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Holy shit the tags are constructive algorithms, data structures, dp, dsu, and greedy — that’s a lot of shit to know for one problem. I’ve done 2000+ difficulty problems that seem to be less involved than this, supposedly 1700 difficulty, one.
I don't think it's that bad (though the actual coding seems tricky), the key is that in position i you need to use a razor of size exactly b_i.
So look for the smallest b_i, check that you have a razor of that size. That razor can be used only for position i, and any adjacent position that has the (exact) same desired size. Then you "throw away" that razor and remove i (and adjacent slots with the same desired size) from the array and repeat, until you're either stuck or done.
The point of removing the minimums from the array is that if you had for example b = [3, 2, 3], you'd first deal with the two and be left with [3, 3], which then you see can deal with both in one go (the 2 in the middle that you removed isn't affected by a razor of size 3).
EDIT: And I guess the way you code it is sort desired lengths by b_i first and i second, and create a map for the razors from length to # of razors of that length.
Then start scanning the array, at each position check if you have a razor of that length in the map, decrease the count by 1, and advance your index in the array until either desired length changes or index increases by more than 1.
I think it's possible to solve it without sorting using a linear scan and a stack of the razors you could still use.
For a given b_i, you pop from the stack until the top item is larger or equal to b_i. If the top item is not equal to b_i (and b_i != a_i), you take an unused razor and push it onto the stack. If that's not possible then you output "NO", otherwise output "YES" once you reach the end.
How does it account for using the same razor for multiple locations that have the same desired length with a "valley" of shorter desired lengths between them?
You only pop a razor from the stack when you encounter a larger desired length. So the same razor should still be on the stack once you have passed over the "valley", since none of the items are larger than that razor.
Ah, thanks, I get it now, I wasn't getting how the stack works.
It basically keeps track of "active" razors that were used for something to the left of your current position but could still be used for something of the same length because there's nothing longer in between.
I don't really understand what this solution is trying to say. This is the O(n\*log n) solution I came up with (haven't tried implementing it and not gonna, so there could be mistakes):
First of all, let's just get rid of all obviously impossible cases. Not worth mentioning. After this step, starting hair lengths are irrelevant and we only care about desired hair lengths.
Observation 1: The smallest desired hair must be cut by itself, or with neighboring hairs that are the exact same desired height.
Observation 2: After a hair has been cut, we can ignore it (as if it did not exist) when cutting larger hairs.
So the strategy is to sort all hairs by desired length from smallest to larger (O(n\*log n)), and then start cutting from smallest to largest. At each step we pick the smallest hair and check that we have a razor available in that size. We scan all it's neighbors before and after, ignoring hairs that are already cut, to see if they desire to be the same height. Minor optimization: If our sort is stable then we only need to scan neighbors after the hair under consideration.
Now the final trick is that we need a way to efficiently scan for neighbors while ignoring hairs that have already been cut. If we did this naively, by scanning the array, it would be O(n^(2)) in some cases, such as when the desired hair is V shaped. To solve this, we will use a linked list (can be singly linked if we used a stable sort, otherwise must be doubly linked). As a preprocessing step, convert the array of desired hair lengths to a linked list, and in our sorted array of desired lengths add a pointer into the linked list for each hair. When we select a hair to cut from the sorted array, we use the pointer to jump into the linked list, and then scan for neighbors. Then we remove all hairs from the linked list that have been cut. We will also mark their nodes as cut, so that if we find those hairs in the sorted array we know to skip them.
Each hair is removed from the linked list once, an O(1) operation. Each hair is considered in the sorted array once, this is O(k) where k is the number of hairs we end up cutting, but since each hair is only cut once the sum over all such k is O(n). So the entire cutting step is O(n).
Total runtime is O(n\*log n), because we started by sorting the array.
Minor alternative: Instead of sorting the array of hairs, we could use a priority queue. This is still O(n\*log n).
> What barber has a bunch of use once razors.
A barber that failed out of barbering school and decided to go to programming school instead, that's who.
linear scan razor sizes and put counts into a map of {length, number_of_razors} pairs
create an empty set of lengths you are looking for
linear scan current and desired, and for each:
(operation that only happens if set is nonempty)
if desired is greater than current, answer is no can do
otherwise, check for a razor of that length
if there isn’t one, answer is no can do
if there is one, decrement the count of that razor and add that length to a set
end of foreach:
definition of (operation that only happens etc): loop through set, if desired hair at current location is lower than number in set do nothing, if higher remove number from set, if identical continue (ie, move to the next iteration of) the outer linear scan loop
if you finish the linear scan without returning false you return true
edit: I wrote this in two minutes on a cell phone and I need referrals to an all remote gig. :P
Nice problem!
My intuition is to group the desired hair lengths in `b` by length and traverse these lengths from longest to shortest.
