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I'm going through the question below.
The sequence [0, 1, ..., N] has been jumbled, and the only clue you have for its order is an array representing whether each number is larger or smaller than the last. Given this information, reconstruct an array that is consistent with it.
For example, given [None, +, +, -, +], you could return [1, 2, 3, 0, 4].
I went through the solution on this post but still unable to understand it as to why this solution works. I don't think I would be able to come up with the solution if I had this in front of me during an interview. Can anyone explain the intuition behind it? Thanks in advance!
This answer tries to give a general strategy to find an algorithm to tackle this type of problems. It is not trying to prove why the given solution is correct, but lying out a route towards such a solution.
A tried and tested way to tackle this kind of problem (actually a wide range of problems), is to start with small examples and work your way up. This works for puzzles, but even so for problems encountered in reality.
First, note that the question is formulated deliberately to not point you in the right direction too easily. It makes you think there is some magic involved. How can you reconstruct a list of N numbers given only the list of plusses and minuses?
Well, you can't. For 10 numbers, there are 10! = 3628800 possible permutations. And there are only 2⁹ = 512 possible lists of signs. It's a very huge difference. Most original lists will be completely different after reconstruction.
Here's an overview of how to approach the problem:
Start with very simple examples
Try to work your way up, adding a bit of complexity
If you see something that seems a dead end, try increasing complexity in another way; don't spend too much time with situations where you don't see progress
While exploring alternatives, revisit old dead ends, as you might have gained new insights
Try whether recursion could work:
given a solution for N, can we easily construct a solution for N+1?
or even better: given a solution for N, can we easily construct a solution for 2N?
Given a recursive solution, can it be converted to an iterative solution?
Does the algorithm do some repetitive work that can be postponed to the end?
....
So, let's start simple (writing 0 for the None at the start):
very short lists are easy to guess:
'0++' → 0 1 2 → clearly only one solution
'0--' → 2 1 0 → only one solution
'0-+' → 1 0 2 or 2 0 1 → hey, there is no unique outcome, though the question only asks for one of the possible outcomes
lists with only plusses:
'0++++++' → 0 1 2 3 4 5 6 → only possibility
lists with only minuses:
'0-------'→ 7 6 5 4 3 2 1 0 → only possibility
lists with one minus, the rest plusses:
'0-++++' → 1 0 2 3 4 5 or 5 0 1 2 3 4 or ...
'0+-+++' → 0 2 1 3 4 5 or 5 0 1 2 3 4 or ...
→ no very obvious pattern seem to emerge
maybe some recursion could help?
given a solution for N, appending one sign more?
appending a plus is easy: just repeat the solution and append the largest plus 1
appending a minus, after some thought: increase all the numbers by 1 and append a zero
→ hey, we have a working solution, but maybe not the most efficient one
the algorithm just appends to an existing list, no need to really write it recursively (although the idea is expressed recursively)
appending a plus can be improved, by storing the largest number in a variable so it doesn't need to be searched at every step; no further improvements seem necessary
appending a minus is more troublesome: the list needs to be traversed with each append
what if instead of appending a zero, we append -1, and do the adding at the end?
this clearly works when there is only one minus
when two minus signs are encountered, the first time append -1, the second time -2
→ hey, this works for any number of minuses encountered, just store its counter in a variable and sum with it at the end of the algorithm
This is in bird's eye view one possible route towards coming up with a solution. Many routes lead to Rome. Introducing negative numbers might seem tricky, but it is a logical conclusion after contemplating the recursive algorithm for a while.
