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find x and y with the help of third number

Suppose that I have three binary numbers, we must Or the first number(x) and the second number(y) together and make the third number(z).
I have the third number, I also have some bits of the first and second number, now I need to get the number of possible states I can make for the first and second numbers with the lowest order.
for example
x = 0 _ (We only know the first bit and the first bit is unknown)
y = _ _ (We do not know any of its bits and we have to fill them in such a way that the condition of the problem is fulfilled)
z = 11
if we bitwise OR operation x and y we find z. The possible states for x , y are [(00,11),(01,10),(01,11)]
Note that the first bit of all states obtained for the number x is zero

What you have is essentially a product where at each position there are between 1 and 3 possibilities for the x and y bits in that position. You can exploit this by using itertools.product. First construct, for each position, a list of the possibilities for that position, and then form their product.
In the following code I am representing numbers as lists of bits, with the blanks being represented by -1:
from itertools import product
#states will be represented by int lists consisting of
#0,1, or -1. -1 will be the flag that the bit is undetermined
def find_states(x,y,z):
bit_pairs = []
for i,j,k in zip(x,y,z):
if k == 0:
if max(i,j) == 1: return [] #no valid states
bit_pairs.append([(0,0)])
else:
if 0 <= min(i,j):
if i == j == 0:
return [] #no valid states
else:
bit_pairs.append([(i,j)])
else: #a -1 is present
match (i,j):
case (-1,-1):
bit_pairs.append([(0,1),(1,0),(1,1)])
case (-1,0):
bit_pairs.append([(1,0)])
case (-1,1):
bit_pairs.append([(0,1),(1,1)])
case (0,-1):
bit_pairs.append([(0,1)])
case (1,-1):
bit_pairs.append([(1,0),(1,1)])
states = []
for p in product(*bit_pairs):
#first elements of the pairs form x, the second ones form y
states.append(([i for i,_ in p],[j for _,j in p]))
return states
For example:
for p in find_states([1, -1, -1, 0], [-1, 1, 0, 0], [1, 1, 0, 0]): print(p)
Output (each row is an x,y pair):
([1, 0, 0, 0], [0, 1, 0, 0])
([1, 1, 0, 0], [0, 1, 0, 0])
([1, 0, 0, 0], [1, 1, 0, 0])
([1, 1, 0, 0], [1, 1, 0, 0])

Move zeroes to end of list

I am working on moving all zeroes to end of list. .. is this approach bad and computationally expensive?
a = [1, 2, 0, 0, 0, 3, 6]
temp = []
zeros = []
for i in range(len(a)):
if a[i] !=0:
temp.append(a[i])
else:
zeros.append(a[i])
print(temp+zeros)
My Program works but not sure if this is a good approach?

A sorted solution that avoids changing the order of the other elements is:
from operator import not_
sorted(a, key=not_)
or without an import:
sorted(a, key=lambda x: not x) # Or x == 0 for specific numeric test
By making the key a simple boolean, sorted splits it into things that are truthy followed by things that are falsy, and since it's a stable sort, the order of things within each category is the same as the original input.

This looks like a list. Could you just use sort?
a = [1, 2, 0, 0, 0, 3, 6]
a.sort(reverse=True)
a
[6, 3, 2, 1, 0, 0, 0]

To move all the zeroes to the end of the list while preserving the order of all the elements in one traversal, we can keep the count of all the non-zero elements from the beginning and swap it with the next element when a non-zero element is encountered after zeroes.
This can be explained as:
arr = [18, 0, 4, 0, 0, 6]
count = 0
for i in range(len(arr):
if arr[i] != 0:
arr[i], arr[count] = arr[count], arr[i]
count += 1
How the loop works:
when i = 0, arr[i] will be 18, so according to the code it will swap with itself, which doesn't make a difference, and count will be incremented by one. When i=1, it will have no affect as till now the list we have traversed is what we want(zero in the end). When i=4, arr[i]= 4 and arr[count(1)]= 0, so we swap them leaving the list as[18, 4, 0, 0, 0, 6] and count becomes 2 signifying two non-zero elements in the beginning. And then the loop continues.

You can try my solution if you like
class Solution:
def moveZeroes(self, nums: List[int]) -> None:
for num in nums:
if num == 0:
nums.remove(num)
nums.append(num)
I have tried this code in leetcode & my submission got accepted using above code.