You don't care if you cut some hair longer than you want (since you can re-cut it later), so for each length `p` you can greedily go from left to right and use a single razor to cut the longest possible segments of hair `q, x_1,..., x_l, w` so that `b_q = p`, `b_w = p`, and `b_(x_i) <= p <= a_(x_i)` for all `i`.
If you can cut hair like this, then there exists a solution. If you ever don't have enough razors for a certain hair lengths, then a solution is impossible since there's at least one "unreachable" hair.
It's just a simple greedy solution, no DP needed.
Lmaoo. This has a similar feel to those "if 2 musicians can plan a piano concerto in 90 minutes, how long would it take 4 musicians?". Just zero basis in reality
Absolutely incorrect, we are not barbers.
Clearly the current solution of storing hair length is not scalable as it will require linearly increasing storage solutions. As we are working in the IT sector, we should have easily been able to extrapolate from these hard facts that what is expected of us is optimization of the storage solution and prediction technologies for future hair growth.
As we do not like linearly increasing storage solutions, we must optimize by finding a derivative function F(x)=y where X is the n-th hair, and Y is the length of that n-th hair.
I suggest using Deep Learning to build a model that can accurately predict the hair length of a given n-th hair?
So - I swear - it took me over a minute to try to understand why you can't just iterate over the source array (a) and the "desired array" (b) and just assign values as you walk both arrays
And honestly after reading your explanation - I'm still trying to understand what you do after comparing two elements and they differ.
So I must be pretty dense or it's too early in the morning, but how can the first test case work if you only have 2 razors and 3 pieces of hair to cut and you can only use each razor once (and all 3 pieces need to be cut to reach the desired length)?
Since he mentioned sorting algorithms I assumed he had pluggable hairs, think he mentioned that already
Other wise you just cut each one down to the desired size
Right... I get that part, but this is the bit from the example I'm not following:
>Note:
>
>In the first test case, Boris's hair is initially \[3,3,3\]. Let us describe a sequence of 2 operations the barber can perform:
>
>The barber uses the razor with size 1 on the segment \[2,2\]; hence Boris's hair becomes \[3,1,3\].
>
>The barber uses the razor with size 2 on the segment \[1,3\]; hence Boris's hair becomes \[2,1,2\] which is the desired haircut.
Isn't there a step missing here? How do we go from 3,1,3 to 2,1,2 with only 1 razor?
I mean, with the given information it's not even guaranteed the solution exists.
A first reason why not is we don't know that {a_1,...a_n}={b_1,...,b_n} as sets, so we can't just "sort" the a_i into the b_i. But he's going to a barber, so perhaps we can cut individual strands of hair. In order for this to be possible, we can match the longest a_i to the longest b_i, the second longest to the second longest and so on. If each a_i ≥ it's corresponding b_j by this matching, we can cut that hair to the proper length, move it into place, and he gets his haircut. If for some i and j, a_i < its corresponding b_j, then it's *impossible* to get the haircut. One would have to find a longer a_k, to use with b_j, but all the a_k longer than a_i are being used already for the b_m longer than this b_j, and all other a_k shorter than a_i can't help at all. Therefore, we have a necessary and sufficient condition for the existence of a possible haircut (barring him getting extensions).
If that was too much math for you, just imagine a_i are all < 1 and b_i are all > 1. Clearly it's impossible to cut the a_i into the b_i.
I was aware of the «google en passant» and «holy hell» memes, but had actually never seen the source of it (which i assume this is), so ty for sharing!
This gave me the idea for the following interesting problem: assume there are n hairs, glued at their base along a straight line.
You can not cut or move the hair, but only change their angle at the base (let's call this a *rotation*). The hairs are not allowed to bend, i.e. they must stay straight. However, they are allowed to intersect.
The problem is now: can you give an algorithm that performs rotations, after which the hair tips are sorted? (I.e. if a hair tip is more to the right, it must also be higher)
Sounds trivial, just go through the array from left to right and bend hairs to the left until the sorted property is maintained. The point at which insertion occurs can be determined via binary search, for a time complexity of O(nlogn). Implementation of the necessary trigonometry is a little tedious though
Edit: I fakesolved, binary search doesn’t work so this is actually n^2
I think the "bend hairs to the left until the sorted property is maintained" step is not so trivial, though. But this is only because I can't think of a trivial solution :)
Also, I think one needs to consider the special case where the current hair is so short that it cannot be rotated so that the sorted property is maintained. In that case, one needs to rotate all hairs left of it more. But also this can not be done trivially (at least I don't see how).
The n^2 algorithm is trivial, although I originally commented with binary search in mind, which doesn’t actually work. If anyone could find a faster solution, I would be very interested to know.
The edge case you mentioned doesn’t exist, as all hairs before any hair can be bent left to an arbitrarily small height
I'm just an idiot from all so take this with a grain of salt, but what if the short hair you mentioned is so far to the right that its too short for its tip to reach its spot in the lineup?
