I am doing some stuff on leetcode and came up with solution it works fine but some cases.
Here is the problem itself:
But in case like this it doesn't:
It doesn't make sense how can I rotate elements if k is bigger than length of array.
If you have any idea how to improve this solution I would be grateful
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
if len(nums) > k:
self.swap(nums, 0, len(nums)-1)
self.swap(nums, 0,k-1)
self.swap(nums, k, len(nums)-1)
def swap(self, nums, start, end):
while start < end:
nums[start], nums[end] = nums[end], nums[start]
start+=1
end-=1
In order to understand why this doesn't work for the cases where k is larger than the array length, let me try to explain some of the logic behind rotating by such values of k.
The modulo operator, % will be useful. For example, if an array is 5 long, and you want to rotate by 5, you end up with the same array. So technically, you'd optimally want to rotate by 0. This is where the % operator comes into play. 5 % 5 = 0. If we want to rotate an array length 5 by 7 spots, we would end up with the same thing as rotating the array by 2, and it turns out that 7 % 5 = 2. Do you see where I am going with this?
This also holds true if the value of k is less than the length of the array. Say we want to rotate an array length 5 by 3, we do 3 % 5 = 3.
So for any rotation of amount k and array length L, optimization rotation amount n is equivalent to n = k % L.
You should modify your code at the beginning of your rotate method to adjust the rotation amount:
k = k % L
and use this value to rotate the correct amount.
The fastest and cleanest solution by far and large is:
def rotate_right(items, shift):
shift = -shift % len(items)
return items[shift:] + items[:shift]
ll = [i + 1 for i in range(7)]
# [1, 2, 3, 4, 5, 6, 7]
rotate_right(ll, 3)
# [5, 6, 7, 1, 2, 3, 4]
rotate_right([1, 2], 3)
# [2, 1]
of course, short of using numpy.roll() or itertools.cycle().
Related
I'm looking for solution to my problem. Say I have a number X, now I want to generate 20 random numbers whose sum would equal to X, but I want those random numbers to have enthropy in them. So for example, if X = 50, the algorithm should generate
3
11
0
6
19
7
etc. The sum of given numbres should equal to 50.
Is there any simple way to do that?
Thanks
Simple way:
Generate random number between 1 and X : say R1;
subtract R1 from X, now generate a random number between 1 and (X - R1) : say R2. Repeat the process until all Ri add to X : i.e. (X-Rn) is zero. Note: each consecutive number Ri will be smaller then the first. If you want the final sequence to look more random, simply permute the resulting Ri numbers. I.e. if you generate for X=50, an array like: 22,11,9,5,2,1 - permute it to get something like 9,22,2,11,1,5. You can also put a limit to how large any random number can be.
One fairly straightforward way to get k random values that sum to N is to create an array of size k+1, add values 0 and N, and fill the rest of the array with k-1 randomly generated values between 1 and N-1. Then sort the array and take the differences between successive pairs.
Here's an implementation in Ruby:
def sum_k_values_to_n(k = 20, n = 50)
a = Array.new(k + 1) { 1 + rand(n - 1) }
a[0] = 0
a[-1] = n
a.sort!
(1..(a.length - 1)).collect { |i| a[i] - a[i-1] }
end
p sum_k_values_to_n(3, 10) # produces, e.g., [2, 3, 5]
p sum_k_values_to_n # produces, e.g., [5, 2, 3, 1, 6, 0, 4, 4, 5, 0, 2, 1, 0, 5, 7, 2, 1, 1, 0, 1]
Given the list of numbers
1 15 2 5 10
I need to obtain
1 2 5 10 15
The only operation I can do is "move the number X at position Y".
In the above example I only need to do "move the number 15 at position 5".
I would like to minimize the number of operations but I can't find/remember a classical algorithm for that, given the operation available.
Some background :
I'm interacting with an API for a kanban-like service.
I have about 600 cards and some actions on our bug-tracker can imply a reordering of these 600 cards in the kanban (multiple cards can move at the same time if the priority of a project is changed)
I can do it in 600 calls to the API but I'm trying to reduce that number as much as possible.
