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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

Fastest way to return duplicate element in list and also find missing element in list?

So my code is as shown below. Input is a list with exactly one duplicate item and one missing item.The answer is a list of two elements long ,first of which is the duplicate element in the list and second the missing element in the list in the range 1 to n.
Example =[1,4,2,5,1] answer=[1,3]
The code below works.
Am , I wrong about the complexity being O(n) and is there any faster way of achieving this in Python?
Also, is there any way I can do this without using extra space.
Note:The elements may be of the order 10^5 or larger
n = max(A)
answer = []
seen = set()
for i in A:
if i in seen:
answer.append(i)
else:
seen.add(i)
for i in xrange(1,n):
if i not in A:
answer.append(i)
print ans

You are indeed correct the complexity of this algorithm is O(n), which is the best you can achieve. You can try to optimize it by aborting the search as soon as you finish the duplicate value. But worst case your duplicate is at the back of the list and you still need to traverse it completely.
The use of hashing (your use of a set) is a good solution. There are a lot other approaches, for instance the use of Counters. But this won't change the assymptotic complexity of the algorithm.
As #Emisor advices, you can leverage the information that you have a list with 1 duplicate and 1 missing value. As you might know if you would have a list with no duplicate and no missing value, summing up all elements of the list would result in 1+2+3+..+n, which can be rewritten in the mathematical equivalent (n*n+1)/2
When you've discovered the duplicate value, you can calculate the missing value, without having to perform:
for i in xrange(1,n):
if i not in A:
answer.append(i)
Since you know the sum if all values would be present: total = (n*n+1)/2) = 15, and you know which value is duplicated. By taking the sum of the array A = [1,4,2,5,1] which is 13 and removing the duplicated value 1, results in 12.
Taking the calculated total and subtracting the calculated 12from it results in 3.
This all can be written in a single line:
(((len(A)+1)*(len(A)+2))/2)-sum(A)-duplicate

Slight optimization (i think)
def lalala2(A):
_max = 0
_sum = 0
seen = set()
duplicate = None
for i in A:
_sum += i
if _max < i:
_max = i
if i in seen:
duplicate = i
elif duplicate is None:
seen.add(i)
missing = -_sum + duplicate + (_max*(_max + 1)/2) # This last term means the sum of every number from 1 to N
return [duplicate , missing]
Looks a bit uglier, and i'm doing stuff like sum() and max() on my own instead of relying on Python's tools. But with this way, we only check every element once. Also, It'll stop adding stuff to the set once it's found the duplicate since it can calculate the missing element from it, once it knows the max

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))

What is a typical way to add a reverse feature to an insertion sort?

I wrote the following insertion sort algorithm
def insertionSort(L, reverse=False):
for j in xrange(1,len(L)):
valToInsert = L[j]
i=j-1
while i>=0 and L[i] > valToInsert:
L[i+1] = L[i]
i-=1
L[i+1] = valToInsert
return L
Edit: All you need to do is change the final > to < to get it to work in reverse.
However, what do most people do in these situations? Write the algorithm twice in two if-statements, one where it's > and the other where it's < instead? What is the "correct" way to typically handle these kinds of scenarios where the change is minor but it simply changes the nature of the loop/code entirely?
I know this question is a little subjective.

You could use a variable for the less-than operator:
import operator
def insertionSort(L, reverse=False):
lt = operator.gt if reverse else operator.lt
for j in xrange(1,len(L)):
valToInsert = L[j]
i = j-1
while 0 <= i and lt(valToInsert, L[i]):
L[i+1] = L[i]
i -= 1
L[i+1] = valToInsert
return L

Option 1:
def insertionSort(L, reverse=False):
# loop is the same...
if reverse:
L.reverse()
return L
Option 2:
def insertionSort(L, reverse=False):
if reverse:
cmpfunc = lambda a, b: cmp(b, a)
else:
cmpfunc = cmp
for j in xrange(1,len(L)):
valToInsert = L[j]
i=j-1
while i>=0 and cmpfunc(L[i], valToInsert) > 0:
L[i+1] = L[i]
i-=1
L[i+1] = valToInsert
return L

