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How to elegantly "min" and then perform an operation with the reference in python

Suppose you have two lists nums1 and nums2, and you want to check which of the two lists first elements is smaller, remove it and then move that into an accumulator list which you return.
You might accomplish it using the following code:
def stackoverflowmethod(nums1, nums2):
accumulator = []
# determine if nums1[0] or nums2[0] is smaller and then remove it
if nums1[0] <= nums2[0]:
accumulator.append(nums1.pop(0))
else:
accumulator.append(nums2.pop(0))
return accumulator
However this might look very unsightly in that you would rather have written something like
def idealMethod(nums1, nums2):
accumulator.append(RemoveFromTheFrontOfOYourList(min(nums1[0], nums2[0])))
return accumulator
The only thing that makes this hard is the mysterious method RemoveFromTheFrontOfOYourList needs to be able to figure out which list an item came from after the min operation was conducted on it,
I.E. if we had a slightly less forgetful min function then something elegant could be done here.
If stylistically what I'm asking for makes sense, how does one accomplish it in python 3?

Use min on the lists, with operator.itemgetter(0) as the key function.
import operator
smaller_0 = min(nums1, nums2, key=operator.itemgetter(0))
accumulator.append(smaller_0.pop(0))
operator.itemgetter(0) is the same as lambda x: x[0] for this purpose.
Or use a ternary that gets the list with a smaller item 0, then pop from it
smaller_0 = nums1 if nums1[0] <= nums2[0] else nums2
accumulator.append(smaller_0.pop(0))
This is like Michael's answer, but a bit DRYer

Use a ternary statement:
acc.append(nums1.pop(0)) if nums1[0] < nums2[0] else acc.append(nums2.pop(0))

Does one for loop mean a time complexity of n in this case?

So, I've run into this problem in the daily coding problem challenge, and I've devised two solutions. However, I am unsure if one is better than the other in terms of time complexity (Big O).
# Given a list of numbers and a number k,
# return whether any two numbers from the list add up to k.
#
# For example, given [10, 15, 3, 7] and k of 17, return true since 10 + 7 is 17.
#
# Bonus: Can you do this in one pass?
# The above part seemed to denote this can be done in O(n).
def can_get_value(lst=[11, 15, 3, 7], k=17):
for x in lst:
for y in lst:
if x+y == k:
return True
return False
def optimized_can_get_value(lst=[10, 15, 3, 7], k=17):
temp = lst
for x in lst:
if k-x in temp:
return True
else:
return False
def main():
print(can_get_value())
print(optimized_can_get_value())
if __name__ == "__main__":
main()
I think the second is better than the first since it has one for loop, but I'm not sure if it is O(n), since I'm still running through two lists. Another solution I had in mind that was apparently a O(n) solution was using the python equivalent of "Java HashSets". Would appreciate confirmation, and explanation of why/why not it is O(n).

The first solution can_get_value() is textbook O(n^2). You know this.
The second solution is as well. This is because elm in list has O(n) complexity, and you're executing it n times. O(n) * O(n) = O(n^2).
The O(n) solution here is to convert from a list into a set (or, well, any type of hash table - dict would work too). The following code runs through the list exactly twice, which is O(n):
def can_get_value(lst, k):
st = set(lst) # make a hashtable (set) where each key is the same as its value
for x in st: # this executes n times --> O(n)
if k-x in st: # unlike for lists, `in` is O(1) for hashtables
return True
return False
This is thus O(n) * O(1) = O(n) in most cases.