Lemma: The minimum number of (delete element, insert element) pairs you can perform to sort a list L (in increasing order) is:
Smin(L) = |L| - |LIC(L)|
Where LIC(L) is the Longest Increasing Subsequence.
Thus, you have to:
Establish the LIC of your list.
Remove the elements not in it and insert them back at the appropriate position (using binary search).
Proof:
By induction.
For a list of size 1, the longest increasing subsequence is of length... 1! The list is already sorted so the number of (del,ins) pairs required is
|L| - |LIC(L)| = 1 - 1 = 0
Now let Ln be a list of length n, 1 ≤ n. Let Ln+1 be the list obtained by adding an element en+1 to the left of Ln.
This element may or may not influence the Longest Increasing Subsequence. Let's try to see how...
Let in,1 and in,2 be the two first elements of LIC(Ln) (*):
If en+1 > in,2, then LIC(Ln+1) = LIC(Ln)
If en+1 ≤ in,1, then LIC(Ln+1) = en+1 || LIC(Ln)
Else, LIC(Ln+1) = LIC(Ln) - in,1 + en+1. We keep the LIC with the highest first element. This is done by removing in,1 from the LIC and replacing it with en+1.
In the first case, we delete en+1, we thus get to sort Ln. By the induction hypothesis, this require n (deletion, insertion) pairs. We then have to insert en+1 at the appropriate position. Thus:
S(Ln+1)min = 1 + S(Ln)min
S(Ln+1)min = 1 + n - |LIC(Ln)|
S(Ln+1)min = |Ln+1| - |LIC(Ln+1|
In the second case, we ignore en+1. We begin by deleting elements not in LIC(Ln). These elements have to be inserted again! There are
S(Ln)min = |Ln| - |LIC(Ln)|
such elements.
Now, we just have to take care and insert them in the right order (relatively to en+1). In the end, it requires:
S(Ln+1)min = |Ln| - |LIC(Ln)|
S(Ln+1)min = |Ln| + 1 - (|LIC(Ln)| + 1)
Since we have |LIC(Ln+1)| = |LIC(Ln)| + 1 and |Ln+1| = |Ln| + 1, we have in the end:
S(Ln+1)min = |Ln+1| - |LIC(Ln+1)|
The last case can be proved by considering the list L'n obtained by removing in,1 from Ln+1. In that case LIC(L'n) = LIC(Ln+1) and thus:
|LIC(L'n)| = |LIC(Ln)| (1)
From there, we can sort L'n (which takes |L'n| - |LIC(L'n| by the induction hypothesis. The previous equality (1) leads to the result.
(*): If LIC(Ln) < 2, then in,2 doesn't exist. Just ignore the comparisons with it. In that case, only case 2 and case 3 apply... The result is still valid
One possible solution is to find the longest increasing subsequence and move only elements that aren't inside it.
I can't prove it's optimal, but it is easy to prove it is correct and better than N swaps.
Here is a proof-of-concept in Python 2. I implemented it as a O(n2) algorithm, but I'm pretty sure it can be reduced to O(n log n).
from operator import itemgetter
def LIS(V):
T = [1]*(len(V))
P = [-1]*(len(V))
for i, v in enumerate(V):
for j in xrange(i-1, -1, -1):
if T[j]+1 > T[i] and V[j] <= V[i]:
T[i] = T[j] + 1
P[i] = j
i, _ = max(enumerate(T), key=itemgetter(1))
while i != -1:
yield i
i = P[i]
def complement(L, n):
for a, b in zip(L, L[1:]+[n]):
for i in range(a+1, b):
yield i
def find_moves(V):
n = len(V)
L = list(LIS(V))[::-1]
SV = sorted(range(n), key=lambda i:V[i])
moves = [(x, SV.index(x)) for x in complement(L, n)]
while len(moves):
a, b = moves.pop()
yield a, b
moves = [(x-(x>a)+(x>b), y) for x, y in moves]
def make_and_print_moves(V):
print 'Initial array:', V
for a, b in find_moves(V):
x = V.pop(a)
V.insert(b, x)
print 'Move {} to {}. Result: {}'.format(a, b, V)
print '***'
make_and_print_moves([1, 15, 2, 5, 10])
make_and_print_moves([4, 3, 2, 1])
make_and_print_moves([1, 2, 4, 3])
It outputs something like:
Initial array: [1, 15, 2, 5, 10]
Move 1 to 4. Result: [1, 2, 5, 10, 15]
***
Initial array: [4, 3, 2, 1]
Move 3 to 0. Result: [1, 4, 3, 2]
Move 3 to 1. Result: [1, 2, 4, 3]
Move 3 to 2. Result: [1, 2, 3, 4]
***
Initial array: [1, 2, 4, 3]
Move 3 to 2. Result: [1, 2, 3, 4]
***
I have an array of non-negative values. I want to build an array of values who's sum is 20 so that they are proportional to the first array.
