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Why do these python and kotlin functions perform differently on hackerrank?

I am trying to solve a contest challenge on Hackerrank (Hack the Interview II - Global Product Distribution) using Kotlin
I started getting annoyed because my code always passed on the test cases with a small number of inputs and failed on the larger ones, even timing out on one.
So I went online and found this python code that solved all test cases neatly. I went as far as converting the Python code line for line into Kotlin. But my Kotlin code always retained the same poor performance as before.
These are the two pieces of code.
Python:
def maxScore(a, m):
a.sort()
print(a)
x=len(a)
if x%m==0:
y=int(x/m)
else:
y=int(x/m)-1
summ=0
count=1
#print(y)
i=0
for _ in range(y):
summ=summ+(sum(a[i:i+m])*count)
count=count+1
i=i+m
print(summ)
summ=summ+sum(a[i:])*count
print(summ)
return summ%1000000007
Kotlin:
fun maxScore(a: Array<Int>, m: Int): Int {
a.sort()
// print(a)
val x = a.size
val y = if (x % m == 0) x / m
else (x / m) - 1
var summ = 0
var count = 1
// print(y)
var i = 0
for (s in 0 until y) {
summ += a.sliceArray(i until (i + m)).sum() * count
count++
i += m
// print(summ)
}
summ += a.sliceArray(i until a.size).sum() * count
// print(summ)
return summ % 1000000007
}
Is there something wrong with the code translation? How can I make the Kotlin code work on the larger test cases?
UPDATE: copyOfRange() performs better than sliceArray(). Code no longer times out on any test case, but still fails on all the large test cases

There's three issues I can see here. I'll point you in the right direction for now.
Both the Python and the Kotlin copy the array each time. This might or might not be a problem. You have up to a million elements and each is copied only once. I'd be surprised if that exceeds your time limits but it might do. It looks like you can avoid the copy with .subList().
It looks like you're treating the leftover items as if they're in a bin of their own. But this bin is smaller than m, which isn't allowed. Check that this is really what you intend.
Kotlin Ints are 32-bit signed integers. You can only store numbers up to about 2 billion before they overflow. You need to avoid this! Look at the constraints - you can have up to a million products with individual values up to a billion each. (This is different from Python ints, which never overflow, and so will always give the right answer, but can use a lot of memory and slow down if you try to do operations on really big numbers, which might well be causing your program to time out.) Here is a hint: (a + b) % n is equal to ((a % n) + (b % n)) % n

Pypy memory usage increases when nothing new is allocated

I'm not sure if this is a duplicate of the other PyPy memory questions, but here I'll provide a concrete example.
from __future__ import division
def mul_inv(a, m):
"""Modular multiplicative inverse, a^-1 mod m. Credit: rosettacode.org"""
m0 = m
x0, x1 = 0, 1
if m == 1: return 1
while a > 1:
assert m != 0, "a and m must be coprime"
q = a // m
a, m = m, a%m
x0, x1 = x1 - q * x0, x0
if x1 < 0: x1 += m0
return x1
M = 1000000009
L = 10**8
bin2 = [0] * L
bin2[0] = 1
for n in range(L-1):
bin2[n+1] = (bin2[n] * (4*n + 2) * mul_inv(n+1, M)) % M
if n % 10**5 == 0: print(n, bin2[n])
print(bin2[:20])
With python 3.6, the program uses 3-4 GB at most and runs to completion (Armin Rigo's list change doesn't change this significantly). With python 2.7.13 running PyPy 5.10.0, the program reaches 8 GB (how much RAM I have) quickly and freezes. Even with gc.collect() calls the program runs out of memory when n is about 3.5 * 10^7.
Where is this memory usage coming from? The only large memory usage should be initializing bin2 as a 10^8 int list. Nothing else should be increasing the memory usage, under the assumption that all the local variables in mul_inv are garbage collected.