It works because all changes are sequential, either adding one or subtracting one, starting both the increasing and the decreasing sequences from the same place. That guarantees we have a sequential list overall. For example, given the arbitrary
[None, +, -, +, +, -]
turned vertically for convenience, we can see
None 0
+ 1
- -1
+ 2
+ 3
- -2
Now just shift them up by two (to account for -2):
2 3 1 4 5 0
+ - + + -
Let's look at first to a solution which (I think) is easier to understand, formalize and demonstrate for correctness (but I will only explain it and not demonstrate in a formal way):
We name A[0..N] our input array (where A[k] is None if k = 0 and is + or - otherwise) and B[0..N] our output array (where B[k] is in the range [0, N] and all values are unique)
At first we see that our problem (find B such that B[k] > B[k-1] if A[k] == + and B[k] < B[k-1] if A[k] == -) is only a special case of another problem:
Find B such that B[k] == max(B[0..k]) if A[k] == + and B[k] == min(B[0..k]) if A[k] == -.
Which generalize from "A value must larger or smaller than the last" to "A value must be larger or smaller than everyone before it"
So a solution to this problem is a solution to the original one as well.
Now how do we approach this problem?
A greedy solution will be sufficient, indeed is easy to demonstrate that the value associated with the last + will be the biggest number in absolute (which is N), the one associated with the second last + will be the second biggest number in absolute (which is N-1) ecc...
And in the same time the value associated with the last - will be the smallest number in absolute (which is 0), the one associated with the second last - will be the second smallest (which is 1) ecc...
So we can start filling B from right to left remembering how many + we have seen (let's call this value X), how many - we have seen (let's call this value Y) and looking at what is the current symbol, if it is a + in B we put N-X and we increase X by 1 and if it is a - in B we put 0+Y and we increase Y by 1.
In the end we'll need to fill B[0] with the only remaining value which is equal to Y+1 and to N-X-1.
An interesting property of this solution is that if we look to only the values associated with a - they will be all the values from 0 to Y (where in this case Y is the total number of -) sorted in reverse order; if we look to only the values associated with a + they will be all the values from N-X to N (where in this case X is the total number of +) sorted and if we look at B[0] it will always be Y+1 and N-X-1 (which are equal).
So the - will have all the values strictly smaller than B[0] and reverse sorted and the + will have all the values strictly bigger than B[0] and sorted.
This property is the key to understand why the solution proposed here works:
It consider B[0] equals to 0 and than it fills B following the property, this isn't a solution because the values are not in the range [0, N], but it is possible with a simple translation to move the range and arriving to [0, N]
The idea is to produce a permutation of [0,1...N] which will follow the pattern of [+,-...]. There are many permutations which will be applicable, it isn't a single one. For instance, look the the example provided:
[None, +, +, -, +], you could return [1, 2, 3, 0, 4].
But you also could have returned other solutions, just as valid: [2,3,4,0,1], [0,3,4,1,2] are also solutions. The only concern is that you need to have the first number having at least two numbers above it for positions [1],[2], and leave one number in the end which is lower then the one before and after it.
So the question isn't finding the one and only pattern which is scrambled, but to produce any permutation which will work with these rules.
This algorithm answers two questions for the next member of the list: get a number who’s both higher/lower from previous - and get a number who hasn’t been used yet. It takes a starting point number and essentially create two lists: an ascending list for the ‘+’ and a descending list for the ‘-‘. This way we guarantee that the next member is higher/lower than the previous one (because it’s in fact higher/lower than all previous members, a stricter condition than the one required) and for the same reason we know this number wasn’t used before.
So the intuition of the referenced algorithm is to start with a referenced number and work your way through. Let's assume we start from 0. The first place we put 0+1, which is 1. we keep 0 as our lowest, 1 as the highest.
l[0] h[1] list[1]
the next symbol is '+' so we take the highest number and raise it by one to 2, and update both the list with a new member and the highest number.
l[0] h[2] list [1,2]
The next symbol is '+' again, and so:
l[0] h[3] list [1,2,3]
The next symbol is '-' and so we have to put in our 0. Note that if the next symbol will be - we will have to stop, since we have no lower to produce.
l[0] h[3] list [1,2,3,0]
Luckily for us, we've chosen well and the last symbol is '+', so we can put our 4 and call is a day.
l[0] h[4] list [1,2,3,0,4]
This is not necessarily the smartest solution, as it can never know if the original number will solve the sequence, and always progresses by 1. That means that for some patterns [+,-...] it will not be able to find a solution. But for the pattern provided it works well with 0 as the initial starting point. If we chose the number 1 is would also work and produce [2,3,4,0,1], but for 2 and above it will fail. It will never produce the solution [0,3,4,1,2].
