In the last optimization, when the elevator was moving up and dropped the passenger of floor 0 at floor 2, and picked the passenger at floor 2 who requested a down , why the elevator changed the direction and dropped them first. Shouldnt the lift continue togo up and attend the request of the passengers going up first?
that is the drawback u would say for this algorithm thats why we have to take care of one direction only by marking any of the direction as true either up or down once the lifts starts to move maintain 3 varibles as bool and mark them as true until it reaches the last pos of requested then mark as false and changes the directin of the lift so that the requested of other can also be taken care of
Just create a button class, there may be tons of button with different functionalityies, i dont get why emergency is special and how it affects your design
Firstly thanks for the comment, it helps create the present and future videos better !!. if you see at 11:07, in the class diagram, there is a class for buttons and we extend from it to create inside and outside buttons. Do you mean somthing else
There might be, also you can make Door class as a base class and then inherit from it as Elevator Door, Emergency Door, Service Door etc. and then create a relationship from Door to Car class imo
Runtime updation of minHeap, as the elevator car crosses a floor, is equivalent to creating a new minHeap everytime? I am not sure there is any straight forward way to update the values in a minHeap/maxHeap.
There is a “decrease key” operation in the Heap which does changing of node’s value, in o(log n) runtime when it is combined with HashMap to maintain the position of node in array (Heap uses behind the scene). You may explore it further.
Hey thanks for this video Sorry this is out of context, but what chair are you using ? I have seen this one at different places but not able to find it
haha, no worries.. I think this one is pretty close www.ikea.com/se/sv/p/markus-kontorsstol-vissle-moerkgra-70261150/ , you can get the same in any country.
Good explanation but a small doubt i.e whenever a position of car changes everytime we have to pop all enteries of both minhips and update it again, so will that be efficient?
yes that might not be efficient but O(nlogn) is not asymptotically tight, search for tighter bound for hepify. It might not be awesome algorithm but can do the work rest is on our innovations.
In the last part of the video , you have used 2 heaps . When you are not using the direction of the car to pickup up the passenger , what's the point in keeping 2 heaps ? You can simply use 1 heap and pickup the closest passenger irrespective of the direction he/she wants to move. This is a bit confusing .
Thanks for pointing out. Yes, if you are not having a constraint on direction then you can use just one min heap, I wanted to give a starting algorithm which anyone can extend and optimized further. e.g.Some elevator systems pick passengers only in the dircetion of travel then the 2 heaps would help.
@@codeTreeMS This was my first thought. Starting at floor 0, it goes up until no one else needs to go up, then it starts at the top and brings down everyone who needs to go down. then start with people going up, rinse, repeat
It depends on how many people work/stay in the building, like an office with a high number of floors with 1000 people working will require more elevators than a smaller building with 300 people, its something easy and you can discuss with your interviewer to decide on this requirement.
Nice explanation of the algorithm. Thank you so much!
You can't know everything. You need to ask requirements and model only those
MinHeap update operation is quite costly since its getting updated at every move of lift in either direction O(NLog(N)) to be precise
It's an ELEVATOR. It only has so many floors. Unless it's being implemented by actual snails, or a 1HZ processor, I think it will be fine.
In the last optimization, when the elevator was moving up and dropped the passenger of floor 0 at floor 2, and picked the passenger at floor 2 who requested a down , why the elevator changed the direction and dropped them first. Shouldnt the lift continue togo up and attend the request of the passengers going up first?
that is the drawback u would say for this algorithm thats why we have to take care of one direction only by marking any of the direction as true either up or down once the lifts starts to move maintain 3 varibles as bool and mark them as true until it reaches the last pos of requested then mark as false and changes the directin of the lift so that the requested of other can also be taken care of
Just create a button class, there may be tons of button with different functionalityies, i dont get why emergency is special and how it affects your design
Firstly thanks for the comment, it helps create the present and future videos better !!. if you see at 11:07, in the class diagram, there is a class for buttons and we extend from it to create inside and outside buttons. Do you mean somthing else
Should change the distance in heap when lift is moving
awesome explanation. Insightful.
Very well explained. Please make more OOD videos
Thanks ! Will get back to making more such vidoes soon.
@@codeTreeMS Thanks :)
Looks like the copy of Educative design
Is there a particular design pattern that you implemented for the UML class diagram? If so what is that pattern called?
isn't there a relationship from door to car? 18:14
There might be, also you can make Door class as a base class and then inherit from it as Elevator Door, Emergency Door, Service Door etc. and then create a relationship from Door to Car class imo
very well explained thankyou for your time.
Very well explained . Thank you so much !
What does the orange diamond mean?
which tool are you using to design this
Runtime updation of minHeap, as the elevator car crosses a floor, is equivalent to creating a new minHeap everytime?
I am not sure there is any straight forward way to update the values in a minHeap/maxHeap.
There is a “decrease key” operation in the Heap which does changing of node’s value, in o(log n) runtime when it is combined with HashMap to maintain the position of node in array (Heap uses behind the scene). You may explore it further.
Any interviewer is going to ask why you picked up min heap of all the data structures.
Thats a resonable question and had a resonable answer, time complexity. in O(1) you can get the closest person to the elevator.
Hey thanks for this video
Sorry this is out of context, but what chair are you using ? I have seen this one at different places but not able to find it
haha, no worries.. I think this one is pretty close www.ikea.com/se/sv/p/markus-kontorsstol-vissle-moerkgra-70261150/ , you can get the same in any country.
Please add timestamps :), btw nice video
Bro code please
Good explanation but a small doubt i.e whenever a position of car changes everytime we have to pop all enteries of both minhips and update it again, so will that be efficient?
yes that might not be efficient but O(nlogn) is not asymptotically tight, search for tighter bound for hepify. It might not be awesome algorithm but can do the work rest is on our innovations.
@@codeTreeMS ok thanks!
code in java?
I like the video because it is succinct and explained well.
In the last part of the video , you have used 2 heaps . When you are not using the direction of the car to pickup up the passenger , what's the point in keeping 2 heaps ? You can simply use 1 heap and pickup the closest passenger irrespective of the direction he/she wants to move. This is a bit confusing .
Thanks for pointing out. Yes, if you are not having a constraint on direction then you can use just one min heap, I wanted to give a starting algorithm which anyone can extend and optimized further. e.g.Some elevator systems pick passengers only in the dircetion of travel then the 2 heaps would help.
@@codeTreeMS This was my first thought. Starting at floor 0, it goes up until no one else needs to go up, then it starts at the top and brings down everyone who needs to go down. then start with people going up, rinse, repeat
16:09 you spelled elevator wrong
Sir, one small doubt, how many elevators are required in buildings
It depends on how many people work/stay in the building, like an office with a high number of floors with 1000 people working will require more elevators than a smaller building with 300 people, its something easy and you can discuss with your interviewer to decide on this requirement.