Nothing wrong with your approach, really depends on how you want to store the resulting values. Here is a way to do it using list.extend() and list.count() that preserves order of the non-zero elements and results in a single list.
a = [1, 2, 0, 0, 0, 3, 6]
result = [n for n in a if n != 0]
result.extend([0] * a.count(0))
print(result)
# [1, 2, 3, 6, 0, 0, 0]

You can try this
a = [1, 2, 0, 0, 0, 3, 6]
x=[i for i in a if i!=0]
y=[i for i in a if i==0]
x.extend(y)
print(x)

There's nothing wrong with your solution, and you should always pick a solution you understand over a 'clever' one you don't if you have to look after it.
Here's an alternative which never makes a new list and only passes through the list once. It will also preserve the order of the items. If that's not necessary the reverse sort solution is miles better.
def zeros_to_the_back(values):
zeros = 0
for value in values:
if value == 0:
zeros += 1
else:
yield value
yield from (0 for _ in range(zeros))
print(list(
zeros_to_the_back([1, 2, 0, 0, 0, 3, 6])
))
# [1, 2, 3, 6, 0, 0, 0]
This works using a generator which spits out answers one at a time. If we spot a good value we return it immediately, otherwise we just count the zeros and then return a bunch of them at the end.
yield from is Python 3 specific, so if you are using 2, just can replace this with a loop yielding zero over and over.

Numpy solution that preserves the order
import numpy as np
a = np.asarray([1, 2, 0, 0, 0, 3, 6])
# mask is a boolean array that is True where a is equal to 0
mask = (a == 0)
# Take the subset of the array that are zeros
zeros = a[mask]
# Take the subset of the array that are NOT zeros
temp = a[~mask]
# Join the arrays
joint_array = np.concatenate([temp, zeros])

I tried using sorted, which is similar to sort().
a = [1, 2, 0, 0, 0, 3, 6]
sorted(a,reverse=True)
ans:
[6, 3, 2, 1, 0, 0, 0]

from typing import List
def move(A:List[int]):
j=0 # track of nonzero elements
k=-1 # track of zeroes
size=len(A)
for i in range(size):
if A[i]!=0:
A[j]=A[i]
j+=1
elif A[i]==0:
A[k]=0
k-=1
since we have to keep the relative order. when you see nonzero element, place that nonzero into the index of jth.
first_nonzero=A[0] # j=0
second_nonzero=A[1] # j=1
third_nonzero=A[2] # j=2
With k we keep track of 0 elements. In python A[-1] refers to the last element of the array.
first_zero=A[-1] # k=-1
second_zero=A[-2] # k=-2
third_zero= A[-3] # k=-3

a = [4,6,0,6,0,7,0]
a = filter (lambda x : x!= 0, a) + [0]*a.count(0)
[4, 6, 6, 7, 0, 0, 0]

How to find peaks of minimal length efficiently?

I have list/array of integers, call a subarray a peak if it goes up and then goes down. For example:
[5,5,4,5,4]
contains
[4,5,4]
which is a peak.
Also consider
[6,5,4,4,4,4,4,5,6,7,7,7,7,7,6]
which contains
[6,7,7,7,7,7,6]
which is a peak.
The problem
Given an input list, I would like to find all the peaks contained in it of minimal length and report them. In the example above, [5,6,7,7,7,7,7,6] is also a peak but we remove the first element and it remains a peak so we don't report it.
So for input list:
L = [5,5,5,5,4,5,4,5,6,7,8,8,8,8,8,9,9,8]
we would return
[4,5,4] and [8,9,9,8] only.
I am having problems devising a nice algorithm for this. Any help would be hugely appreciated.

Using itertools
Here is a short solution using itertools.groupby to detect peaks. The groups identifying peaks are then unpacked to yield the actual sequence.
from itertools import groupby, islice
l = [1, 2, 1, 2, 2, 0, 0]
fst, mid, nxt = groupby(l), islice(groupby(l), 1, None), islice(groupby(l), 2, None)
peaks = [[f[0], *m[1], n[0]] for f, m, n in zip(fst, mid, nxt) if f[0] < m[0] > n[0]]
print(peaks)
Output
[[1, 2, 1], [1, 2, 2, 0]]
Using a loop (faster)
The above solution is elegant but since three instances of groupby are created, the list is traversed three times.
Here is a solution using a single traversal.
def peaks(lst):
first = 0
last = 1
while last < len(lst) - 1:
if lst[first] < lst[last] == lst[last+1]:
last += 1
elif lst[first] < lst[last] > lst[last+1]:
yield lst[first:last+2]
first = last + 1
last += 2
else:
first = last
last += 1
l = [1, 2, 1, 2, 2, 0, 0]
print(list(peaks(l)))
Output
[[1, 2, 1], [1, 2, 2, 0]]
Notes on benchmark
Upon benchmarking with timeit, I noticed an increase in performance of about 20% for the solution using a loop. For short lists the overhead of groupby could bring that number up to 40%. The benchmark was done on Python 3.6.