This would be an easy problem, except that I want the proportional array to sum to exactly
20, compensating for any rounding error.
For example, the array
input = [400, 400, 0, 0, 100, 50, 50]
would yield
output = [8, 8, 0, 0, 2, 1, 1]
sum(output) = 20
However, most cases are going to have a lot of rounding errors, like
input = [3, 3, 3, 3, 3, 3, 18]
naively yields
output = [1, 1, 1, 1, 1, 1, 10]
sum(output) = 16 (ouch)
Is there a good way to apportion the output array so that it adds up to 20 every time?
There's a very simple answer to this question: I've done it many times. After each assignment into the new array, you reduce the values you're working with as follows:
Call the first array A, and the new, proportional array B (which starts out empty).
Call the sum of A elements T
Call the desired sum S.
For each element of the array (i) do the following:
a. B[i] = round(A[i] / T * S). (rounding to nearest integer, penny or whatever is required)
b. T = T - A[i]
c. S = S - B[i]
That's it! Easy to implement in any programming language or in a spreadsheet.
The solution is optimal in that the resulting array's elements will never be more than 1 away from their ideal, non-rounded values. Let's demonstrate with your example:
T = 36, S = 20. B[1] = round(A[1] / T * S) = 2. (ideally, 1.666....)
T = 33, S = 18. B[2] = round(A[2] / T * S) = 2. (ideally, 1.666....)
T = 30, S = 16. B[3] = round(A[3] / T * S) = 2. (ideally, 1.666....)
T = 27, S = 14. B[4] = round(A[4] / T * S) = 2. (ideally, 1.666....)
T = 24, S = 12. B[5] = round(A[5] / T * S) = 2. (ideally, 1.666....)
T = 21, S = 10. B[6] = round(A[6] / T * S) = 1. (ideally, 1.666....)
T = 18, S = 9. B[7] = round(A[7] / T * S) = 9. (ideally, 10)
Notice that comparing every value in B with it's ideal value in parentheses, the difference is never more than 1.
It's also interesting to note that rearranging the elements in the array can result in different corresponding values in the resulting array. I've found that arranging the elements in ascending order is best, because it results in the smallest average percentage difference between actual and ideal.
Your problem is similar to a proportional representation where you want to share N seats (in your case 20) among parties proportionnaly to the votes they obtain, in your case [3, 3, 3, 3, 3, 3, 18]
There are several methods used in different countries to handle the rounding problem. My code below uses the Hagenbach-Bischoff quota method used in Switzerland, which basically allocates the seats remaining after an integer division by (N+1) to parties which have the highest remainder:
def proportional(nseats,votes):
"""assign n seats proportionaly to votes using Hagenbach-Bischoff quota
:param nseats: int number of seats to assign
:param votes: iterable of int or float weighting each party
:result: list of ints seats allocated to each party
"""
quota=sum(votes)/(1.+nseats) #force float
frac=[vote/quota for vote in votes]
res=[int(f) for f in frac]
n=nseats-sum(res) #number of seats remaining to allocate
if n==0: return res #done
if n<0: return [min(x,nseats) for x in res] # see siamii's comment
#give the remaining seats to the n parties with the largest remainder
remainders=[ai-bi for ai,bi in zip(frac,res)]
limit=sorted(remainders,reverse=True)[n-1]
#n parties with remainter larger than limit get an extra seat
for i,r in enumerate(remainders):
if r>=limit:
res[i]+=1
n-=1 # attempt to handle perfect equality
if n==0: return res #done
raise #should never happen
However this method doesn't always give the same number of seats to parties with perfect equality as in your case:
proportional(20,[3, 3, 3, 3, 3, 3, 18])
[2,2,2,2,1,1,10]
You have set 3 incompatible requirements. An integer-valued array proportional to [1,1,1] cannot be made to sum to exactly 20. You must choose to break one of the "sum to exactly 20", "proportional to input", and "integer values" requirements.