In order to analyze the asymptotic runtime of your code, you need to know the runtime of each of the functions which you call as well. We generally think of arithmetic expressions like addition as being constant time (O(1)), so your first function has two for loops over n elements and the loop body only takes constant time, coming out to O(n * n * 1) = O(n^2).
The second function has only one for loop, but checking membership for a list is an O(n) function in the length of the list, so you still have O(n * n) = O(n^2). The latter option may still be faster (Python probably has optimized code for checking list membership), but it won't be asymptotically faster (the runtime still increases quadratically in n).
EDIT - as #Mark_Meyer pointed out, your second function is actually O(1) because there's a bug in it; sorry, I skimmed it and didn't notice. This answer assumes a corrected version of the second function like
def optimized_can_get_value(lst, k=17):
for x in lst:
if k - x in lst:
return True
return False
(Note - don't have a default value for you function which is mutable. See this SO question for the troubles that can bring. I also removed the temporary list because there's no need for that; it was just pointing to the same list object anyway.)
EDIT 2: for fun, here are a couple of O(n) solutions to this (both use that checking containment for a set is O(1)).
A one-liner which still stops as soon as a solution is found:
def get_value_one_liner(lst, k):
return any(k - x in set(lst) for x in lst)
EDIT 3: I think this is actually O(n^2) because we call set(lst) for each x. Using Python 3.8's assignment expressions could, I think, give us a one-liner that is still efficient. Does anybody have a good Python <3.8 one-liner?
And a version which tries not to do extra work by building up a set as it goes (not sure if this is actually faster in practice than creating the whole set at the start; it probably depends on the actual input data):
def get_value_early_stop(lst, k):
values = set()
for x in lst:
if x in values:
return True
values.add(k - x)
return False

O(n) list subtraction

When working on an AoC puzzle, I found I wanted to subtract lists (preserving ordering):
def bag_sub(list_big, sublist):
result = list_big[:]
for n in sublist:
result.remove(n)
return result
I didn't like the way the list.remove call (which is itself O(n)) is contained within the loop, that seems needlessly inefficient. So I tried to rewrite it to avoid that:
def bag_sub(list_big, sublist):
c = Counter(sublist)
result = []
for k in list_big:
if k in c:
c -= Counter({k: 1})
else:
result.append(k)
return result
Is this now O(n), or does the Counter.__isub__ usage still screw things up?
This approach requires that elements must be hashable, a restriction which the original didn't have. Is there an O(n) solution which avoids creating this additional restriction? Does Python have any better "bag" datatype than collections.Counter?
You can assume sublist is half the length of list_big.

I'd use a Counter, but I'd probably do it slightly differently, and I'd probably do this iteratively...
def bag_sub(big_list, sublist):
sublist_counts = Counter(sublist)
result = []
for item in big_list:
if sublist_counts[item] > 0:
sublist_counts[item] -= 1
else:
result.append(item)
return result
This is very similar to your solution, but it's probably not efficient to create an entire new counter every time you want to decrement the count on something.1
Also, if you don't need to return a list, then consider a generator function...
This works as long as all of the elements in list_big and sublist can be hashed. This solution is O(N + M) where N and M are the lengths of list_big and sublist respectively.
If the elements cannot be hashed, you are out of luck unless you have other constraints (e.g. the inputs are sorted using the same criterion). If your inputs are sorted, you could do something similar to the merge stage of merge-sort to determine which elements from bag_sub are in sublist.
1Note that Counters also behave a lot like a defaultdict(int) so it's perfectly fine to look for an item in a counter that isn't there already.

Is this now O(n), or does the Counter.__isub__ usage still screw things up?
This would be expected-case O(n), except that when Counter.__isub__ discards nonpositive values, it goes through every key to do so. You're better off just subtracting 1 from the key value the "usual" way and checking c[k] instead of k in c. (c[k] is 0 for k not in c, so you don't need an in check.)
if c[k]:
c[k] -= 1
else:
result.append(k)
Is there an O(n) solution which avoids creating this additional restriction?
Only if the inputs are sorted, in which case a standard variant of a mergesort merge can do it.
Does Python have any better "bag" datatype than collections.Counter?
collections.Counter is Python's bag.

Removing an item from a list of length N is O(N) if the list is unordered, because you have to find it.
Removing k items from a list of length N, therefore, is O(kN) if we focus on "reasonable" cases where k << N.
So I don't see how you could get it down to O(N).
A concise way to write this:
new_list = [x for x in list_big if x not in sublist]
But that's still O(kN).