Find all combination that sum to N when multiplies by index

Given a number of items (n), what is the most efficient way to generate all possible lists [a1, a2, ..., an] of non-negative integers under the condition that:
1*a1 + 2*a2 + 3*a3 + ... + n*an = n
using Python?
So for example, given an n of 5, the following combinations are:
[0,0,0,0,1]
[1,0,0,1,0]
[0,1,1,0,0]
[2,0,1,0,0]
[1,2,0,0,0]
[3,1,0,0,0]
[5,0,0,0,0]
I've implemented a brute-force method that generates all permutations and then checks if the list meets the above requirement, but is there a more efficient way to do this?

A "greedy" algorithm works well for this. I'm using Python 3 here:
def pick(total):
def inner(highest, total):
if total == 0:
yield result
return
if highest == 1:
result[0] = total
yield result
result[0] = 0
return
for i in reversed(range(total // highest + 1)):
result[highest - 1] = i
newtotal = total - i * highest
yield from inner(min(highest - 1, newtotal),
newtotal)
result = [0] * total
yield from inner(total, total)
Then, e.g.,
for x in pick(5):
print(x)
displays:
[0, 0, 0, 0, 1]
[1, 0, 0, 1, 0]
[0, 1, 1, 0, 0]
[2, 0, 1, 0, 0]
[1, 2, 0, 0, 0]
[3, 1, 0, 0, 0]
[5, 0, 0, 0, 0]
Like most recursive algorithms, it does a more-or-less obvious thing, then recurses to solve the (sub)problem that remains.
Here inner(highest, total) means to find all the decompositions of total using integers no larger than highest. How many copies of highest can we use? The more-than-less obvious answer is that we can use 0, 1, 2, ..., up to (and including) total // highest copies, but no more than that. Unless highest is 1 - then we have to use exactly total copies of 1.
However many copies we use of highest, the subproblem remaining is to decompose whatever remains of the total using integers no larger than highest - 1. Passing min(highest - 1, newtotal) instead of highest - 1 is an optimization, since it's pointless trying any integer larger than the new total.

Python Itertools questions [duplicate]

I've got a the following "bars and stars" algorithm, implemented in Python, which prints out all decomposition of a sum into 3 bins, for sums going from 0 to 5.
I'd like to generalise my code so it works with N bins (where N less than the max sum i.e 5 here).
The pattern is if you have 3 bins you need 2 nested loops, if you have N bins you need N-1 nested loops.
Can someone think of a generic way of writing this, possibly not using loops?
# bars and stars algorithm
N=5
for n in range(0,N):
x=[1]*n
for i in range(0,(len(x)+1)):
for j in range(i,(len(x)+1)):
print sum(x[0:i]), sum(x[i:j]), sum(x[j:len(x)])

If this isn't simply a learning exercise, then it's not necessary for you to roll your own algorithm to generate the partitions: Python's standard library already has most of what you need, in the form of the itertools.combinations function.
From Theorem 2 on the Wikipedia page you linked to, there are n+k-1 choose k-1 ways of partitioning n items into k bins, and the proof of that theorem gives an explicit correspondence between the combinations and the partitions. So all we need is (1) a way to generate those combinations, and (2) code to translate each combination to the corresponding partition. The itertools.combinations function already provides the first ingredient. For the second, each combination gives the positions of the dividers; the differences between successive divider positions (minus one) give the partition sizes. Here's the code:
import itertools
def partitions(n, k):
for c in itertools.combinations(range(n+k-1), k-1):
yield [b-a-1 for a, b in zip((-1,)+c, c+(n+k-1,))]
# Example usage
for p in partitions(5, 3):
print(p)
And here's the output from running the above code.
[0, 0, 5]
[0, 1, 4]
[0, 2, 3]
[0, 3, 2]
[0, 4, 1]
[0, 5, 0]
[1, 0, 4]
[1, 1, 3]
[1, 2, 2]
[1, 3, 1]
[1, 4, 0]
[2, 0, 3]
[2, 1, 2]
[2, 2, 1]
[2, 3, 0]
[3, 0, 2]
[3, 1, 1]
[3, 2, 0]
[4, 0, 1]
[4, 1, 0]
[5, 0, 0]

