So, Code Jam 2018 Qualification round ended today and the for Second problem 'Trouble Sort', I got the right program but in the judging process, I didn't get it as it said that I exceeded the time limit, but there were no time limit specifications in the question.
Can someone help me with the proper code?
Question: https://codejam.withgoogle.com/2018/challenges/
but there were no time limit specifications in the question
You were not careful enough reading the problem statement.
Test set 1 (Visible)
3 ≤ N ≤ 100.
Time limit (for the entire test set): 10 seconds.
Test set 2 (Hidden)
3 ≤ N ≤ 105.
Time limit (for the entire test set): 20 seconds.
You can also read the analysis of the problem
Related
I was asked in a job interview to build a 'simple' unlimited integer.
I was given endless space and time, so for 'simplicity' I just used a linked list of digits (0-9).
An 'increment' operation increments the first digit and cascades an increment to the next one if it's 0 (was 9 initially).
So far so good.
Then I was asked what is the run time for such an increment operation, given a random initial state.
I got kinda stuck on it, started talking about sums of logarithms on bases 10^n and such, when the interviewer got frustrated and told me it's constant time.
I can't figure it out by myself, but it seems to me that it can't be constant.
Please, can anyone explain this to me?
I am working on an assignment where I am supposed to compute the density of an event. Let's say that a certain event happens 5 times within seconds, it would mean that it would have a higher density than if it were to happen 5 times within hours.
I have in my possession, the time at which the event happens.
I was first thinking about computing the elapsed time between each two successive events and then play with the average and mean of these values.
My problem is that I do not know how to accurately represent this notion of density through mathematics. Let's say that I have 5 events happening really close to each other, and then a long break, and then again 5 events happening really close to each other. I would like to be able to represent this as high density. How should I go about it?
In the last example, I understand that my mean won't be truly representative but that my standard deviation will show that. However, how could I have a single density value (let's say between 0 and 1) with which I could rank different events?
Thank you for your help!
I would try the harmonic mean, which represents the rate at which your events happen, by still giving you an averaged time value. It is defined by :
I think its behaviour is close to what you expect as it measures what you want, but not between 0 and 1 and with inverse tendencies (small values mean dense, large values mean sparse). Let us go through a few of your examples :
~5 events in an hour. Let us suppose for simplicity there is 10 minutes between each event. Then we have H = 6 /(6 * 1/10) = 10
~5 events in 10 minutes, then nothing until the end of the hour (50 minutes). Let us suppose all short intervals are 2.5 minutes, then H = 6 / (5/2.5 + 1/50) = 6 * 50 / 101 = 2.97
~5 events in 10 minutes, but this cycle restarts every half hour thus we have 20 minutes as the last interval instead of 50. Then we get H = 6 / (5/2.5 + 1/20) = 6 * 20 / 41 = 2.92
As you can see the effect of the longer and rarer values in a set is diminished by the fact that we use inverses, thus less weight to the "in between bursts" behaviour. Also you can compare behaviours with the same "burst density" but that do not happen at the same frequency, and you will get numbers that are close but whose ordering still reflects this difference.
For density to make sense you need to define 2 things:
the range where you look at it,
and the unit of time
After that you can say for example, that from 12:00 to 12:10 the density of the event was an average of 10/minute.
What makes sense in your case obviously depends on what your input data is. If your measurement lasts for 1 hour and you have millions of entries then probably seconds or milliseconds are better choice for unit. If you measure for a week and have a few entries then day is a better unit.
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Plain English explanation of Big O
The wikipedia article on logarithmic growth is a stub. Many of the answers I read on stackoverflow give clarification of how efficient a process or function is based on a logarithmic function using 0 (I assume[see below] it is a 0[zero] and not an O[letter as M,N,O,P,Q], but please correct my assumption if it is wrong) and an n or N.
Can someone explain the logarithmic explanations pertaining to the common computational explanations better? maybe in terms of time in seconds(milliseconds are also welcome, just trying to conceptualize it in real life time differences...), terms of size, and/or in terms of weight?
I have seen the following frequently: (please feel free to include other ones as well)
O(1)
O(N)
My assumption is based that a 0 [zero] outside of a code block does not have a slash through it, while inside a code block a 0 does have a slash through it.