You'll probably notice that sorted and list.sort and all other functions that do any kind of potentially-decorated processing have a key parameter, and those that specifically do ordering also have a reverse parameter. (The Sorting Mini-HOWTO covers this.)
So, you can look at how they're implemented. Unfortunately, in CPython, all of this stuff is implemented in C. Plus, it uses a custom algorithm called "timsort" (described in listsort.txt). But I think can explain the key parts here, since it's blindingly simple. The list.sort code is separate from the sorted code, and they're both spread out over a slew of functions. But if you just look at the top-level function listsort, you can see how it handles the reverse flag:
1982 /* Reverse sort stability achieved by initially reversing the list,
1983 applying a stable forward sort, then reversing the final result. */
1984 if (reverse) {
1985 if (keys != NULL)
1986 reverse_slice(&keys[0], &keys[saved_ob_size]);
1987 reverse_slice(&saved_ob_item[0], &saved_ob_item[saved_ob_size]);
1988 }
Why reverse the list at the start as well as the end? Well, in the case where the list is nearly-sorted in the first place, many sort algorithms—including both timsort and your insertion sort—will do a lot better starting in the right order than in backward order. Yes, it wastes an O(N) reverse call, but you're already doing one of those—and, since any sort algorithm is at least O(N log N), and yours is specifically O(N^2), this doesn't make it algorithmically worse. Of course for smallish N, and a better sort, and a list in random order, this wasted 2N is pretty close to N log N, so it can make a difference in practice. It'll be a difference that vanishes as N gets huge, but if you're sorting millions of smallish lists, rather than a few huge ones, it might be worth worrying about.
Second, notice that it does the reversing by creating a reverse slice. This, at least potentially, could be optimized by referencing the original list object with __getitem__ in reverse order, meaning the two reversals are actually O(1). The simplest way to do this is to literally create a reverse slice: lst[::-1]. Unfortunately, this actually creates a new reversed list, so timsort includes its own custom reverse-slice object. But you can do the same thing in Python by creating a ReversedList class.
This probably won't actually be faster in CPython, because the cost of the extra function calls is probably high enough to swamp the differences. But you're complaining about the algorithmic cost of the two reverse calls, and this solves the problem, in effectively the same way that the built-in sort functions do.
You can also look at how PyPy does it. Its list is implemented in listobject.py. It delegates to one of a few different Strategy classes depending on what the list contains, but if you look over all of the strategies (except the ones that have nothing to do), they basically do the same thing: sort the list, then reverse it.
So, it's good enough for CPython, and for PyPy… it's probably good enough for you.

Question on a solution from Google python class day

Hey,
I'm trying to learn a bit about Python so I decided to follow Google's tutorial. Anyway I had a question regarding one of their solution for an exercise.
Where I did it like this way.
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
return sorted(list1 + list2)
However they did it in a more complicated way. So is Google's solution quicker? Because I noticed in the comment lines that the solution should work in "linear" time, which mine probably isn't?
This is their solution
def linear_merge(list1, list2):
# +++your code here+++
# LAB(begin solution)
result = []
# Look at the two lists so long as both are non-empty.
# Take whichever element [0] is smaller.
while len(list1) and len(list2):
if list1[0] < list2[0]:
result.append(list1.pop(0))
else:
result.append(list2.pop(0))
# Now tack on what's left
result.extend(list1)
result.extend(list2)
return result

this could be another soln?
#
def linear_merge(list1, list2):
tmp = []
while len(list1) and len(list2):
#print list1[-1],list2[-1]
if list1[-1] > list2[-1]:
tmp.append(list1.pop())
else:
tmp.append(list2.pop())
#print "tmp = ",tmp
#print list1,list2
tmp = tmp + list1
tmp = tmp + list2
tmp.reverse()
return tmp

Yours is not linear, but that doesn't mean it's slower. Algorithmic complexity ("big-oh notation") is often only a rough guide and always only tells one part of the story.
However, theirs isn't linear either, though it may appear to be at first blush. Popping from a list requires moving all later items, so popping from the front requires moving all remaining elements.
It is a good exercise to think about how to make this O(n). The below is in the same spirit as the given solution, but avoids its pitfalls while generalizing to more than 2 lists for the sake of exercise. For exactly 2 lists, you could remove the heap handling and simply test which next item is smaller.
import heapq
def iter_linear_merge(*args):
"""Yield non-decreasing items from sorted a and b."""
# Technically, [1, 1, 2, 2] isn't an "increasing" sequence,
# but it is non-decreasing.
nexts = []
for x in args:
x = iter(x)
for n in x:
heapq.heappush(nexts, (n, x))
break
while len(nexts) >= 2:
n, x = heapq.heappop(nexts)
yield n
for n in x:
heapq.heappush(nexts, (n, x))
break
if nexts: # Degenerate case of the heap, not strictly required.
n, x = nexts[0]
yield n
for n in x:
yield n
Instead of the last if-for, the while loop condition could be changed to just "nexts", but it is probably worthwhile to specially handle the last remaining iterator.
If you want to strictly return a list instead of an iterator:
def linear_merge(*args):
return list(iter_linear_merge(*args))

With mostly-sorted data, timsort approaches linear. Also, your code doesn't have to screw around with the lists themselves. Therefore, your code is possibly just a bit faster.
But that's what timing is for, innit?

I think the issue here is that the tutorial is illustrating how to implement a well-known algorithm called 'merge' in Python. The tutorial is not expecting you to actually use a library sorting function in the solution.
sorted() is probably O(nlgn); then your solution cannot be linear in the worst case.
It is important to understand how merge() works because it is useful in many other algorithms. It exploits the fact the input lists are individually sorted, moving through each list sequentially and selecting the smallest option. The remaining items are appended at the end.
The question isn't which is 'quicker' for a given input case but about which algorithm is more complex.
There are hybrid variations of merge-sort which fall back on another sorting algorithm once the input list size drops below a certain threshold.