Another recursive variant, using a generator function, i.e. instead of right away printing the results, it yields them one after another, to be printed by the caller.
The way to convert your loops into a recursive algorithm is as follows:
identify the "base case": when there are no more bars, just print the stars
for any number of stars in the first segment, recursively determine the possible partitions of the rest, and combine them
You can also turn this into an algorithm to partition arbitrary sequences into chunks:
def partition(seq, n, min_size=0):
if n == 0:
yield [seq]
else:
for i in range(min_size, len(seq) - min_size * n + 1):
for res in partition(seq[i:], n-1, min_size):
yield [seq[:i]] + res
Example usage:
for res in partition("*****", 2):
print "|".join(res)

Take it one step at a time.
First, remove the sum() calls. We don't need them:
N=5
for n in range(0,N):
x=[1]*n
for i in range(0,(n+1)): # len(x) == n
for j in range(i,(n+1)):
print i, j - i, n - j
Notice that x is an unused variable:
N=5
for n in range(0,N):
for i in range(0,(n+1)):
for j in range(i,(n+1)):
print i, j - i, n - j
Time to generalize. The above algorithm is correct for N stars and three bars, so we just need to generalize the bars.
Do this recursively. For the base case, we have either zero bars or zero stars, which are both trivial. For the recursive case, run through all the possible positions of the leftmost bar and recurse in each case:
from __future__ import print_function
def bars_and_stars(bars=3, stars=5, _prefix=''):
if stars == 0:
print(_prefix + ', '.join('0'*(bars+1)))
return
if bars == 0:
print(_prefix + str(stars))
return
for i in range(stars+1):
bars_and_stars(bars-1, stars-i, '{}{}, '.format(_prefix, i))
For bonus points, we could change range() to xrange(), but that will just give you trouble when you port to Python 3.

This can be solved recursively in the following approach:
#n bins, k stars,
def F(n,k):
#n bins, k stars, list holds how many elements in current assignment
def aux(n,k,list):
if n == 0: #stop clause
print list
elif n==1: #making sure all stars are distributed
list[0] = k
aux(0,0,list)
else: #"regular" recursion:
for i in range(k+1):
#the last bin has i stars, set them and recurse
list[n-1] = i
aux(n-1,k-i,list)
aux(n,k,[0]*n)
The idea is to "guess" how many stars are in the last bin, assign them, and recurse to a smaller problem with less stars (as much that were assigned) and one less bin.
Note: It is easy to replace the line
print list
with any output format you desire when the number of stars in each bin is set.

Here is a nonrecursive algorithm that replicates the "bars and stars" nested loop approach. This assumes the bars all start on the right, and finish on the left (bins going from [x,0,0,...] to [0,0,..,x]). There will always be a zero in the first bin when a loop finishes, so you can follow the logic and match it to "bars and stars."
def combos(nbins, qty):
bins = [0]*nbins
bins[0] = qty #starting bin quantities
while True:
yield bins
if bins[-1] == qty:
return #last combo, we're done!
#leftmost bar movement (inner loop)
if bins[0] > 0:
bins[0] -= 1
bins[1] += 1
else:
#bump next bar in nested loops
#i.e., find first nonzero entry, and split it
nz = 1
while bins[nz] == 0:
nz +=1
bins[0]=bins[nz]-1
bins[nz+1] += 1
bins[nz] = 0
Here is the result of 4 bins, quantity 3:
for m in combos(4, 3):
print(m)
[3, 0, 0, 0]
[2, 1, 0, 0]
[1, 2, 0, 0]
[0, 3, 0, 0]
[2, 0, 1, 0]
[1, 1, 1, 0]
[0, 2, 1, 0]
[1, 0, 2, 0]
[0, 1, 2, 0]
[0, 0, 3, 0]
[2, 0, 0, 1]
[1, 1, 0, 1]
[0, 2, 0, 1]
[1, 0, 1, 1]
[0, 1, 1, 1]
[0, 0, 2, 1]
[1, 0, 0, 2]
[0, 1, 0, 2]
[0, 0, 1, 2]
[0, 0, 0, 3]