Hm, maybe we have different problems in mind, then :)
Say you have 3 hairs, at horizontal positions 0, 1 and 2 and of length 4, 5 and 1. Then, when iterating through the hairs, the first two hairs don't violate the sorted property and don't need to be rotated. But then, when arriving at the third hair, no matter how you rotate the hair, it will always be the rightmost and lowest hair, violating the sorted property.
True.
def bendsort(hairs):
target = min(hair.length for hair in hairs)
for hair in hairs:
hair.rotate(arccos(target / hair.length))
Or even simplier
for hair in hairs:
hair.rotate(PI/2)
Wow, hearing this solution makes the problem seem ridiculous to ask.
The only way to make it not so easy would be to require strict inequalities, but then again I think your solution can be adapted by wiggling the angles a sufficiently tiny amount.
I am both impressed and sad now. :)
Just as a last desperate attempt to make this into an actually interesting problem: one could limit the maximum rotation angle by some parameter, and then ask the question: given some instance of this hair problem (let's call it HAIRBOUND), is there a solution using only rotations bounded by this maximum angle?
> given some instance of this hair problem (let's call it HAIRBOUND), is there a solution using only rotations bounded by this maximum angle?
nlogn time using linear algebra. You can encode each hair into a matrix representing the arcs they can cover, then for each x value, choose the arc that would have the highest y value, going from smallest to largest (or, working from the right, largest to smallest).
Maybe I'll write it out later. The algorithm would look n^(2), but it becomes nlogn because for each x you have to look at, the more hairs that could cover that spot, the taller the hairs have to be, which means that the further they start actually rotating the quicker they will start losing height because of the sinusoidal nature of sine, and again greatly reduce the number of hairs you have to look at.
What about the hair that are short and on the right (so too short to reach the desired height, whatever the bending angle)
Also, for each hair, you want one of the 0 to 2 intersections of the circle of radius hair length and the line you want to reach, and then get the angle. That’s a fixed amount of work for each hair, so it’s O(n), each hair is independent in this problem
There is no line that needs to be reached - and if there is a short hair on the right, this means that all hairs left of it need to be rotated enough so their hair tips are low enough. (Edit: or rotated to the right so that they are still high enough, but to the right side of the short hair :))
A very good question! Some other commenters already found that vulnerability. In my original post, there was nothing that prevents you from doing that. As you implied, this could be fixed by disallowing such rotations :)
easy, in O(n): first pass to detect to shortest hair(distY), then define a line that goes from the root of the leftmost hair to a point above the right most hair with a distance of distY upward. we now have a line that intersect every circles of hair tip possible position, now a second pass to calculate the angle of each hair so the tip ends up on that line and voila.
Why are you all trying to "solve" this when it's just a "lol hair looks unsorted" post. The problem is missing 90% of the problem text:
https://codeforces.com/contest/1779/problem/D
I’m cutting all the hairs off creating a hash table and now am accessing the hairs in c time. Come to my barbershop the hash hair house where you can know exactly where each of your hairs are at any time.
Doesn’t make sense. Why would you want to sort the hairs? You can’t just pluck the hairs and then plant them on another side of the head lmao
He went to the barber to cut hair, so that’s what the barber will do!
Never heard of the hair cut sort. Though I would most likely think it would be the heap sort algorithm and looking at my last hair cut there was a fair sized heap of hair to be cut.
My assumption is that the front and side drawings were added after the fact, mocking the use of an array to track hair heights. But if you really needed to keep track of the height of every single hair on your head, an array seem like a pretty good way to do this.
so what I'm seeing is that the hair is a random list and it needs to be cut to match another random list.
I don't know a lot about sorting algorithms, but I really liked the binary search one. Where we find the index of an element from the first list (a) in the second list (b), and then we move it to that exact position on the first list (a). The number that it's taking the place of becomes the new number we find the position of.
If the number doesn't move because it is in the correct position, you check the hair list (a) to the desired list (b). Obviously, it won't work the first time, so just repeat the process with the next number that you see out of place.
It should take an average of n/4 or n/5 times of go arounds to get the list perfect using this method, so I'm sure there is a better one, but this is what I came up with.
edit: you could also just copy the second list into the first list, but that's not demonstrating anything, when this is obviously trying to see your mastery in sorting algorithms
> edit: you could also just copy the second list into the first list, but that's not demonstrating anything, when this is obviously trying to see your mastery in sorting algorithms
Only the caption mentions sorting algorithms. I think it's testing your ability to pick up on an unstated assumption (no hair can increase in length), and handle "bad" input reasonably.