What this means is that the execution time (or other resource) is some function of the amount of data. Let's say it takes you 5 minutes to blow up 10 baloons. If the function is O(1), then blowing up 50 baloons also takes 5 minutes. If it is O(n), then blowing up 50 baloons will take 25 minutes.
O(log n) means that as things scale up, managing larger n becomes easier (it follows logorithmic growth). Let's say you want to find an address in a directory of f(n) items. Suppose it takes 6.64 seconds to search a directory of 100 entries. [6.64 = log_2(100)] Then it might only take 7.64 seconds to search 200 entries. That is logorithmic growth.
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Closed 10 years ago.
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Failure rate of a system
If a system has a 10% independent chance of failing in any given hour, what are the chances of it failing in a given 2 hour period or n-hours period?
Note: 10% failure probability in 1 hour has nothing to do with 10% of the time . It's just that a system has a 10% independent chance of failing in any given hour
Let Pfail be the probability that the system fails in any given hour.
Then Pnofail, the probability that the system does not fail in any given hour, is 1 - Pfail.
The chance of it not failing in 2 hours is (Pnofail)2, since it must independently not-fail in each of those hours, and the joint probability of two independent events is the product of the probability of each event (that is, P(A ∩ B) = P(A)*P(B)).
More generally, then, the chance of it not failing in n hours is (Pnofail)n .
The chance of it failing in n hours is 1 - (chance of not failing in n hours).
You should be able to work it out from there.
Here is my scenario,
I run a Massage Place which offers various type of massages. Say 30 min Massage, 45 min massage, 1 hour massage, etc. I have 50 rooms, 100 employees and 30 pieces of equipment.When a customer books a massage appointment, the appointment requires 1 room, 1 employee and 1 piece of equipment to be available.
What is a good algorithm to find available resources for 10 guests for a given day
Resources:
Room – 50
Staff – 100
Equipment – 30
Business Hours : 9AM - 6PM
Staff Hours: 9AM- 6PM
No of guests: 10
Services
5 Guests- (1 hour massages)
3 Guests - (45mins massages)
2 Guests - (1 hour massage).
They are coming around the same time. Assume there are no other appointment on that day
What is the best way to get ::
Top 10 result - Fastest search which meets all conditions gets the top 10 result set. Top ten is defined by earliest available time. 9 – 11AM is best result set. 9 – 5pm is not that good.
Exhaustive search (Find all combinations) - all sets – Every possible combination
First available met (Only return the first match) – stop after one of the conditions have been met
I would appreciate your help.
Thanks
Nick
First, it seems the number of employees, rooms, and equipment are irrelevant. It seems like you only care about which of those is the lowest number. That is your inventory. So in your case, inventory = 30.
Next, it sounds like you can service all 10 people at the same time within the first hour of business. In fact, you can service 30 people at the same time.
So, no algorithm is necessary to figure that out, it's a static solution. If you take #Mario The Spoon's advice and weight the different duration massages with their corresponding profits, then you can start optimizing when you have more than 30 customers at a time.
Looks like you are trying to solve a problem for which there are quite specialized software applications. If your problem is small enough, you could try to do a brute force approach using some looping and backtracking, but as soon as the problem becomes too big, it will take too much time to iterate through all possibilities.
If the problem starts to get big, look for more specialized software. Things to look for are "constraint based optimization" and "constraint programming".
E.g. the ECLIPSe tool is an open-source constraint programming environment. You can find some examples on http://eclipseclp.org/examples/index.html. One nice example you can find there is the SEND+MORE=MONEY problem. In this problem you have the following equation:
S E N D
+ M O R E
-----------
= M O N E Y
Replace every letter by a digit so that the sum is correct.
This also illustrates that although you can solve this brute-force, there are more intelligent ways to solve this (see http://eclipseclp.org/examples/sendmore.pl.txt).
Just an idea to find a solution:
You might want to try to solve it with a constraint satisfaction problem (CSP) algorithm. That's what some people do if they have to solve timetable problems in general (e.g. room reservation at the University).
There are several tricks to improve CSP performance like forward checking, building a DAG and then do a topological sort and so on...
Just let me know, if you need more information about CSP :)