If any two items in a list is equal to a given number
I just subscribed to the daily coding problems and received my first today.
The problem is:
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.
This is what I came up with:
def anyequalto(x, y):
for i in x:
if y - i in x:
return True
anyequalto([10, 15, 3, 7], 17)
I was wondering if anyone had any notes or alternatives as I'm very new to Python and programming.
python beginner k-sum
add a comment |
I just subscribed to the daily coding problems and received my first today.
The problem is:
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.
This is what I came up with:
def anyequalto(x, y):
for i in x:
if y - i in x:
return True
anyequalto([10, 15, 3, 7], 17)
I was wondering if anyone had any notes or alternatives as I'm very new to Python and programming.
python beginner k-sum
Your revised code is actually less efficient. There is no point in creating a new list each time.
– Solomon Ucko
Dec 3 at 20:11
This puzzle must be making the rounds, because someone posted it on this site a few days ago: codereview.stackexchange.com/questions/208138/…
– Eric Lippert
Dec 3 at 22:33
4
Please do not add new code to your question. Your new code has a few aspects that deserve to be reviewed. You can do so in a different follow-up question. Please refer to the rules for that.
– Josay
Dec 3 at 22:43
1
Your revised code doesn't work correctly with duplicate values. contains_pair_totalling([5, 5], 10) returns false. Sets don't contain duplicates, so your remove call breaks this case.
– jaxad0127
Dec 3 at 22:50
1
The original code didn’t work with the sum double any list value, either.anyequalto([5], 10)
returnsTrue
!
– AJNeufeld
Dec 3 at 22:59
add a comment |
I just subscribed to the daily coding problems and received my first today.
The problem is:
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.
This is what I came up with:
def anyequalto(x, y):
for i in x:
if y - i in x:
return True
anyequalto([10, 15, 3, 7], 17)
I was wondering if anyone had any notes or alternatives as I'm very new to Python and programming.
python beginner k-sum
I just subscribed to the daily coding problems and received my first today.
The problem is:
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.
This is what I came up with:
def anyequalto(x, y):
for i in x:
if y - i in x:
return True
anyequalto([10, 15, 3, 7], 17)
I was wondering if anyone had any notes or alternatives as I'm very new to Python and programming.
python beginner k-sum
python beginner k-sum
edited Dec 4 at 7:02
200_success
128k15150412
128k15150412
asked Dec 3 at 17:44
SimonJ
413
413
Your revised code is actually less efficient. There is no point in creating a new list each time.
– Solomon Ucko
Dec 3 at 20:11
This puzzle must be making the rounds, because someone posted it on this site a few days ago: codereview.stackexchange.com/questions/208138/…
– Eric Lippert
Dec 3 at 22:33
4
Please do not add new code to your question. Your new code has a few aspects that deserve to be reviewed. You can do so in a different follow-up question. Please refer to the rules for that.
– Josay
Dec 3 at 22:43
1
Your revised code doesn't work correctly with duplicate values. contains_pair_totalling([5, 5], 10) returns false. Sets don't contain duplicates, so your remove call breaks this case.
– jaxad0127
Dec 3 at 22:50
1
The original code didn’t work with the sum double any list value, either.anyequalto([5], 10)
returnsTrue
!
– AJNeufeld
Dec 3 at 22:59
add a comment |
Your revised code is actually less efficient. There is no point in creating a new list each time.
– Solomon Ucko
Dec 3 at 20:11
This puzzle must be making the rounds, because someone posted it on this site a few days ago: codereview.stackexchange.com/questions/208138/…
– Eric Lippert
Dec 3 at 22:33
4
Please do not add new code to your question. Your new code has a few aspects that deserve to be reviewed. You can do so in a different follow-up question. Please refer to the rules for that.
– Josay
Dec 3 at 22:43
1
Your revised code doesn't work correctly with duplicate values. contains_pair_totalling([5, 5], 10) returns false. Sets don't contain duplicates, so your remove call breaks this case.
– jaxad0127
Dec 3 at 22:50
1
The original code didn’t work with the sum double any list value, either.anyequalto([5], 10)
returnsTrue
!
– AJNeufeld
Dec 3 at 22:59
Your revised code is actually less efficient. There is no point in creating a new list each time.
– Solomon Ucko
Dec 3 at 20:11
Your revised code is actually less efficient. There is no point in creating a new list each time.