Oops, it's a bad case of the optimization for lists of integers. The problem is that this starts as a list of ints:
bin2 = [0] * L
This is internally stored as an array of ints. It's usually much more compact, even though in this case it doesn't change anything---because on CPython it's a list containing L copies of the same object 0.
But the problem is that pretty soon, we store a long in the list. At this point, we need to turn the whole list into the generic kind that can store anything. But! The problem is that we see 100 million zeroes, and so we create 100 million 0 objects. This creates instantly 3 GB of memory pressure for nothing, in addition to 800MB for the list itself.
We can check that the problem doesn't occur if we initialize the list like this, so that it really contains 100 million times the same object:
bin2 = [0L] * L # Python 2.x
bin2[0] = 1
That said, in your example you don't need the list to contain 100 million elements in the first place. You can initialize it as:
bin2 = [1]
and use bin2.append(). This lets the program start much more quickly and without any large memory usage near the beginning.
Note that PyPy3 still uses more memory than CPython3.

AFAICT the issue here is that you're assigning longs to the array, and despite your modulo, PyPy doesn't seem to notice that the number still fits into a machine word.
I can think of two ways to fix this:
Pass the value assigned to bin2[n+1] through int().
Use array.array().
The former only affects PyPy2, and results in what appears to be a stable memory footprint of ~800MB on my Mac, whereas the latter appears to stabilise at ~1.4GB regardless of whether I run it in PyPy2 or PyPy3.
I haven't run the program fully to completion, though, so YMMV…

Efficient random generator for very large range (in python)

I am trying to create a generator that returns numbers in a given range that pass a particular test given by a function foo. However I would like the numbers to be tested in a random order. The following code will achieve this:
from random import shuffle
def MyGenerator(foo, num):
order = list(range(num))
shuffle(order)
for i in order:
if foo(i):
yield i
The Problem
The problem with this solution is that sometimes the range will be quite large (num might be of the order 10**8 and upwards). This function can become slow, having such a large list in memory. I have tried to avoid this problem, with the following code:
from random import randint
def MyGenerator(foo, num):
tried = set()
while len(tried) <= num - 1:
i = randint(0, num-1)
if i in tried:
continue
tried.add(i)
if foo(i):
yield i
This works well most of the time, since in most cases num will be quite large, foo will pass a reasonable number of numbers and the total number of times the __next__ method will be called will be relatively small (say, a maximum of 200 often much smaller). Therefore its reasonable likely we stumble upon a value that passes the foo test and the size of tried never gets large. (Even if it only passes 10% of the time, we wouldn't expect tried to get larger than about 2000 roughly.)
However, when num is small (close to the number of times that the __next__ method is called, or foo fails most of the time, the above solution becomes very inefficient - randomly guessing numbers until it guesses one that isn't in tried.
My attempted solution...
I was hoping to use some kind of function that maps the numbers 0,1,2,..., n onto themselves in a roughly random way. (This isn't being used for any security purposes and so doesn't matter if it isn't the most 'random' function in the world). The function here (Create a random bijective function which has same domain and range) maps signed 32-bit integers onto themselves, but I am not sure how to adapt the mapping to a smaller range. Given num I don't even need a bijection on 0,1,..num just a value of n larger than and 'close' to num (using whatever definition of close you see fit). Then I can do the following:
def mix_function_factory(num):
# something here???
def foo(index):
# something else here??
return foo
def MyGenerator(foo, num):
mix_function = mix_function_factory(num):
for i in range(num):
index = mix_function(i)
if index <= num:
if foo(index):
yield index
(so long as the bijection isn't on a set of numbers massively larger than num the number of times index <= num isn't True will be small).
My Question
Can you think of one of the following:
A potential solution for mix_function_factory or even a few other potential functions for mix_function that I could attempt to generalise for different values of num?
A better way of solving the original problem?
Many thanks in advance....