Python Non-recursive Permutation

Does anyone understand the following iterative algorithm for producing all permutations of a list of numbers?
I do not understand the logic within the while len(stack) loop. Can someone please explain how it works?
# Non-Recursion
#param nums: A list of Integers.
#return: A list of permutations.
def permute(self, nums):
if nums is None:
return []
nums = sorted(nums)
permutation = []
stack = [-1]
permutations = []
while len(stack):
index = stack.pop()
index += 1
while index < len(nums):
if nums[index] not in permutation:
break
index += 1
else:
if len(permutation):
permutation.pop()
continue
stack.append(index)
stack.append(-1)
permutation.append(nums[index])
if len(permutation) == len(nums):
permutations.append(list(permutation))
return permutations
I'm just trying to understand the code above.

As mentioned in the comments section to your question, debugging may provide a helpful way to understand what the code does. However, let me provide a high-level perspective of what your code does.
First of all, although there are no recursive calls to the function permute, the code your provided is effectively recursive, as all it does is keeping its own stack, instead of using the one provided by the memory manager of your OS. Specifically, the variable stack is keeping the recursive state, so to speak, that is passed from one recursive call to another. You could, and perhaps should, consider each iteration of the outer while loop in the permute function as a recursive call. If you do so, you will see that the outer while loop helps 'recursively' traverse each permutation of nums in a depth-first manner.
Noticing this, it's fairly easy to figure out what each 'recursive call' does. Basically, the variable permutation keeps the current permutation of nums which is being formed as while loop progresses. Variable permutations store all the permutations of nums that are found. As you may observe, permutations are updated only when len(permutation) is equal to len(nums) which can be considered as the base case of the recurrence relation that is being implemented using a custom stack. Finally, the inner while loop picks which element of nums to add to the current permutation(i.e. stored in variable permutation) being formed.
So that is about it, really. You can figure out what is exactly being done on the lines relevant to the maintenance of stack using a debugger, as suggested. As a final note, let me repeat that I, personally, would not consider this implementation to be non-recursive. It just so happens that, instead of using the abstraction provided by the OS, this recursive solution keeps its own stack. To provide a better understanding of how a proper non-recursive solution would be, you may observe the difference in recursive and iterative solutions to the problem of finding nth Fibonacci number provided below. As you can see, the non-recursive solution keeps no stack, and instead of dividing the problem into smaller instances of it(recursion) it builds up the solution from smaller solutions. (dynamic programming)
def recursive_fib(n):
if n == 0:
return 0
elif n == 1:
return 1
return recursive_fib(n-1) + recursive_fib(n-2)
def iterative_fib(n):
f_0 = 0
f_1 = 1
for i in range(3, n):
f_2 = f_1 + f_0
f_0 = f_1
f_1 = f_2
return f_1

The answer from #ilim is correct and should be the accepted answer but I just wanted to add another point that wouldn't fit as a comment. Whilst I imagine you are studying this algorithm as an exercise it should be pointed out that a better way to proceed, depending on the size of the list, may be to user itertools's permutations() function:
print [x for x in itertools.permutations([1, 2, 3])]
Testing on my machine with a list of 11 items (39m permutations) took 1.7secs with itertools.permutations(x) but took 76secs using the custom solution above. Note however that with 12 items (479m permutations) the itertools solution blows up with a memory error. If you need to generate permutations of such size efficiently you may be better dropping to native code.

Python lists, dictionary optimization

I have been attending a couple of hackathons. I am beginning to understand that writing code is not enough. The code has to be optimized. That brings me to my question. Here are two questions that I faced.
def pairsum(numbers, k)
"""Write a function that returns two values in numbers whose sum is K"""
for i, j in numbers:
if i != j:
if i+j == k
return i, j
I wrote this function. And I was kind of stuck with optimization.
Next problem.
string = "ksjdkajsdkajksjdalsdjaksda"
def dedup(string):
""" write a function to remove duplicates in the variable string"""
output = []
for i in string:
if i not in output:
output.append(i)
These are two very simple programs that I wrote. But I am stuck with optimization after this. More on this, when we optimize code, how does the complexity reduce? Any pointers will help. Thanks in advance.

