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hey hey everybody this is larry this is day 26 of the january league daily challenge um hit the like button hit the subscribe button join me on discord let me let me know what you think about today's farm i hope everyone's doing well uh it's january um i don't know i think i maybe i'm trying to mix it up a little bit today's farmer's path of minimum effort i record and do these prompts live um yeah i hope everyone's doing well you know um you know stay healthy you know um if you have the opportunity you know take care of yourself and if you you know if yeah and if um you know you're fortunate that um that's good then you know try to take care of other people around you um okay so today's problem is path with minimum effort you're a hiker for paying for an upcoming hype pike i actually do hike a lot so that would be me but it is very cold here in new york in the winter so not anytime soon um i did a bit of hiking in hong kong though but um that was a while ago time for ice a given height anyway sorry um yeah you start the upper left cell and you're going to bottom right cell okay up down left oh i've done that right okay i mean i thought it was going to be one of those that you could only go down and write to be honest uh you could wish to find a route that requires the minimum effort throughout its minimum effort is the maximum absolute difference in heights between two consecutive cells effect of the route of the route or wow depending on where you are return the minimum effort required to travel from the top left cell to the bottom right cell um so i mean i do know this one immediately um i think this might have been a recent i don't know if it's recent contest or not by the way i think i would have gotten it the same way i think just guessing the pharma i would have been like oh it's just you know dynamic programming going from um upper corner to the right corner but then here you can go up down left right and the general thing is that in dynamic programming um what you're looking for is some sort of a dag and in and what i mean by that is a direct directed asynchronous graph which means that you know you don't have any cycles just a short answer um and in this case clearly if you could go down left right then you definitely have that um yeah and here the effort is the maximum absolute difference between heights between two consecutive cells and here there it makes it um a sort of a reachability problem right um which means that okay um you know you look at it a little bit backwards you have this example one where okay um you know uh the maximum effort is going to be two because of you know adjacentness and stuff like that and you can actually you know maybe part or attempt a part of the graph but it doesn't go all the way um well yeah and actually mentions it here um so the idea is try to for me and this maybe it's just a lot of experience or a little bit of experience and a lot of practice is just and and you know one common question that i get asked very often is um how do you know right and the short answer is i don't always um the short answer is that i think about it in my head i go can this be solved uh this way and then i give it like maybe you know you know how long it takes um you know if you watch my contest videos like sometimes i spend a lot of time just thinking about it trying to prove to myself and sometimes i go immediately because i solved it before or i have like a similar uh property that i saw before um so that i'm like oh yeah this this is true because i've solved this other thing then that was true so here the the key observation the key thing to note is that um okay if there's a of if there's an absolute uh maximum or uh if there's a if there's an effort of well one is that it's a minimization problem right and some some part of this minimization and it's not always true but um when the problem asks you to minimize um you the first thing i would ask myself uh maybe not the first thing but like top three things say is i can ask um if k works does k plus one work um if k plus one work does k plus two work um you know and so forth right um and you have this thing and then you you ask yourself maybe the opposite question as well which is if k doesn't work um this k minus one also doesn't work right um and so forth so you kind of try to figure out that way and and from that you basically have a boolean array of okay you have force first force first first first force and then at some point it becomes true and then it becomes true all the way through afterwards right um and from here you probably finally guess what i was hinting at which is now you have a monotonic uh amount of tonicity property uh that you know you which you can just basically do binary search on um and that's basically what we're going to do uh i mean you know after you come up with that of course um i i think this one i was a little confident i didn't even do that but after after that of course is that you look at constraints to make sure that this is true um here you know rows times columns you go uh a hundred so this is my after bad thing so i do 100 times 100 uh that's 10 000 and then you have height as you go to 10 to the six and in this case that means that's the max you know 10 to the 6 is always going to be a good answer because that's the maximum delta of the height right so then you know you do log of that which is i don't know what was it 2 2 let's say 15ish no that's not even close my math is terrible 2 to the 20ish yeah so 2 to the 20 so that means that um log of that is about 20. so here this means is equal to about e roughly 200 something operations so that's gonna be good enough that's my off the bat uh thingy um like you know obviously it's it's well it's gonna be a little bit rough but that's that's good enough for me to be confident to move on um so okay so now let's let's get um started with the code um and also yeah i i think i skipped some steps in explanation but what i mean is that okay so how can you prove show that k is good enough right for a certain k well then it becomes just a connectivity