The problem is basically generating a random permutation of the integers in the range 0..n-1.
Luckily for us, these numbers have a very useful property: they all have a distinct value modulo n. If we can apply some mathemical operations to these numbers while taking care to keep each number distinct modulo n, it's easy to generate a permutation that appears random. And the best part is that we don't need any memory to keep track of numbers we've already generated, because each number is calculated with a simple formula.
Examples of operations we can perform on every number x in the range include:
Addition: We can add any integer c to x.
Multiplication: We can multiply x with any number m that shares no prime factors with n.
Applying just these two operations on the range 0..n-1 already gives quite satisfactory results:
>>> n = 7
>>> c = 1
>>> m = 3
>>> [((x+c) * m) % n for x in range(n)]
[3, 6, 2, 5, 1, 4, 0]
Looks random, doesn't it?
If we generate c and m from a random number, it'll actually be random, too. But keep in mind that there is no guarantee that this algorithm will generate all possible permutations, or that each permutation has the same probability of being generated.
Implementation
The difficult part about the implementation is really just generating a suitable random m. I used the prime factorization code from this answer to do so.
import random
# credit for prime factorization code goes
# to https://stackoverflow.com/a/17000452/1222951
def prime_factors(n):
gaps = [1,2,2,4,2,4,2,4,6,2,6]
length, cycle = 11, 3
f, fs, next_ = 2, [], 0
while f * f <= n:
while n % f == 0:
fs.append(f)
n /= f
f += gaps[next_]
next_ += 1
if next_ == length:
next_ = cycle
if n > 1: fs.append(n)
return fs
def generate_c_and_m(n, seed=None):
# we need to know n's prime factors to find a suitable multiplier m
p_factors = set(prime_factors(n))
def is_valid_multiplier(m):
# m must not share any prime factors with n
factors = prime_factors(m)
return not p_factors.intersection(factors)
# if no seed was given, generate random values for c and m
if seed is None:
c = random.randint(n)
m = random.randint(1, 2*n)
else:
c = seed
m = seed
# make sure m is valid
while not is_valid_multiplier(m):
m += 1
return c, m
Now that we can generate suitable values for c and m, creating the permutation is trivial:
def random_range(n, seed=None):
c, m = generate_c_and_m(n, seed)
for x in range(n):
yield ((x + c) * m) % n
And your generator function can be implemented as
def MyGenerator(foo, num):
for x in random_range(num):
if foo(x):
yield x

That may be a case where the best algorithm depends on the value of num, so why not using 2 selectable algorithms wrapped in one generator ?
you could mix your shuffle and set solutions with a threshold on the value of num. That's basically assembling your 2 first solutions in one generator:
from random import shuffle,randint
def MyGenerator(foo, num):
if num < 100000 # has to be adjusted by experiments
order = list(range(num))
shuffle(order)
for i in order:
if foo(i):
yield i
else: # big values, few collisions with random generator
tried = set()
while len(tried) < num:
i = randint(0, num-1)
if i in tried:
continue
tried.add(i)
if foo(i):
yield i
The randint solution (for big values of num) works well because there aren't so many repeats in the random generator.

Getting the best performance in Python is much trickier than in lower-level languages. For example, in C, you can often save a little bit in hot inner loops by replacing a multiplication by a shift. The overhead of python bytecode-orientation erases this. Of course, this changes again when you consider which variant of "python" you're targetting (pypy? numpy? cython?)- you really have to write your code based on which one you're using.
But even more important is arranging operations to avoid serialized dependencies, since all CPUs are superscalar these days. Of course, real compilers know about this, but it still matters when choosing an algorithm.
One of the easiest ways to gain a little bit over existing answers would be by by generating numbers in chunks using numpy.arange() and applying the ((x + c) * m) % n to the numpy ndarray directly. Every python-level loop that can be avoided helps.
If the function can be applied directly to numpy ndarrays, that might even better. Of course, a sufficiently-small function in python will be dominated by function-call overhead anyway.
The best fast random-number-generator today is PCG. I wrote a pure-python port here but concentrated on flexibility and ease-of-understanding rather than speed.
Xoroshiro128+ is second-best-quality and faster, but less informative to study.
Python's (and many others') default choice of Mersenne Twister is among the worst.
(there's also something called splitmix64 which I don't know enough about to place - some people say it's better than xoroshiro128+, but it has a period problem - of course, you might want that here)
Both default-PCG and xoroshiro128+ use a 2N-bit state to generate N-bit numbers. This is generally desirable, but means numbers will be repeated. PCG has alternate modes that avoid this, however.
Of course, much of this depends on whether num is (close to) a power of 2. In theory, PCG variants can be created for any bit width, but currently only various word sizes are implemented since you'd need explicit masking. I'm not sure exactly how to generate the parameters for new bit sizes (perhaps it's in the paper?), but they can be tested simply by doing a period/2 jump and verifying that the value is different.
Of course, if you're only making 200 calls to the RNG, you probably don't actually need to avoid duplicates on the math side.
Alternatively, you could use an LFSR, which does exist for every bit size (though note that it never generates the all-zeros value (or equivalently, the all-ones value)). LFSRs are serial and (AFAIK) not jumpable, and thus can't be easily split across multiple tasks. Edit: I figured out that this is untrue, simply represent the advance step as a matrix, and exponentiate it to jump.
Note that LFSRs do have the same obvious biases as simply generating numbers in sequential order based on a random start point - for example, if rng_outputs[a:b] all fail your foo function, then rng_outputs[b] will be much more likely as a first output regardless of starting point. PCG's "stream" parameter avoids this by not generating numbers in the same order.
Edit2: I have completed what I thought was a "brief project" implementing LFSRs in python, including jumping, fully tested.