I needed to solve the same problem and found this post, but I really wanted a non-recursive general-purpose algorithm that didn't rely on itertools and couldn't find one, so came up with this.
By default, the generator produces the sequence in either lexical order (as the earlier recursive example) but can also produce the reverse-order sequence by setting the "reversed" flag.
def StarsAndBars(bins, stars, reversed=False):
if bins < 1 or stars < 1:
raise ValueError("Number of bins and objects must both be greater than or equal to 1.")
if bins == 1:
yield stars,
return
bars = [ ([0] * bins + [ stars ], 1) ]
if reversed:
while len(bars)>0:
b = bars.pop()
if b[1] == bins:
yield tuple(b[0][y] - b[0][y-1] for y in range(1, bins+1))
else:
bar = b[0][:b[1]]
for x in range(b[0][b[1]], stars+1):
newBar = bar + [ x ] * (bins - b[1]) + [ stars ]
bars.append( (newBar, b[1]+1) )
bars = [ ([0] * bins + [ stars ], 1) ]
else:
while len(bars)>0:
newBars = []
for b in bars:
for x in range(b[0][-2], stars+1):
newBar = b[0][1:bins] + [ x, stars ]
if b[1] < bins-1 and x > 0:
newBars.append( (newBar, b[1]+1) )
yield tuple(newBar[y] - newBar[y-1] for y in range(1, bins+1))
bars = newBars

This problem can also be solved somewhat less verbosely than the previous answers with a list comprehension:
from numpy import array as ar
from itertools import product
number_of_stars = M
number_of_bins = N
decompositions = ar([ar(i) for i in product(range(M+1), repeat=N) if sum(i)==M])
Here the itertools.product() produces a list containing the Cartesian product of the list range(M+1) with itself, where the product has been applied (repeats=)N times. The if statement removes the combinations where the number don't add up to the number of stars, for example one of the combinations is of 0 with 0 with 0 or [0,0,0].
If we're happy with a list of lists then we can simply remove the np.array()'s (just ar for brevity in the example). Here's an example output for 3 stars in 3 bins:
array([[0, 0, 3],
[0, 1, 2],
[0, 2, 1],
[0, 3, 0],
[1, 0, 2],
[1, 1, 1],
[1, 2, 0],
[2, 0, 1],
[2, 1, 0],
[3, 0, 0]])
I hope this answer helps!

Since I found the code in most answers quite hard to follow i.e. asking myself how the shown algorithms relate to the actual problem of stars and bars let's do this step by step:
First we define a function to insert a bar | into a string stars at a given position p:
def insert_bar(stars, p):
head, tail = stars[:p], stars[p:]
return head + '|' + tail
Usage:
insert_bar('***', 1) # returns '*|**'
To insert multiple bars at different positions e.g. (1,3) a simple way is to use reduce (from functools)
reduce(insert_bar, (1,3), '***') # returns '*|*|*'
If we branch the definition of insert_bar to handle both cases we get a nice and reusable function to insert any number of bars into a string of stars
def insert_bars(stars, p):
if type(p) is int:
head, tail = stars[:p], stars[p:]
return head + '|' + tail
else:
return reduce(insert_bar, p, stars)
As #Mark Dickinson explaind in his answer itertools.combinations lets us produce the n+k-1 choose k-1 combinations of bar positions.
What is now left to do is to create a string of '*' of length n, insert the bars at the given positions, split the string at the bars and calculate the length of each resulting bin. The implementation below is thus literally a verbatim translation of the problem statement into code
def partitions(n, k):
for positions in itertools.combinations(range(n+k-1), k-1):
yield [len(bin) for bin in insert_bars(n*"*", positions).split('|')]

anyone looking for the specific case of k=2 can save ALOT of time by simply creating a range and stacking it with the reverse. Comparing versus accepted answer.
n = 500000
%timeit np.array([[i,j] for i,j in partitions(n,2)])
>>> 396 ms ± 13.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
rng = np.arange(n+1)
np.vstack([rng, rng[::-1]]).T
>>> 2.91 ms ± 190 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
And they are indeed equivalent.
it2k = np.array([[i,j] for i,j in partitions(n,2)])
rng = np.arange(n+1)
np2k = np.vstack([rng, rng[::-1]]).T
(np2k == it2k).all()
>>> True