problem right because let's say you have a graph uh that's you know you have a graph that looks like this um yeah let's say you have a graph that looks like this you know you can think of it as a well yeah uh you think about a graph that is connected if given a certain k um let's say k is equal to well 10 just some big number well 1 and 2 is going to be connected because it's less than 10 and so forth right and then now you change it to a smaller number let's just say 2. well 1 and 2 are connected 2 and 2 are connected but 2 and 8 are not connected because it is greater than two right so that's basically the graph that we're gonna or you know an implicit graph that we're gonna keep on building for each one and that's basically what we are going to do and that's where the the hundred times a hundred times uh 20 came from earlier because um you know we're going to do 20 binary search and each breakfast search is going to roughly take you know of r times c which is 100 times 100 i know that yours are times like four for directions and their checks and stuff like that but that's the big o roughly um estimate right and also this is like where we said 200 000 so like we have a lot of room of our margin you know just a lot of room to work with um so yeah um i would also say at this point i know that there are conversations about binary search and templates and stuff like that so definitely um follow with me on how to kind of go to it i don't usually do a template i just ask myself the same questions every time and that makes it uh easier the only thing that you get to choose is whether the bands are inclusive or exclusive for me i always choose to make them inclusive and what that means is that um you know we have a left and a right and when it's inclusive it means that both this and this are still in the consideration and when they're too at the same place when these are on top of each other in the middle somewhere like i guess this one actually um that means that you know your range has one element which means that um which means that that's your answer right because your range is one element that means that you're now with the scope all the way down right so that's basically it um yeah so left we can let's just start with zero because i think that's the most um oh sorry that's the minimum like if all if the entire graph has the same number then you won't left to be zero or i think even in this case uh but definitely when the graph is uh in the same number and then on the right side you can actually it's just probably actually the max of um the heights but in terms of just keeping it easy i'm just going to do 10 to the six which is because that's the height of the bow and technically you can also say it's 10 to the 6 minus 1 if you want to be tight but you know um there are a lot of general like the balance are pretty generous so i wouldn't worry about it and again when i say as i said when the left and the right collapses to one place because they're inclusive um you know we're in the right place so we have left less than right because when this is not true anymore that means that um that means that the left is equal to right and with left is equal to right that means that's the answer so we can even write just return left at the very end um so then we have just like standard ish left plus right over the two um i know that there's some overflow considerations so in other languages you might have to write differently in python that's how we write it and i may change this up a little bit depending on the stuff but i just started writing this and then now i write if good is equal to mid um and i'll write this good function later but um what how do we want to define it though right well we don't define it as um you know if mid if we can get from uh top left to bottom right uh with effort mid right so this is good what does this mean right so let's go back to you know let's say we have this range and we have you know this range all the way out here you know middle is somewhere here right say um so this is good what does this mean well that means that we don't have to consider everything to the right because we know that this is already good um so that at the very least like if everything to the left is wrong this is the right answer right so that means that we want to set the right to the mid for this reason and exactly the same way um if we have oops if we're considering this point um we know that this is force so everything to the left of this is force and even the mid is force because by definition is force so we could get rid of it so then we could you know the next iteration we want to move it to the right of that uh uh it's hard for me to do ascii alright but yeah so but that's why you do left is you're gonna make plus one right um this is not a template i think about this on every problem every binary search problem um and yeah and that's basically the idea and you could write like inverse functions and stuff like that to kind of get this to always be kind of similar um if not perfectly uh the same um but um but yeah um so yeah so now we have to write the good function right oops fingers in the wrong place um yeah let's just say target or effort say that's the name so as we said again i'm gonna copy and paste this with some little type lazy tires whatever uh with minimum effort of oh i will target because i'm usually called target or x but yeah right so this is just you know and now this is more fundamental which is what we talked about implicit graph um so we're going to do a breakfast search you can also do it with depth first search and stuff like that um do i want to do i guess it doesn't really matter but i'm going to do a breath first search with um um [Music] yeah and there's no cost in this one um because it is binary right like either you get to the end and you don't so i'm just going to put everything in the queue actually i don't even have to put it i could even use a stack um but but let's just keep it consistent with that because i think i you know i think the optimizations are changes that you can make but um i usually try to do it the same like similar