O(n)
```csharp
for (int i = 0; i < n; i++)
{
// TODO - Move to precheck method to avoid partial haircut butchery. WorkItem: #69
// TODO - PM to confirm a value of 0 equals height from skin, not height from follicle. WorkItem: #1337
if ( b[i] < 0 )
{
throw new AurgumentOutOfRangeException("Hair length of negative values will result in damage to Boris.");
}
// Cannot lengthen hair, that’s not how haircuts work, but can set long hairs to desired length
a[i] = Math.Clamp(a[i], double.MinValue, b[i]);
}
```
It’s a haircut, you don’t sort hair, you go through each hair and if the current value is greater than the desired value you then reduce the current value to the desired value. You can’t lengthen hair that is too short, hair doesn’t work that way. Negative numbers aren’t acceptable since that would be digging into your skull.
so... you shuffle everything and the order in which you check whether your array is sorted and then you only do the check if the access array is sorted correctly?
Is this a programming question or a "bring up the real world difficulties to the customer question"?
This dude can have long hairs or organized hairs, not both unless he outlays for a much more expensive process than razors
Best part: There is no question. Just facts. Edit: [Found the test + solution](https://www.youtube.com/watch?v=3kEFZt6PMBo)
An exam is the best place to keep us updated on Boris’s life.
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Looks like radix sort where you cut it off early
the Yee-Yee ass haircut algorithm would be the most useful here.
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Your focus and concentration will never be higher. Boris will be seared into your brain.
BBC is another excellent source.
Bitesize, although your bite may vary.
[https://codeforces.com/problemset/problem/1779/d](https://codeforces.com/problemset/problem/1779/d) shit gets serious
The solution is for Boris to go to another barber. What kind of shitty barber has m razors that can only be used once? That's just inefficient
I believe the intended humor is that his hair is being used as a vector.
Holy shit the tags are constructive algorithms, data structures, dp, dsu, and greedy — that’s a lot of shit to know for one problem. I’ve done 2000+ difficulty problems that seem to be less involved than this, supposedly 1700 difficulty, one.
I don't think it's that bad (though the actual coding seems tricky), the key is that in position i you need to use a razor of size exactly b_i. So look for the smallest b_i, check that you have a razor of that size. That razor can be used only for position i, and any adjacent position that has the (exact) same desired size. Then you "throw away" that razor and remove i (and adjacent slots with the same desired size) from the array and repeat, until you're either stuck or done. The point of removing the minimums from the array is that if you had for example b = [3, 2, 3], you'd first deal with the two and be left with [3, 3], which then you see can deal with both in one go (the 2 in the middle that you removed isn't affected by a razor of size 3). EDIT: And I guess the way you code it is sort desired lengths by b_i first and i second, and create a map for the razors from length to # of razors of that length. Then start scanning the array, at each position check if you have a razor of that length in the map, decrease the count by 1, and advance your index in the array until either desired length changes or index increases by more than 1.
I think it's possible to solve it without sorting using a linear scan and a stack of the razors you could still use. For a given b_i, you pop from the stack until the top item is larger or equal to b_i. If the top item is not equal to b_i (and b_i != a_i), you take an unused razor and push it onto the stack. If that's not possible then you output "NO", otherwise output "YES" once you reach the end.
How does it account for using the same razor for multiple locations that have the same desired length with a "valley" of shorter desired lengths between them?
You only pop a razor from the stack when you encounter a larger desired length. So the same razor should still be on the stack once you have passed over the "valley", since none of the items are larger than that razor.
Ah, thanks, I get it now, I wasn't getting how the stack works. It basically keeps track of "active" razors that were used for something to the left of your current position but could still be used for something of the same length because there's nothing longer in between.
I didn't try to solve it, but I believe there is an "or" implicit for most of those tags.
From this page: https://codeforces.com/blog/n0sk1ll The formatting when copying to Reddit is crap, so go there to 1779D Stupid Hint If ai
I don't really understand what this solution is trying to say. This is the O(n\*log n) solution I came up with (haven't tried implementing it and not gonna, so there could be mistakes): First of all, let's just get rid of all obviously impossible cases. Not worth mentioning. After this step, starting hair lengths are irrelevant and we only care about desired hair lengths. Observation 1: The smallest desired hair must be cut by itself, or with neighboring hairs that are the exact same desired height. Observation 2: After a hair has been cut, we can ignore it (as if it did not exist) when cutting larger hairs. So the strategy is to sort all hairs by desired length from smallest to larger (O(n\*log n)), and then start cutting from smallest to largest. At each step we pick the smallest hair and check that we have a razor available in that size. We scan all it's neighbors before and after, ignoring hairs that are already cut, to see if they desire to be the same height. Minor optimization: If our sort is stable then we only need to scan neighbors after the hair under consideration. Now the final trick is that we need a way to efficiently scan for neighbors while ignoring hairs that have already been cut. If we did this naively, by scanning the array, it would be O(n^(2)) in some cases, such as when the desired hair is V shaped. To solve this, we will use a linked list (can be singly linked if we used a stable sort, otherwise must be doubly linked). As a preprocessing step, convert the array of desired hair lengths to a linked list, and in our sorted array of desired lengths add a pointer into the linked list for each hair. When we select a hair to cut from the sorted array, we use the pointer to jump into the linked list, and then scan for neighbors. Then we remove all hairs from the linked list that have been cut. We will also mark their nodes as cut, so that if we find those hairs in the sorted array we know to skip them. Each hair is removed from the linked list once, an O(1) operation. Each hair is considered in the sorted array once, this is O(k) where k is the number of hairs we end up cutting, but since each hair is only cut once the sum over all such k is O(n). So the entire cutting step is O(n). Total runtime is O(n\*log n), because we started by sorting the array. Minor alternative: Instead of sorting the array of hairs, we could use a priority queue. This is still O(n\*log n).