– Solomon Ucko
Dec 3 at 20:11
This puzzle must be making the rounds, because someone posted it on this site a few days ago: codereview.stackexchange.com/questions/208138/…
– Eric Lippert
Dec 3 at 22:33
This puzzle must be making the rounds, because someone posted it on this site a few days ago: codereview.stackexchange.com/questions/208138/…
– Eric Lippert
Dec 3 at 22:33
4
4
Please do not add new code to your question. Your new code has a few aspects that deserve to be reviewed. You can do so in a different follow-up question. Please refer to the rules for that.
– Josay
Dec 3 at 22:43
Please do not add new code to your question. Your new code has a few aspects that deserve to be reviewed. You can do so in a different follow-up question. Please refer to the rules for that.
– Josay
Dec 3 at 22:43
1
1
Your revised code doesn't work correctly with duplicate values. contains_pair_totalling([5, 5], 10) returns false. Sets don't contain duplicates, so your remove call breaks this case.
– jaxad0127
Dec 3 at 22:50
Your revised code doesn't work correctly with duplicate values. contains_pair_totalling([5, 5], 10) returns false. Sets don't contain duplicates, so your remove call breaks this case.
– jaxad0127
Dec 3 at 22:50
1
1
The original code didn’t work with the sum double any list value, either.
anyequalto([5], 10)
returns True
!– AJNeufeld
Dec 3 at 22:59
The original code didn’t work with the sum double any list value, either.
anyequalto([5], 10)
returns True
!– AJNeufeld
Dec 3 at 22:59
add a comment |
6 Answers
6
active
oldest
votes
There's a failing case you might have missed. If the target number is exactly twice the value of one of the entries, then you'll wrongly return true.
Test case:
anyequalto([5], 10)
Besides that, try to think of a better name for the function. Perhaps contains_pair_totalling()
as a start?
Didn't think of that. I'd fix this by taking the number it's checking out of the list, then checking it and adding it back for the next run. Any thoughts?
– SimonJ
Dec 3 at 18:27
Yes, that would be a correct fix. It might not be efficient for very large inputs; I don't know whether that's a concern.
– Toby Speight
Dec 4 at 8:11
1
@SimonJ you wouldn't need to add it back - if it is part of a pair summing tok
then there won't be a next run, and if it isn't then it won't affect future runs.
– Especially Lime
Dec 4 at 8:52
add a comment |
Return value
Adding a few tests, we have:
print(anyequalto([10, 15, 3, 7], 17))
print(anyequalto([10, 15, 3, 7], 18))
print(anyequalto([10, 15, 3, 7], 19))
giving
True
True
None
The None
value seems a bit unexpected to me. We'd probably want False
to be returned in that particular case.
Also, even if you expect None
to be returned in that case, the Python Style Guide recommends being explicit for that (emphasis is mine):
Be consistent in return statements. Either all return statements in a
function should return an expression, or none of them should. If any
return statement returns an expression, any return statements where no
value is returned should explicitly state this as return None, and an
explicit return statement should be present at the end of the function
(if reachable).
Names, documentation and tests
The variables names could be clearer.
Also, the function behavior can be described in a docstring.
Finally, it could be worth writing tests for it.
You'd get something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
for i in num_lst:
if n - i in num_lst:
return True
return False
TESTS = [
# Random tests cases
([10, 15, 3, 7], 17, True),
([10, 15, 3, 7], 18, True),
([10, 15, 3, 7], 19, False),
# Edge case
(, 0, False),
(, 1, False),
# Same value
([5, 5], 10, True),
([5], 10, True),
# Zero
([5, 0], 0, True),
([5, 0], 5, True),
([5, 0], 2, False),
]
for (lst, n, expected_res) in TESTS:
res = anyequalto(lst, n)
if res != expected_res:
print("Error with ", lst, n, "got", res, "expected", expected_res)
Data structure
At the moment, you can iterate on the list (via the in
check) for each element of the list. This leads to an O(n²)
behavior.
You can makes t hings more efficient by building a set to perform the in
test in constant time and have an overall O(n)
behavior.
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
for i in num_set:
if n - i in num_set:
return True
return False
Using the Python toolbox
The solution you are using could be written in a more concise and efficient way using the all
or any
builtin.
You have something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
return any(n - i in num_set for i in num_set)
The any builtin is something I hadn't seen yet. Since someone pointed out that if the value of n is the exact double of an item in num_lst. How would that work out with any? If possible.
– SimonJ
Dec 3 at 18:37
anyequalto([5], 10)
should beFalse
, so you cannot simply createnum_set
like this. You need to iterate over the numbers, check ifn-i
is in the set and only then addi
to the set.