problems the same way as much as possible so that i don't have to think about it even if it's not necessarily like i mean in this case it'll be optimal in the big o sense but maybe you know performance wise you could shave off like a little bit of a constant factor but i'm not gonna worry about it it's fine right like it's you know especially on an interview you could if you're able to talk about it later or even during you know that's fine as well um so yeah so then now we appen uh zero zero this is the top left corner of course um so then now we just do exact you know this is a breath first search um actually we also want to put a scene i'm going to use a set i know that is a little bit inefficient technically it's all one but there are higher constants what i would also recommend if you're trying this at home is um doing a boolean array or a boolean matrix in this case because you know the rows and the columns and so forth um so you don't you know so you could create a matrix already actually i forgot about doing this um so i so that's two rows is equal to the length of this and columns is equal to the length of heights of zero um double checking that it has to be greater than zero otherwise we should just check for it it's not a big deal um but yeah and what i always do is i have a directional uh directions uh array um yeah uh and i write that roughly similar way it's easy to see all the four directions that we practice if you don't um but yeah but now we get to you know x and the y is equal to q dot pop left um and then if x is equal to rows minus one and y is equal to columns minus one uh we return true actually i don't think i need to put this here i can put this elsewhere so um so i might do this later because um you'll see why so actually let me just delete this um but yeah and then now i write just roughly the same way you may um you might find other ways of writing and that's fine you have to work on your own ways of you know finding what's what's good for you but this for me is just you know getting the d x d y of the directions and then here i just set x plus t x y plus t y um so that i get the next x and y um and then if uh the next x and y is equal to or it's uh within the bounds uh i write it like this in python anyway um yeah and so now we know it's inside and also we haven't seen it yet right um actually let's do the other thing first uh where um we check the the f right part right so height and heights of um uh x y and we do an absolute value to check minus heights of nx and y if this is less than or equal to effort and we we also now want to check that we haven't seen it before because we've done it before then uh then we don't do it again so then after if all this is true uh we put it in scene uh and then we also put it in uh in the queue um oh and here we can do one more thing which is what we talked about earlier which is if n x is equal to rows minus one and and y is equal to columns minus one uh we can return true early termination and this is very standard breakfast search stuff um so yeah so that's basically it um i you know if you if you ran out of places to check then you know you would just return force i hope this is right i mean i think this is roughly right um you know maybe i have silly mistakes here and there but um but yeah so let's put in oops type copy and paste the same one i suppose oh no this is slightly different but uh let me see if i copy the white one oh yeah this is i did copy and paste the right one but uh yeah let's run the code uh looks good let's give us some mid close our fingers oh is it too slow a wrong answer i mean yeah this is obvious why this is the wrong answer uh i guess this is actually it's funny because usually i do write the if x y here and that would have fixed it um but um yeah i guess the other thing to learn is uh always check keep on checking but um oh well it's okay because the trade-off here is that um if i write it here then you know obviously it affects this case which i just haven't considered to be honest um but if i write it here then it terminates a little bit earlier because you don't put them on the queue um you know so that it it went a little bit faster for that reason um but i missed the natural case as a result and that was the trade-off that i made you know uh as i was doing it sometimes i do it both ways sometimes i don't i don't know let's give it a submit again yeah um cool uh yeah so what is the complexity of this as i said well just binary search is going to be log of max height time let's call it uh h um so this is about roughly 20 uh say because max h is like a million as we'd said and for each iteration we do a breadth first search which is linear as you might know which is uh r times c uh yeah all of our times c for uh um uh for the complexity um and also of r times c space um and you can maybe make some argument that we you know if we reuse the q in a good way um we could conserve the space so that it'll always be over or times c um you can also debate about like garbage collection and stuff like that um so i'll let that let you to decide how that factors in but in terms of so this is the for the good function so in total time um so total complexity or time complexity is equal to be um r times c times log h um with total space uh i'm just going to quote all r times save you can make some arguments on binary search and garbage collection and all that stuff um yeah so that's all i have for this problem um you know uh yeah um you know my new sign off is you know actually i've been doing this on twitch um but for whatever reason i guess i just never did this on um on lead code which is you know um i hope everyone has a great day have a great week um had a great month it's challenging times obviously still the light hopefully is coming soon at the end of the tunnel you know stay strong i hope everyone stay healthy to good mental health know do good not do well i mean do well too but do good and you know take care of yourself and if you're fortunate enough to be in a good place take care of the people around you treasure them hug them loved them and and i'll see you later bye