Knowing the most efficient Python idioms and also designing code that can reduce iterations and bail out early with an answer is a major part of optimization. Here are a few examples:
List list comprehensions and generators are usually fastest:
With a straightforward nested approach, a generator is faster than a for loop:
def pairsum(numbers, k):
"""Returns two unique values in numbers whose sum is k"""
return next((i, j) for i in numbers for j in numbers if i+j == k and i != j)
This is probably faster on average since it only goes though one iteration at most and does not check if a possible result is in numbers unless k-i != i:
def pairsum(numbers, k):
"""Returns two unique values in numbers whose sum is k"""
return next((k-i, i) for i in numbers if k-i != i and k-i in numbers)
Ouput:
>>> pairsum([1,2,3,4,5,6], 8)
(6, 2)
Note: I assumed numbers was a flat list since the doc string did not mention tuples and it makes the problem more difficult which is what I would expect in a competition.
For the second problem, if you are to create your own function as opposed to just using ''.join(set(s)) you were close:
def dedup(s):
"""Returns a string with duplicate characters removed from string s"""
output = ''
for c in s:
if c not in output:
output += c
return output
Tip: Do not use string as a name
You can also do:
def dedup(s):
for c in s:
s = c + s.replace(c, '')
return s
or a much faster recursive version:
def dedup(s, out=''):
s0, s = s[0], s.replace(s[0], '')
return dedup(s, n + s0) if s else out + s0
but not as fast as set for strings without lots of duplicates:
def dedup(s):
return ''.join(set(s))
Note: set() will not preserve the order of the remaining characters while the other approaches will preserve the order based on first occurrence.

Your first program is a little vague. I assume numbers is a list of tuples or something? Like [(1,2), (3,4), (5,6)]? If so, your program is pretty good, from a complexity standpoint - it's O(n). Perhaps you want a little more Pythonic solution? The neatest way to clean this up would be to join your conditions:
if i != j and i + j == k:
But this simply increases readability. I think it may also add an additional boolean operation, so it might not be an optimization.
I am not sure if you intended for your program to return the first pair of numbers which sum to k, but if you wanted all pairs which meet this requirement, you could write a comprehension:
def pairsum(numbers, k):
return list(((i, j) for i, j in numbers if i != j and i + j == k))
In that example, I used a generator comprehension instead of a list comprehension so as to conserve resources - generators are functions which act like iterators, meaning that they can save memory by only giving you data when you need it. This is called lazy iteration.
You can also use a filter, which is a function which returns only the elements from a set for which a predicate returns True. (That is, the elements which meet a certain requirement.)
import itertools
def pairsum(numbers, k):
return list(itertools.ifilter(lambda t: t[0] != t[1] and t[0] + t[1] == k, ((i, j) for i, j in numbers)))
But this is less readable in my opinion.
Your second program can be optimized using a set. If you recall from any discrete mathematics you may have learned in grade school or university, a set is a collection of unique elements - in other words, a set has no duplicate elements.
def dedup(mystring):
return set(mystring)
The algorithm to find the unique elements of a collection is generally going to be O(n^2) in time if it is O(1) in space - if you allow yourself to allocate more memory, you can use a Binary Search Tree to reduce the time complexity to O(n log n), which is likely how Python sets are implemented.
Your solution took O(n^2) time but also O(n) space, because you created a new list which could, if the input was already a string with only unique elements, take up the same amount of space - and, for every character in the string, you iterated over the output. That's essentially O(n^2) (although I think it's actually O(n*m), but whatever). I hope you see why this is. Read the Binary Search Tree article to see how it improves your code. I don't want to re-implement one again... freshman year was so grueling!