I hope this helps understanding the approach.
This is not an explanation for the question put forward by OP.
Just want to share a possible approach.
Given: N = 7
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Go from 0 to N
[1] fill all '-' starting from right going left.
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Answer: 2 1 0
[2] fill all the vacant places i.e [X & +] starting from left going right.
Index: 0 1 2 3 4 5 6 7
Pattern: X + - + - + - + //X = None
Answer: 3 4 5 6 7
Final:
Pattern: X + - + - + - + //X = None
Answer: 3 4 2 5 1 6 0 7
My answer definitely is too late for your problem but if you need a simple proof, you probably would like to read it:
+min_last or min_so_far is a decreasing value starting from 0.
+max_last or max_so_far is an increasing value starting from 0.
In the input, each value is either "+" or "-" and for each increase the value of max_so_far or decrease the value of min_so_far by one respectively, excluding the first one which is None. So, abs(min_so_far, max_so_far) is exactly equal to N, right? But because you need the range [0, n] but max_so_far and min_so_far now are equal to the number of "+"s and "-"s with the intersection part with the range [0, n] being [0, max_so_far], what you need to do is to pad it the value equal to min_so_far for the final solution (because min_so_far <= 0 so you need to take each value of the current answer to subtract by min_so_far or add by abs(min_so_far)).
I have the following question:
I am building a model when I first test for stationarity. Then I have an if loop, saying:
if p>0.05:
x=y['boxcox']
else:
x=y['Normal']
If the pvalue is bigger than 0.05, then I do the boxcox transformation, if not, then I use my original values. This works.
I then have a large code, that is working.
However, in the end, I want to transform my values back.
Again with the if loop.
But how do I get the if loop started?
I first wanted to do:
if any (x==y['BoxCox']):
.....transform back
This works if I orginially have transformed my values, but not if I didn't, which makes sense, because the code does not know y['BoxCox'].
But how do I get the if loop initiated?
Thanks a lot!
If I understand your question correctly, you do not transform anything back, rather you remember the initial state. "Transforming back" sounds like a potential source of bugs. What if you alter your transformation algorithm and forget to update the transforming back part?
here is a simplified example, to illustrate my understanding of your problem:
x = 4
if x > 2:
x = x + 1
else:
x = x + 100
print("Result = ", x)
print("Initial value was ???") // you cannot tell what was the initial x
You can simply do not touch the initial values, to be accessible at any time:
x = 4
if x > 2:
result = x + 1
else:
result = x + 100
print("Result = ", result)
print("Initial value = , x)
I'm trying to generate coordinates in a mulidimensional array.
the range for each digit in the coords is -1 to 1. <=> seems like the way to go comparing two random numbers. I'm having trouble because randomizing it takes forever, coords duplicate and sometimes don't fill all the way through. I've tried uniq! which only causes the initialization to run forever while it tries to come up with the different iterations.
the coords look something like this. (-1, 0, 1, 0, 0)
5 digits give position. I could write them out but I'd like to generate the coords each time the program is initiated. The coords would then be assigned to a hash tied to a key. 1 - 242.
I could really use some advice.
edited to add code. It does start to iterate but it doesn't fill out properly. Short of just writing out an array with all possible combos and randomizing before merging it with the key. I can't figure out how.
room_range = (1..241)
room_num = [*room_range]
p room_num
$rand_loc_cords = []
def Randy(x)
srand(x)
y = (rand(100) + 1) * 1500
z = (rand(200) + 1) * 1000
return z <=> y
end
def rand_loc
until $rand_loc_cords.length == 243 do
x = Time.new.to_i
$rand_loc_cords.push([Randy(x), Randy(x), Randy(x), Randy(x), Randy(x)])
$rand_loc_cords.uniq!
p $rand_loc_cords
end
#p $rand_loc_cords
end
rand_loc
You are trying to get all possible permutations of -1, 0 and 1 with a length of 5 by sheer luck, which can take forever. There are 243 of them (3**5) indeed:
coords = [-1,0,1].repeated_permutation(5).to_a
Shuffle the array if the order should be randomized.