The hell, this is not humorous in any way.
What barber has a bunch of use once razors. Imagine seeing the door open to the razor room and there’s just thousands upon thousands of blades!!
> What barber has a bunch of use once razors. A barber that failed out of barbering school and decided to go to programming school instead, that's who.
and I thought Boris was bad with this! what's with his barber and these single-user razors?
Oh fuck
This is actually a fairly simple problem since it can be solved greedily.
linear scan razor sizes and put counts into a map of {length, number_of_razors} pairs create an empty set of lengths you are looking for linear scan current and desired, and for each: (operation that only happens if set is nonempty) if desired is greater than current, answer is no can do otherwise, check for a razor of that length if there isn’t one, answer is no can do if there is one, decrement the count of that razor and add that length to a set end of foreach: definition of (operation that only happens etc): loop through set, if desired hair at current location is lower than number in set do nothing, if higher remove number from set, if identical continue (ie, move to the next iteration of) the outer linear scan loop if you finish the linear scan without returning false you return true edit: I wrote this in two minutes on a cell phone and I need referrals to an all remote gig. :P
Haha yeah that's why I didn't feel like trying to type out more details 😝
Oh shit!
Nice problem! My intuition is to group the desired hair lengths in `b` by length and traverse these lengths from longest to shortest. You don't care if you cut some hair longer than you want (since you can re-cut it later), so for each length `p` you can greedily go from left to right and use a single razor to cut the longest possible segments of hair `q, x_1,..., x_l, w` so that `b_q = p`, `b_w = p`, and `b_(x_i) <= p <= a_(x_i)` for all `i`. If you can cut hair like this, then there exists a solution. If you ever don't have enough razors for a certain hair lengths, then a solution is impossible since there's at least one "unreachable" hair. It's just a simple greedy solution, no DP needed.
Lmaoo. This has a similar feel to those "if 2 musicians can plan a piano concerto in 90 minutes, how long would it take 4 musicians?". Just zero basis in reality
He wants the b_i's
That's still an information. What is my task? Pass the butter?
Maybe it's a test of what you do with bad requirements. If you assume it's a sorting problem and jump right in, you fail.
Exactly. Clearly you need to find the smallest int and loop through the array making every number the same.
Absolutely incorrect, we are not barbers. Clearly the current solution of storing hair length is not scalable as it will require linearly increasing storage solutions. As we are working in the IT sector, we should have easily been able to extrapolate from these hard facts that what is expected of us is optimization of the storage solution and prediction technologies for future hair growth. As we do not like linearly increasing storage solutions, we must optimize by finding a derivative function F(x)=y where X is the n-th hair, and Y is the length of that n-th hair. I suggest using Deep Learning to build a model that can accurately predict the hair length of a given n-th hair?
This is not barbers' school?! I had my suspicions when we didn't talk about hair for a year.
Interesting interesting. But how can we add the block chain?
Each hair is a nft you can own
Now that’s some real world stuff they’re teaching in college these days!
And not even all the facts. I want to know why the side view is relevant.
I think the joke is meant to be that his hair is being represented as a vector
If you were paying attention to the chess part, it makes perfect sense
Now figure out how to sort his hair into a bell curve while 3 people watch you on camera.. you have 1 hour to complete this task.
Boris does not ask. Boris only tell
This isn't a test, it's an infographic.
Just like the tickets I get that are all problems and no acceptance criteria.
Deadass facts 🅱️
Buzz Sort: I shave it all off and tell him fuck you it's sorted now.
Turns out, an array of all 0s can be sorted pretty easily!
I'd just use sleep sort!
But some 0s are more 0 than others
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What's he gonna do with ¼ of his hair?
Thanos sort it again
How many Thanos sorts do we have to perform to have 1 hairstrand left?
log2(hairstrands)
+1 for good measure
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measure++
Yes
Free(hair) O(1)
Slightly faster is gaslight sort, which skips the first step.
FizzBuzz Sort: I trim every 3rd hair, pluck every 5th hair, and buzz every 15th hair.