The key to optimization is basically to figure out a way to make the code do less work, in terms of the total number of primitive steps that needs to be performed. Code that employs control structures like nested loops quickly contributes to the number of primitive steps needed. Optimization is therefore often about replacing loops iterating over the a full list with something more clever.
I had to change the unoptimized pairsum() method sligtly to make it usable:
def pairsum(numbers, k):
"""
Write a function that returns two values in numbers whose sum is K
"""
for i in numbers:
for j in numbers:
if i != j:
if i+j == k:
return i,j
Here we see two loops, one nested inside the other. When describing the time complexity of a method like this, we often say that it is O(n²). Since when the length of the numbers array passed in grows proportional to n, then the number of primitive steps grows proportional to n². Specifically, the i+j == k conditional is evaluated exactly len(number)**2 times.
The clever thing we can do here is to presort the array at the cost of O(n log(n)) which allows us to hone in on the right answer by evaluating each element of the sorted array at most one time.
def fast_pairsum(numbers, k):
sortedints = sorted(numbers)
low = 0
high = len(numbers) - 1
i = sortedints[0]
j = sortedints[-1]
while low < high:
diff = i + j - k
if diff > 0:
# Too high, let's lower
high -= 1
j = sortedints[high]
elif diff < 0:
# Too low, let's increase.
low += 1
i = sortedints[low]
else:
# Just right
return i, j
raise Exception('No solution')
These kinds of optimization only begin to really matter when the size of the problem becomes large. On my machine the break-even point between pairsum() and fast_pairsum() is with a numbers array containing 13 integers. For smaller arrays pairsum() is faster, and for larger arrays fast_pairsum() is faster. As the size grows fast_pairsum() becomes drastically faster than the unoptimized pairsum().
The clever thing to do for dedup() is to avoid having to linearly scan through the output list to find out if you've already seen a character. This can be done by storing information about which characters you've seen in a set, which has O(log(n)) look-up cost, rather than the O(n) look-up cost of a regular list.
With the outer loop, the total cost becomes O(n log(n)) rather than O(n²).
def fast_dedup(string):
# if we didn't care about the order of the characters in the
# returned string we could simply do
# return set(string)
seen = set()
output = []
seen_add = seen.add
output_append = output.append
for i in string:
if i not in seen:
seen_add(i)
output_append(i)
return output
On my machine the break-even point between dedup() and fast_dedup() is with a string of length 30.
The fast_dedup() method also shows another simple optimization trick: Moving as much of the code out of the loop bodies as possible. Since looking up the add() and append() members in the seen and output objects takes time, it is cheaper to do it once outside the loop bodies and store references to the members in variables that is used repeatedly inside the loop bodies.

To properly optimize Python, one needs to find a good algorithm for the problem and a Python idiom close to that algorithm. Your pairsum example is a good case. First, your implementation appears wrong — numbers is most likely a sequence of numbers, not a sequence of pairs of numbers. Thus a naive implementation would look like this:
def pairsum(numbers, k)
"""Write a function that returns two values in numbers whose sum is K"""
for i in numbers:
for j in numbers:
if i != j and i + j != k:
return i, j
This will perform n^2 iterations, n being the length of numbers. For small ns this is not a problem, but once n gets into hundreds, the nested loops will become visibly slow, and once n gets into thousands, they will become unusable.
An optimization would be to recognize the difference between the inner and the outer loops: the outer loop traverses over numbers exactly once, and is unavoidable. The inner loop, however, is only used to verify that the other number (which has to be k - i) is actually present. This is a mere lookup, which can be made extremely fast by using a dict, or even better, a set:
def pairsum(numbers, k)
"""Write a function that returns two values in numbers whose sum is K"""
numset = set(numbers)
for i in numbers:
if k - i in numset:
return i, k - i
This is not only faster by a constant because we're using a built-in operation (set lookup) instead of a Python-coded loop. It actually does less work because set has a smarter algorithm of doing the lookup, it performs it in constant time.
Optimizing dedup in the analogous fashion is left as an excercise for the reader.

Your string one, order preserving is most easily and should be fairly efficient written as:
from collections import OrderedDict
new_string = ''.join(OrderedDict.fromkeys(old_string))