– Eric Duminil
Dec 4 at 10:30
And[5, 0], 0
should beFalse
.
– Eric Duminil
Dec 4 at 10:41
I relied on the original code behavior and made sure I didn't change it while making it more explicit via the corresponding test cases. Maybe I should have hilighted it as a bug (I do not quite remember how the problem was expressed initially).
– Josay
Dec 4 at 10:46
Also, if I were to re-implement it based on that new behavior, I'd probably just use a Counter...
– Josay
Dec 4 at 10:51
|
show 1 more comment
Here's a fix to the edge case. We want to check explicitly that we have a valid solution, which means that we want something akin to any2
instead of any
(since if we have just a single match then we have the problem case).
def anyequalto(numbers, k):
number_set = set(numbers)
solutions = [num for num in numbers
if k - num in number_set]
return len(solutions) > 1
This fixes the edge case and still retains O(n)
runtime since it uses a set
.
>>> anyequalto([5], 10)
False
Aside
Right now this produces a list with the list comprehension used to generate solutions
which one might argue is needlessly inefficient. I couldn't think of a stylistically good way to assert that a generator has more than one element aside from checking for StopIteration
which I think is kinda clunky.
next(iterator, sentinel)
will returnsentinel
if the iterator is exhausted.
– spectras
Dec 4 at 3:16
Nice, it also works withanyequalto([5, 5], 10) # True
.
– Eric Duminil
Dec 4 at 19:25
add a comment |
- As others have mentioned, your code could be more efficient with set lookup.
- It should return
False
with[5]
and10
as input parameters. - Python method names are written in
snake_case
. - Python is a dynamic language, which means method names and parameter names are very important. They make it easier to read your code and understand it again 6 months after having written it.
x
sounds like an unknown object or a float, not like a list of integers.
Here's a possible way to write the method:
def contains_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return True
unique_numbers.add(number)
return False
It outputs:
contains_pair_with_given_sum([5, 10, 7], 12)
# True
contains_pair_with_given_sum([5], 10)
# False
You could also use the fact that a pair of integers is truthy in Python and return the pair when found:
def find_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return (number, desired_sum - number)
unique_numbers.add(number)
It outputs:
find_pair_with_given_sum([5, 10, 7], 12)
# (7, 5)
find_pair_with_given_sum([5], 10)
# None
add a comment |
for a much faster way to check if a number is in some container, use a
set()
:
anyequalto({10, 15, 3, 7}, 17)
. even if you have to convert from a list first,
it will still be faster to do that once rather than linearly search the list each
iteration. This will make your function O(n) rather than O(n^2)choose more expressive variable names that
x
andy
. something like
numbers
andk
j
If you want to write this in a more functional style, you could do:
def anyequalto(numbers, k):
numbers = set(numbers)
return any((k-num) in numbers
for num in numbers)
Because any will terminate early if the generator produces a true
value, this will be very similar in speed to an iterative solution.
1
It should beany(k in numbers for num in numbers)
, notany(for num in numbers k in numbers)
.
– Graipher
Dec 3 at 18:10
anyequalto([5], 10)
should beFalse
.
– Eric Duminil
Dec 4 at 10:32
add a comment |
A simple solution would use any
and itertools.combinations
from itertools import combinations
def anytwoequalto(numbers, k):
return any(((i + j) == k) for i, j in combinations(numbers, 2))
itertools.combinations
iterates over the given object returning tuples with the given number of items in them with no repetitions. eg.
combinations('ABCD', 2) =>
(('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D'))
any
returns True
if any of the expressions ((i + j) == k)
evaluate to True
If you wish to consider that an item can be added to itself, use combinations_with_replacement
instead.
Note: depending on the version of Python you are using, you may need to add extra parentheses (()
) around the expression inside the any
call.
Hope this makes sense.
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
There's a failing case you might have missed. If the target number is exactly twice the value of one of the entries, then you'll wrongly return true.
Test case:
anyequalto([5], 10)
Besides that, try to think of a better name for the function. Perhaps contains_pair_totalling()
as a start?
Didn't think of that. I'd fix this by taking the number it's checking out of the list, then checking it and adding it back for the next run. Any thoughts?
– SimonJ
Dec 3 at 18:27
Yes, that would be a correct fix. It might not be efficient for very large inputs; I don't know whether that's a concern.
– Toby Speight
Dec 4 at 8:11
1
@SimonJ you wouldn't need to add it back - if it is part of a pair summing tok
then there won't be a next run, and if it isn't then it won't affect future runs.