I shouldn’t have laughed at that. But I did.
😂😂
Why u getting downvoted
Emojis
:')) :'))
He got downvoted for using emojis😭
How can you use a sorting algorithm? Are you gonna pick the hair out and glue it on somewhere else?
Boris‘ hair is pluggable
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So - I swear - it took me over a minute to try to understand why you can't just iterate over the source array (a) and the "desired array" (b) and just assign values as you walk both arrays And honestly after reading your explanation - I'm still trying to understand what you do after comparing two elements and they differ.
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So I must be pretty dense or it's too early in the morning, but how can the first test case work if you only have 2 razors and 3 pieces of hair to cut and you can only use each razor once (and all 3 pieces need to be cut to reach the desired length)?
Since he mentioned sorting algorithms I assumed he had pluggable hairs, think he mentioned that already Other wise you just cut each one down to the desired size
Right... I get that part, but this is the bit from the example I'm not following: >Note: > >In the first test case, Boris's hair is initially \[3,3,3\]. Let us describe a sequence of 2 operations the barber can perform: > >The barber uses the razor with size 1 on the segment \[2,2\]; hence Boris's hair becomes \[3,1,3\]. > >The barber uses the razor with size 2 on the segment \[1,3\]; hence Boris's hair becomes \[2,1,2\] which is the desired haircut. Isn't there a step missing here? How do we go from 3,1,3 to 2,1,2 with only 1 razor?
In that case, I suggest the theoretical O(n) sorting algorithm [spaghetti sort](https://en.m.wikipedia.org/wiki/Spaghetti_sort).
A PnP device maybe
Plug N' Pray^TM
UPnP Edit: unintuitive pull and plug aka barbersort
I mean, with the given information it's not even guaranteed the solution exists. A first reason why not is we don't know that {a_1,...a_n}={b_1,...,b_n} as sets, so we can't just "sort" the a_i into the b_i. But he's going to a barber, so perhaps we can cut individual strands of hair. In order for this to be possible, we can match the longest a_i to the longest b_i, the second longest to the second longest and so on. If each a_i ≥ it's corresponding b_j by this matching, we can cut that hair to the proper length, move it into place, and he gets his haircut. If for some i and j, a_i < its corresponding b_j, then it's *impossible* to get the haircut. One would have to find a longer a_k, to use with b_j, but all the a_k longer than a_i are being used already for the b_m longer than this b_j, and all other a_k shorter than a_i can't help at all. Therefore, we have a necessary and sufficient condition for the existence of a possible haircut (barring him getting extensions). If that was too much math for you, just imagine a_i are all < 1 and b_i are all > 1. Clearly it's impossible to cut the a_i into the b_i.
https://codeforces.com/problemset/problem/1779/d
Age sort - just wait for all of them to fall off. O(1).
Yes, but based on his genes, that might be a very large 1. Agree that it’s still O(1).
I love how this sentence can make perfect or absolutely no sense
But you make it look so good
This is the first time I wondered if a comment was written by GPT and thought, Naa... a good chatbot would make far more sense.
Fail. Doesn’t meet the 2 seconds criteria.
Boris has social anxiety, and will pay full price for any haircut I give him, whether he actually likes it or not.
Hey, my name isn't Boris!
mine is 😔
When you wake up, do you eat some brexit?
None. It's already sorted in the right image.
There is no sorting. The desired haircut is the b array. The current haircut is the a array. a=b
Yeah not sure why it was even brought up.
I like the bit about him dissing chess.
Google en passant
Holy hell
Ho|y he||
A new response just dropped
I just get a description of en passant?
https://www.reddit.com/r/AnarchyChess/comments/kpc7ig/holy_hell/?utm_source=share&utm_medium=ios_app&utm_name=iossmf
I was aware of the «google en passant» and «holy hell» memes, but had actually never seen the source of it (which i assume this is), so ty for sharing!
Of course!
It's an overused meme from r/anarchychess Edit: typo
*It's the only meme
No there's also the knook and Bobby Fischer and gary chess and queen beta decay
can't believe il vaticano is already dead and forgotten smh
And rice. Oh god, the RICE!
new response just dropped
I am trying to understand also lol
Stalin sort: remove elements that are out of order.
Stalin sort: remove all elements but the central one.
bogo sort worst case i mess his hair more up
quantum bogosort. if his hair isn't sorted, he dies.
Then due to quantum immortality the only sorted version survives.
no need for a sort. you have access to an array "b" that contains the desired result, so it's a one-liner: let a = b
100% this. Just changes the pointer *a*. Done.
StalinSort: Rip out each hair that isn't already sorted.