If you choose to break the requirement for integer values, then use floating point or rational numbers. If you choose to break the exact sum requirement, then you've already solved the problem. Choosing to break proportionality is a little trickier. One approach you might take is to figure out how far off your sum is, and then distribute corrections randomly through the output array. For example, if your input is:
[1, 1, 1]
then you could first make it sum as well as possible while still being proportional:
[7, 7, 7]
and since 20 - (7+7+7) = -1, choose one element to decrement at random:
[7, 6, 7]
If the error was 4, you would choose four elements to increment.
A naïve solution that doesn't perform well, but will provide the right result...
Write an iterator that given an array with eight integers (candidate) and the input array, output the index of the element that is farthest away from being proportional to the others (pseudocode):
function next_index(candidate, input)
// Calculate weights
for i in 1 .. 8
w[i] = candidate[i] / input[i]
end for
// find the smallest weight
min = 0
min_index = 0
for i in 1 .. 8
if w[i] < min then
min = w[i]
min_index = i
end if
end for
return min_index
end function
Then just do this
result = [0, 0, 0, 0, 0, 0, 0, 0]
result[next_index(result, input)]++ for 1 .. 20
If there is no optimal solution, it'll skew towards the beginning of the array.
Using the approach above, you can reduce the number of iterations by rounding down (as you did in your example) and then just use the approach above to add what has been left out due to rounding errors:
result = <<approach using rounding down>>
while sum(result) < 20
result[next_index(result, input)]++
So the answers and comments above were helpful... particularly the decreasing sum comment from #Frederik.
The solution I came up with takes advantage of the fact that for an input array v, sum(v_i * 20) is divisible by sum(v). So for each value in v, I mulitply by 20 and divide by the sum. I keep the quotient, and accumulate the remainder. Whenever the accumulator is greater than sum(v), I add one to the value. That way I'm guaranteed that all the remainders get rolled into the results.
Is that legible? Here's the implementation in Python:
def proportion(values, total):
# set up by getting the sum of the values and starting
# with an empty result list and accumulator
sum_values = sum(values)
new_values = []
acc = 0
for v in values:
# for each value, find quotient and remainder
q, r = divmod(v * total, sum_values)
if acc + r < sum_values:
# if the accumlator plus remainder is too small, just add and move on
acc += r
else:
# we've accumulated enough to go over sum(values), so add 1 to result
if acc > r:
# add to previous
new_values[-1] += 1
else:
# add to current
q += 1
acc -= sum_values - r
# save the new value
new_values.append(q)
# accumulator is guaranteed to be zero at the end
print new_values, sum_values, acc
return new_values
(I added an enhancement that if the accumulator > remainder, I increment the previous value instead of the current value)
Given an array like [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], I want to get a random value that takes into consideration the position.
I want the likelihood of 1 popping up to be way bigger than 10.
Is something like this possible?
For the sake of simplicity let's assume an array arr = [x, y, z] from which we will be sampling values. We'd like to see following relative frequencies of x, y and z:
frequencies = [5, 2, 1]
Preprocess these frequencies to calculate margins for our subsequent dice roll:
thresholds = frequencies.clone
1.upto(frequencies.count - 1).each { |i| thresholds[i] += thresholds[i - 1] }
Let's sum them up.
max = frequencies.reduce :+
Now choose a random number
roll = 1 + rand max
index = thresholds.find_index { |x| roll <= x }
Return arr[index] as a result. To sum up:
def sample arr, frequencies
# assert arr.count == frequencies.count
thresholds = frequencies.clone
1.upto(frequencies.count - 1).each { |i| thresholds[i] += thresholds[i - 1] }
max = frequencies.reduce :+
roll = 1 + rand(max)
index = thresholds.find_index { |x| roll <= x }
arr[index]
end
Let's see how it works.