Time Complexity - Codility - Ladder - Python

The question is available here. My Python code is
def solution(A, B):
if len(A) == 1:
return [1]
ways = [0] * (len(A) + 1)
ways[1], ways[2] = 1, 2
for i in xrange(3, len(ways)):
ways[i] = ways[i-1] + ways[i-2]
result = [1] * len(A)
for i in xrange(len(A)):
result[i] = ways[A[i]] & ((1<<B[i]) - 1)
return result
The detected time complexity by the system is O(L^2) and I can't see why. Thank you in advance.

First, let's show that the runtime genuinely is O(L^2). I copied a section of your code, and ran it with increasing values of L:
import time
import matplotlib.pyplot as plt
def solution(L):
if L == 0:
return
ways = [0] * (L+5)
ways[1], ways[2] = 1, 2
for i in xrange(3, len(ways)):
ways[i] = ways[i-1] + ways[i-2]
points = []
for L in xrange(0, 100001, 10000):
start = time.time()
solution(L)
points.append(time.time() - start)
plt.plot(points)
plt.show()
The result graph is this:
To understand why this O(L^2) when the obvious "time complexity" calculation suggests O(L), note that "time complexity" is not a well-defined concept on its own since it depends on which basic operations you're counting. Normally the basic operations are taken for granted, but in some cases you need to be more careful. Here, if you count additions as a basic operation, then the code is O(N). However, if you count bit (or byte) operations then the code is O(N^2). Here's the reason:
You're building an array of the first L Fibonacci numbers. The length (in digits) of the i'th Fibonacci number is Theta(i). So ways[i] = ways[i-1] + ways[i-2] adds two numbers with approximately i digits, which takes O(i) time if you count bit or byte operations.
This observation gives you an O(L^2) bit operation count for this loop:
for i in xrange(3, len(ways)):
ways[i] = ways[i-1] + ways[i-2]
In the case of this program, it's quite reasonable to count bit operations: your numbers are unboundedly huge as L increases and addition of huge numbers is linear in clock time rather than O(1).
You can fix the complexity of your code by computing the Fibonacci numbers mod 2^32 -- since 2^32 is a multiple of 2^B[i]. That will keep a finite bound on the numbers you're dealing with:
for i in xrange(3, len(ways)):
ways[i] = (ways[i-1] + ways[i-2]) & ((1<<32) - 1)
There are some other issues with the code, but this will fix the slowness.