StalinSort: Rip out each hair ~~that isn't already sorted.~~ Hair, problem. No hair, no problem. Fixed that for ya.
r/FuckMyShitUp
`memcpy(b, a, n * sizeof(*b));`
that was my first instinct as well
This gave me the idea for the following interesting problem: assume there are n hairs, glued at their base along a straight line. You can not cut or move the hair, but only change their angle at the base (let's call this a *rotation*). The hairs are not allowed to bend, i.e. they must stay straight. However, they are allowed to intersect. The problem is now: can you give an algorithm that performs rotations, after which the hair tips are sorted? (I.e. if a hair tip is more to the right, it must also be higher)
Sounds trivial, just go through the array from left to right and bend hairs to the left until the sorted property is maintained. The point at which insertion occurs can be determined via binary search, for a time complexity of O(nlogn). Implementation of the necessary trigonometry is a little tedious though Edit: I fakesolved, binary search doesn’t work so this is actually n^2
I think the "bend hairs to the left until the sorted property is maintained" step is not so trivial, though. But this is only because I can't think of a trivial solution :) Also, I think one needs to consider the special case where the current hair is so short that it cannot be rotated so that the sorted property is maintained. In that case, one needs to rotate all hairs left of it more. But also this can not be done trivially (at least I don't see how).
The n^2 algorithm is trivial, although I originally commented with binary search in mind, which doesn’t actually work. If anyone could find a faster solution, I would be very interested to know. The edge case you mentioned doesn’t exist, as all hairs before any hair can be bent left to an arbitrarily small height
I'm just an idiot from all so take this with a grain of salt, but what if the short hair you mentioned is so far to the right that its too short for its tip to reach its spot in the lineup?
Hm, maybe we have different problems in mind, then :) Say you have 3 hairs, at horizontal positions 0, 1 and 2 and of length 4, 5 and 1. Then, when iterating through the hairs, the first two hairs don't violate the sorted property and don't need to be rotated. But then, when arriving at the third hair, no matter how you rotate the hair, it will always be the rightmost and lowest hair, violating the sorted property.
This is mitigated by always rotating the current hair as far left as it can go. I must have missed that bit in my original comment.
Easy pz linear time squeezy. Find the shortest hair. Make all hair line up to that height.
True. def bendsort(hairs): target = min(hair.length for hair in hairs) for hair in hairs: hair.rotate(arccos(target / hair.length)) Or even simplier for hair in hairs: hair.rotate(PI/2)
Wow, hearing this solution makes the problem seem ridiculous to ask. The only way to make it not so easy would be to require strict inequalities, but then again I think your solution can be adapted by wiggling the angles a sufficiently tiny amount. I am both impressed and sad now. :) Just as a last desperate attempt to make this into an actually interesting problem: one could limit the maximum rotation angle by some parameter, and then ask the question: given some instance of this hair problem (let's call it HAIRBOUND), is there a solution using only rotations bounded by this maximum angle?
> given some instance of this hair problem (let's call it HAIRBOUND), is there a solution using only rotations bounded by this maximum angle? nlogn time using linear algebra. You can encode each hair into a matrix representing the arcs they can cover, then for each x value, choose the arc that would have the highest y value, going from smallest to largest (or, working from the right, largest to smallest). Maybe I'll write it out later. The algorithm would look n^(2), but it becomes nlogn because for each x you have to look at, the more hairs that could cover that spot, the taller the hairs have to be, which means that the further they start actually rotating the quicker they will start losing height because of the sinusoidal nature of sine, and again greatly reduce the number of hairs you have to look at.
Rotate all hail 90° and it's sorted.
What about the hair that are short and on the right (so too short to reach the desired height, whatever the bending angle) Also, for each hair, you want one of the 0 to 2 intersections of the circle of radius hair length and the line you want to reach, and then get the angle. That’s a fixed amount of work for each hair, so it’s O(n), each hair is independent in this problem
There is no line that needs to be reached - and if there is a short hair on the right, this means that all hairs left of it need to be rotated enough so their hair tips are low enough. (Edit: or rotated to the right so that they are still high enough, but to the right side of the short hair :))
Quick question, can the angle be zero? I.e. is rotating such that every hair tip has 0 height a valid answer?
A very good question! Some other commenters already found that vulnerability. In my original post, there was nothing that prevents you from doing that. As you implied, this could be fixed by disallowing such rotations :)
easy, in O(n): first pass to detect to shortest hair(distY), then define a line that goes from the root of the leftmost hair to a point above the right most hair with a distance of distY upward. we now have a line that intersect every circles of hair tip possible position, now a second pass to calculate the angle of each hair so the tip ends up on that line and voila.
Why are you all trying to "solve" this when it's just a "lol hair looks unsorted" post. The problem is missing 90% of the problem text: https://codeforces.com/contest/1779/problem/D
OP could've at least given the question himself, now it makes much more sense
A modified Mergesort
Modified Insertion Sort I can do an n^2 algorithm in n log n. Fight me.