data = 80_000.times.map { sample [:x, :y, :z], [5, 2, 1] }
A histogram for data shows that sample works as we've intended.
def coin_toss( arr )
arr.detect{ rand(2) == 0 } || arr.last
end
a = (1..10).to_a
10.times{ print coin_toss( a ), ' ' } #=> 1 1 1 9 1 5 4 1 1 3
This takes the first element of the array, flips a coin, returns the element and stops if the coinflip is 'tails'; the same with the next element otherwise. If it is 'heads' all the way, return the last element.
A simple way to implement this with an logarithmic probabilistic of being selected is to simulate coin flips. Generate a random integer 0 and 1, the index to that array to choose is the number of consecutive 1s you get. With this method, the chance of selecting 2 is 1/2 as likely as 1, 3 is 1/4th as likely, etc. You can vary the probability slightly say by generating random numbers between 0 and 5 and count the number of consecutive rounds above 1, which makes each number in the array 4/5th as likely to appear as the one before.
A better and more general way to solve this problem is to use the alias method. See the answer to this question for more information:
Data structure for loaded dice?
I have a program (a fractal) that draws lines in an interlaced order. Originally, given H lines to draw, it determines the number of frames N, and draws every Nth frame, then every N+1'th frame, etc.
For example, if H = 10 and N = 3, it draws them in order:
0, 3, 6, 9,
1, 4, 7,
2, 5, 8.
However I didn't like the way bands would gradually thicken, leaving large areas between undrawn for a long time. So the method was enhanced to recursively draw midpoint lines in each group instead of the immediately sebsequent lines, for example:
0, (32) # S (step size) = 32
8, (24) # S = 16
4, (12) # S = 8
2, 6, (10) # S = 4
1, 3, 5, 7, 9. # S = 2
(The numbers in parentheses are out of range and not drawn.) The algorithm's pretty simple:
Set S to a power of 2 greater than N*2, set F = 0.
While S > 1:
Draw frame F.
Set F = F + S.
If F >= H, then set S = S / 2; set F = S / 2.
When the odd numbered frames are drawn on the last step size, they are drawn in simple order just as an the initial (annoying) method. The same with every fourth frame, etc. It's not as bad because some intermediate frames have already been drawn.
But the same permutation could recursively be applied to the elements for each step size. In the example above, the last line would change to:
1, # the 0th element, S' = 16
9, # 4th, S' = 8
5, # 2nd, S' = 4
3, 7. # 1st and 3rd, S' = 2
The previous lines have too few elements for the recursion to take effect. But if N was large enough, some lines might require multiple levels of recursion. Any step size with 3 or more corresponding elements can be recursively permutated.
Question 1. Is there a common name for this permutation on N elements, that I could use to find additional material on it? I am also interested in any similar examples that may exist. I would be surprised if I'm the first person to want to do this.
Question 2. Are there some techniques I could use to compute it? I'm working in C but I'm more interested at the algorithm level at this stage; I'm happy to read code other language (within reason).
I have not yet tackled its implemention. I expect I will precompute the permutation first (contrary to the algorithm for the previous method, above). But I'm also interested if there is a simple way to get the next frame to draw without having to precomputing it, similar in complexity to the previous method.
It sounds as though you're trying to construct one-dimensional low-discrepancy sequences. Your permutation can be computed by reversing the binary representation of the index.
def rev(num_bits, i):
j = 0
for k in xrange(num_bits):
j = (j << 1) | (i & 1)
i >>= 1
return j
Example usage:
>>> [rev(4,i) for i in xrange(16)]
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
A variant that works on general n:
def rev(n, i):
j = 0
while n >= 2:
m = i & 1
if m:
j += (n + 1) >> 1
n = (n + 1 - m) >> 1
i >>= 1
return j
>>> [rev(10,i) for i in xrange(10)]
[0, 5, 3, 8, 2, 7, 4, 9, 1, 6]