I've taken the relevant parts of the function:
def solution(A, B):
for i in xrange(3, len(A) + 1): # replaced ways for clarity
# ...
for i in xrange(len(A)):
# ...
return result
Observations:
A is an iterable object (e.g. a list)
You're iterating over the elements of A in sequence
The behavior of your function depends on the number of elements in A, making it O(A)
You're iterating over A twice, meaning 2 O(A) -> O(A)
On point 4, since 2 is a constant factor, 2 O(A) is still in O(A).
I think the page is not correct in its measurement. Had the loops been nested, then it would've been O(A²), but the loops are not nested.
This short sample is O(N²):
def process_list(my_list):
for i in range(0, len(my_list)):
for j in range(0, len(my_list)):
# do something with my_list[i] and my_list[j]
I've not seen the code the page is using to 'detect' the time complexity of the code, but my guess is that the page is counting the number of loops you're using without understanding much of the actual structure of the code.
EDIT1:
Note that, based on this answer, the time complexity of the len function is actually O(1), not O(N), so the page is not incorrectly trying to count its use for the time-complexity. If it were doing that, it would've incorrectly claimed a larger order of growth because it's used 4 separate times.
EDIT2:
As #PaulHankin notes, asymptotic analysis also depends on what's considered a "basic operation". In my analysis, I've counted additions and assignments as "basic operations" by using the uniform cost method, not the logarithmic cost method, which I did not mention at first.
Most of the time simple arithmetic operations are always treated as basic operations. This is what I see most commonly being done, unless the algorithm being analysed is for a basic operation itself (e.g. time complexity of a multiplication function), which is not the case here.
The only reason why we have different results appears to be this distinction. I think we're both correct.
EDIT3:
While an algorithm in O(N) is also in O(N²), I think it's reasonable to state that the code is still in O(N) b/c, at the level of abstraction we're using, the computational steps that seem more relevant (i.e. are more influential) are in the loop as a function of the size of the input iterable A, not the number of bits being used to represent each value.
Consider the following algorithm to compute an:
def function(a, n):
r = 1
for i in range(0, n):
r *= a
return r
Under the uniform cost method, this is in O(N), because the loop is executed n times, but under logarithmic cost method, the algorithm above turns out to be in O(N²) instead due to the time complexity of the multiplication at line r *= a being in O(N), since the number of bits to represent each number is dependent on the size of the number itself.

Codility Ladder competition is best solved in here:
It is super tricky.
We first compute the Fibonacci sequence for the first L+2 numbers. The first two numbers are used only as fillers, so we have to index the sequence as A[idx]+1 instead of A[idx]-1. The second step is to replace the modulo operation by removing all but the n lowest bits

Python Random Array of 0s and 1s [duplicate]

This question already has answers here:
Binary random array with a specific proportion of ones?
(6 answers)
Closed 4 years ago.
I want to randomly produce an array of n ones and m zeros.
I thought of this solution:
produce the ones array (np.ones)
produce the zeros array (np.zeros)
combine them to one array (np.hstack)
shuffle the resulting array (np.random.shuffle)
Seems to be not natural as a solution. Some pythonic ideas?

Your solution seems reasonable. It states exactly what it's doing, and does it clearly.
Let's compare your implementation:
a = np.hstack((np.ones(n), np.zeros(m)))
np.random.shuffle(a)
… with an obvious alternative:
a = np.ones(n+m)
a[:m] = 0
np.random.shuffle(a)
That might save a bit of time not allocating and moving hunks of data around, but it takes a bit more thought to understand.
And doing it in Python instead of in NumPy:
a = np.array([1]*n + [0]*m)
np.random.shuffle(a)
… might be a little more concise, but it seems less idiomatically NumPy (in the same way that np.array([1]*n) is less idiomatic than np.ones(n)), and it's going to be slower and use more memory for no good reason. (You could improve the memory by using np.fromiter, but then it's pretty clearly not going to be more concise.)
Of course if you're doing this more than once, the real answer is to factor it out into a function. Then the function's name will explain what it does, and almost any solution that isn't too tortured will be pretty easy to understand…

I'd make an array of n ones and m zeros as
a = np.array([1] * n + [0] * m)
Then I'd call np.random.shuffle() on it.

Use numpy.random.permutation:
a = numpy.random.permutation([1] * n + [0] * m)
or, using arrays instead of an initial list:
a = numpy.random.permutation(numpy.concatenate(np.ones(n), np.zeros(m)))
(I don't know enough about numpy to comment on the difference between concatenate and hstack; they seem to produce the same results here.)

I think your solution is suitable, in that it's readable and pythonic. You didn't say whether memory or performance are considerations. It's possible that np.random.shuffle is as good as O(m + n), but the other answers suggest that it does more than a single pass to shuffle the values. You could do it in O(m + n) with only a single pass and no memory overhead like this:
import random
m = 600 # zeros
n = 400 # ones
result = []
while m + n > 0:
if (m > 0 and random.random() < float(m)/float(m + n)):
result.append(0)
m -= 1
else:
result.append(1)
n -= 1