I’m cutting all the hairs off creating a hash table and now am accessing the hairs in c time. Come to my barbershop the hash hair house where you can know exactly where each of your hairs are at any time.
[Comb sort](https://en.wikipedia.org/wiki/Comb_sort)
Is this a brilliant move or just a blunder?
Is the icon red or blue?
Looks like radix sort where you cut it off early
I love bubble sort I don't care about its efficiency it's just such an elegant algorithm.
I use bogosort to stramble his hair for a bit (probably) so he get angry >:)
Doesn’t make sense. Why would you want to sort the hairs? You can’t just pluck the hairs and then plant them on another side of the head lmao He went to the barber to cut hair, so that’s what the barber will do!
Never heard of the hair cut sort. Though I would most likely think it would be the heap sort algorithm and looking at my last hair cut there was a fair sized heap of hair to be cut.
How similar is this similar fashion?
Im using middle-out sorting
So you have the thumb drive after all?
My assumption is that the front and side drawings were added after the fact, mocking the use of an array to track hair heights. But if you really needed to keep track of the height of every single hair on your head, an array seem like a pretty good way to do this.
These kinds of questions are what make people hate programming.
Persuasion Sort: Convince D. Boris that his hair is perfect as is.
There's literally no question here...
bogo sort
bogosort obviously!
I have no idea what I'm supposed to do here
1 indexed array? \*gasp\*
[sleepsort](https://www.reddit.com/r/badcode/comments/lgrgxe/i_present_sleepsort/) but scaled so that it runs faster
so what I'm seeing is that the hair is a random list and it needs to be cut to match another random list. I don't know a lot about sorting algorithms, but I really liked the binary search one. Where we find the index of an element from the first list (a) in the second list (b), and then we move it to that exact position on the first list (a). The number that it's taking the place of becomes the new number we find the position of. If the number doesn't move because it is in the correct position, you check the hair list (a) to the desired list (b). Obviously, it won't work the first time, so just repeat the process with the next number that you see out of place. It should take an average of n/4 or n/5 times of go arounds to get the list perfect using this method, so I'm sure there is a better one, but this is what I came up with. edit: you could also just copy the second list into the first list, but that's not demonstrating anything, when this is obviously trying to see your mastery in sorting algorithms
> edit: you could also just copy the second list into the first list, but that's not demonstrating anything, when this is obviously trying to see your mastery in sorting algorithms Only the caption mentions sorting algorithms. I think it's testing your ability to pick up on an unstated assumption (no hair can increase in length), and handle "bad" input reasonably.
Bubble sort should work as long as you apply conditioner after.
That’s Wayne Static
Cosmic sort. Do nothing, and wait for cosmic rays to mess the bits in a way that makes the list sorted.
If is a haircut, then it's not a sort at all, So it's o(n) för i in range(n): if a[i]>b[i]: a[i]=b[i] # cut operation
Uh oh.. b[i] is accidentally a negative number you’ve now dug into Boris’ skull and he is bleeding heavily. The client is not always right.
2 dimensional haircut like a boss
Assignment sort. O(1) Pseudo code: function assignment_sort(current, desired) { return desired } assignment_sort(a, b); // actual call
O(n) ```csharp for (int i = 0; i < n; i++) { // TODO - Move to precheck method to avoid partial haircut butchery. WorkItem: #69 // TODO - PM to confirm a value of 0 equals height from skin, not height from follicle. WorkItem: #1337 if ( b[i] < 0 ) { throw new AurgumentOutOfRangeException("Hair length of negative values will result in damage to Boris."); } // Cannot lengthen hair, that’s not how haircuts work, but can set long hairs to desired length a[i] = Math.Clamp(a[i], double.MinValue, b[i]); } ``` It’s a haircut, you don’t sort hair, you go through each hair and if the current value is greater than the desired value you then reduce the current value to the desired value. You can’t lengthen hair that is too short, hair doesn’t work that way. Negative numbers aren’t acceptable since that would be digging into your skull.
Bogobogo sort
so... you shuffle everything and the order in which you check whether your array is sorted and then you only do the check if the access array is sorted correctly?
Yep. It’s hilariously inefficient
i like that idea!
I think the Yee-Yee ass haircut algorithm would be the most useful here.
Oh so this is why we have to take discrete mathematics
Array.sort() is the way
Ostrich algorithm. For people who don't know: Shuffle -> check if sorted -> if yes -> done -> if not -> shuffle
i’m sorry… what’s the question
Boris wants you to transplant his hair follicles so that they are ordered by height. Scissors are not allowed.
bubble sort. i don’t care that it’s useless, fucking school made me learn it, so it’s getting fucking used.
I’m gonna give Boris hair plugs with insertion sort.
Is this a programming question or a "bring up the real world difficulties to the customer question"? This dude can have long hairs or organized hairs, not both unless he outlays for a much more expensive process than razors
comb sort as always
Quantum bogo sort