I have two sets x [xmin,xmax] and y [ymin, ymax] and I would like to execute a function going stepwise from the min to the max values of x and y. So I want to apply a function to the Cartesian product of x and y. I would then like to save each combination as a row to a CSV file. I've been trying for some time with a do loop, but got a bit stuck on how to create the list in the end. For instance:
for x: 1 thru 2 step 1 do
for y: 1 thru 2 step 1 do
print([x,y,find_root (exp(a*x) = y, a, 0, 1)])
I'd get the values of x and y and the function of all combinations, but I struggle to save it and export it to a CSV, because I don't know how to create the list with [[1,1,function(1,1)],[1,2,function(1,2)],[2,1,function(2,1)],[2,2,function(2,2)]], that I could export with write_data.
Alternatively, I'd like to use:
xlist:makelist(x,x,1,2,1);
ylist:makelist(y,y,1,2,1);
create_list([x,y,x^y],x,xlist,y,ylist);
In this case I don't know how to include the function in create list or how to use map.
How do I do the above or is there a better way?
About speeding up the construction of the 1 million item list, how about solving the equation just once and then substituting values of x and y? E.g.:
solve (exp(a*x) = y, a);
my_solution : rhs (first (%));
create_list ([x, y, ev (my_solution)], x, xlist, y, ylist);
Here ev evaluates my_solution with the current values of the variables it contains (namely x and y).
About writing a CSV file, try this:
write_data (my_list, "my_output_file", 'comma);
The task is to calculate a sum of natural numbers from 0 to M. I wrote the following code using SWI-Prolog:
my_sum(From, To, _) :- From > To, !.
my_sum(From, To, S) :-
From = 0,
Next is 1,
S is 1,
my_sum(Next, To, S).
my_sum(From, To, S) :-
From > 0,
Next is From + 1,
S is S + Next,
my_sum(Next, To, S).
But when I try to calculate:
my_sum(0,10,S), writeln(S).
I got False instead of correct number. What is going wrong with this example?
this is surely false for Next \= 0: S is S + Next. Another more fundamental problem is that you're doing the computation in 'reverse' order. That is, when From > To and the program stop, you don't 'get back' the result. Then you should add an accumulator (another parameter, to be propagated to all recursive calls) and unify it with the partial sum at that last step...
Anyway, should be simpler:
my_sum(From, To, S) :-
From < To,
Next is From + 1,
my_sum(Next, To, T),
S is T + From.
my_sum(N, N, N).
| ?- my_sum(2, 4, N).
N = 9
I'd write the predicate along these lines, using a worker predicate with an additional accumulator:
sum(X,Y,Z) :-
integer(X) ,
integer(Y) ,
sum(X,Y,0,Z)
.
sum(X,X,T,Z) :- Z is T+X .
sum(X,Y,T,Z) :- X < Y , X1 is X+1 , T1 is T+X , sum(X1,Y,T1,Z) .
sum(X,Y,T,Z) :- X > Y , X1 is X-1 , T1 is T+X , sum(X1,Y,T1,Z) .
This implementation is simple, bi-directional, meaning that sum(1,3,X) and sum(3,1,X) both yield 6 as a result (1+2+3), and tail recursive, meaning that it should be able to handle a range of any size without a stack overflow.
As it happens, there's a purely analytic solution as well:
sum(N, Sum) :- Sum is N * (N+1) / 2.
In use:
?- sum(100, N).
N = 5050.
You used the loop tag so this probably isn't an answer you desire, but it's good to prefer this kind of solution when one exists.
predicates
sum(integer,integer)
clauses
sum(0,0).
sum(N,R):-
N1=N-1,
sum(N1,R1),
R=R1+N.