. You can subtract the length of the longer edge coming out from W to have better estimate (so a car will always take the shorter edge coming out of W). If TSP() would happen to be optimal, then the whole algorithm presented here would be also optimal. The overall algorithm is as good as TSP() implementation is, it highly depends on good estimation.
Let earliestOrderTime be a time of the earliest not handled yet order.
Repeat every minute:
If there is a new order: If s is empty, set earliestOrderTime to current time. Add it to a set s. Calculate t = TSP(s + W).
If (current time + t >= earliestOrderTime + 2 hours): send a car for a TSP(s + W) trip. Make s an empty set.
Example
For your exemplary data it will work like this:
10:00. earliestOrderTime = 10:00. s = {O1}. t = TSP({01, W}) = 10 - 5 = 5.
10:00 + 0:05 < 10:00 + 2:00, so we don't send a car yet, we wait.
...
10:20. s = {O1, O2}. t = 46 - 20 = 26.
10:20 + 0:26 < 10:00 + 2:00, so we wait.
...
10:25. s = {O1, O2, O3}. t = 2 + 7 + 21 + 20 - 20 = 30.
10:25 + 0:30 < 10:00 + 2:00, so we wait.
...
11.30.
11:30 + 0:30 >= 10:00 + 2:00, so we send a car to go to O3, O1, O2 and back to W. He visits orders at 11:32, 11:39, 12:00 and come backs at 12:20. Guys where waiting 67, 99 and 100 minutes.
There is one very famous problem. I am asking the same here.
There is number of elephants time span given, here time span means, year of birth to year of death.
You have to calculate the period where maximum number of elephants are alive.
Example:
1990 - 2013
1995 - 2000
2010 - 2020
1992 - 1999
Answer is 1995 - 1999
I tried hard to solve this, but I am unable to do so.
How can I solve this problem?
I got approach for when a user asks to find the number of elephants in any year. I solved that by using segment tree, whenever any elephants time span given, increase every year of that time span by 1. We can solve that in this way. Can this be used to solve the above problem?
For above question, I only need the high-level approach, I will code it myself.
Split each date range into start date and end date.
Sort the dates. If a start date and an end date are the same, put the end date first (otherwise you could get an empty date range as the best).
Start with a count of 0.
Iterate through the dates using a sweep-line algorithm:
If you get a start date:
Increment the count.
If the current count is higher than the last best count, set the count, store this start date and set a flag.
If you get an end date:
If the flag is set, store the stored start date and this end date with the count as the best interval so far.
Reset the flag.
Decrement the count.
Example:
For input:
1990 - 2013
1995 - 2000
2010 - 2020
1992 - 1999
Split and sorted: (S = start, E = end)
1990 S, 1992 S, 1995 S, 1999 E, 2000 E, 2010 S, 2013 E, 2020 E
Iterating through them:
count = 0
lastStart = N/A
1990: count = 1
count = 1 > 0, so set flag
and lastStart = 1990
1992: count = 2
count = 2 > 0, so set flag
and lastStart = 1992
1995: count = 3
count = 3 > 0, so set flag
and lastStart = 1995
1999: flag is set, so
record [lastStart (= 1995), 1999] with a count of 3
reset flag
count = 2
2000: flag is not set
reset flag
count = 1
2010: count = 2
since count = 2 < 3, don't set flag
2013: flag is not set
reset flag
count = 1
2020: flag is not set
reset flag
count = 0
How about this?
Say I have all the above data stored in a file. Read it into two arrays separated by the " - ".
Hence, now I have birthYear[] which contains all the birth years and deathYear[] containing all the death years.
so birthYear[] = [1990, 1995, 2010, 1992]
deathYear[] = [2013, 2000, 2020, 1999]
Get the min birth year and the max death year. Create a Hashtable with the Key as a year, and the Value as the count.
Hence,
HashTable<String, Integer> numOfElephantsAlive = new HashTable<String, Integer>();
Now, from the min(BirthYear) to the max(BirthYear), do the following :
Iterate through the Birth Year Array and do an add to the HashTable all the years in between the BirthYear and Corresponding DeathYear with the count being 1. If the key already exists, add 1 to it. Hence, for the last case :
1992 - 1999
HashTable.put(1992, 1)
HashTable.put(1993, 1)
and so on for every year.
Say, for example, you have a Hashtable that looks like this at the end of it:
Key Value
1995 3
1996 3
1992 2
1993 1
1994 3
1998 1
1997 2
1999 2
Now, you need the range of the Years when the number of elephants were maximum. Hence, let's iterate and find the year with the max value. This is pretty easy. Iterate over the keySet() and get the year.
Now, you need a contiguous range of years. You can either do this in two ways:
Do Collections.sort() over the keySet() and when you hit the max value, save all contiguous locations.
Hence, on hitting 3 for our example at 1994, we would check for all the following years with a 3. This will return you your range which is the min-year, max-year combo.
One approach maybe:
Iterate through the periods. Keep track of a list of periods up to now. Note: At each step, the number of periods increases by 2 (or 1 if there is no overlap with the existing list of periods).
For example
1990 - 2013
Period List contains 1 period { (1990,2013) }
Count List contains 1 entry { 1 }
1995 - 2000
Period List contains 3 periods { (1990,1995), (1995,2000), (2000,2013) }
Count List contains 3 entries { 1, 2, 1 }
2010 - 2020
Period List contains 5 periods { (1990,1995), (1995,2000), (2000,2010), (2010, 2013), (2013, 2020) }
Count List contains 5 entries { 1, 2, 1, 2, 1 }
1992 - 1999
Period List contains 7 periods { (1990,1992), (1992,1995), (1995,1999), (1999,2000), (2000,2010), (2010, 2013), (2013, 2020) }
Count List contains 7 entries { 1, 2, 3, 2, 1, 2, 1 }
1) arrange in assending order year wise starting from the largest series.
2) count the years for largest series for whole data set
3) then identify the largest count.
4) the largest count is your answer for years... this can be done in Algo.