– Especially Lime
Dec 4 at 8:52
add a comment |
There's a failing case you might have missed. If the target number is exactly twice the value of one of the entries, then you'll wrongly return true.
Test case:
anyequalto([5], 10)
Besides that, try to think of a better name for the function. Perhaps contains_pair_totalling()
as a start?
Didn't think of that. I'd fix this by taking the number it's checking out of the list, then checking it and adding it back for the next run. Any thoughts?
– SimonJ
Dec 3 at 18:27
Yes, that would be a correct fix. It might not be efficient for very large inputs; I don't know whether that's a concern.
– Toby Speight
Dec 4 at 8:11
1
@SimonJ you wouldn't need to add it back - if it is part of a pair summing tok
then there won't be a next run, and if it isn't then it won't affect future runs.
– Especially Lime
Dec 4 at 8:52
add a comment |
There's a failing case you might have missed. If the target number is exactly twice the value of one of the entries, then you'll wrongly return true.
Test case:
anyequalto([5], 10)
Besides that, try to think of a better name for the function. Perhaps contains_pair_totalling()
as a start?
There's a failing case you might have missed. If the target number is exactly twice the value of one of the entries, then you'll wrongly return true.
Test case:
anyequalto([5], 10)
Besides that, try to think of a better name for the function. Perhaps contains_pair_totalling()
as a start?
answered Dec 3 at 18:16
Toby Speight
23.3k639111
23.3k639111
Didn't think of that. I'd fix this by taking the number it's checking out of the list, then checking it and adding it back for the next run. Any thoughts?
– SimonJ
Dec 3 at 18:27
Yes, that would be a correct fix. It might not be efficient for very large inputs; I don't know whether that's a concern.
– Toby Speight
Dec 4 at 8:11
1
@SimonJ you wouldn't need to add it back - if it is part of a pair summing tok
then there won't be a next run, and if it isn't then it won't affect future runs.
– Especially Lime
Dec 4 at 8:52
add a comment |
Didn't think of that. I'd fix this by taking the number it's checking out of the list, then checking it and adding it back for the next run. Any thoughts?
– SimonJ
Dec 3 at 18:27
Yes, that would be a correct fix. It might not be efficient for very large inputs; I don't know whether that's a concern.
– Toby Speight
Dec 4 at 8:11
1
@SimonJ you wouldn't need to add it back - if it is part of a pair summing tok
then there won't be a next run, and if it isn't then it won't affect future runs.
– Especially Lime
Dec 4 at 8:52
Didn't think of that. I'd fix this by taking the number it's checking out of the list, then checking it and adding it back for the next run. Any thoughts?
– SimonJ
Dec 3 at 18:27
Didn't think of that. I'd fix this by taking the number it's checking out of the list, then checking it and adding it back for the next run. Any thoughts?
– SimonJ
Dec 3 at 18:27
Yes, that would be a correct fix. It might not be efficient for very large inputs; I don't know whether that's a concern.
– Toby Speight
Dec 4 at 8:11
Yes, that would be a correct fix. It might not be efficient for very large inputs; I don't know whether that's a concern.
– Toby Speight
Dec 4 at 8:11
1
1
@SimonJ you wouldn't need to add it back - if it is part of a pair summing to
k
then there won't be a next run, and if it isn't then it won't affect future runs.– Especially Lime
Dec 4 at 8:52
@SimonJ you wouldn't need to add it back - if it is part of a pair summing to
k
then there won't be a next run, and if it isn't then it won't affect future runs.– Especially Lime
Dec 4 at 8:52
add a comment |
Return value
Adding a few tests, we have:
print(anyequalto([10, 15, 3, 7], 17))
print(anyequalto([10, 15, 3, 7], 18))
print(anyequalto([10, 15, 3, 7], 19))
giving
True
True
None
The None
value seems a bit unexpected to me. We'd probably want False
to be returned in that particular case.
Also, even if you expect None
to be returned in that case, the Python Style Guide recommends being explicit for that (emphasis is mine):
Be consistent in return statements. Either all return statements in a
function should return an expression, or none of them should. If any
return statement returns an expression, any return statements where no
value is returned should explicitly state this as return None, and an
explicit return statement should be present at the end of the function
(if reachable).
Names, documentation and tests
The variables names could be clearer.
Also, the function behavior can be described in a docstring.
Finally, it could be worth writing tests for it.
You'd get something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
for i in num_lst:
if n - i in num_lst:
return True
return False
TESTS = [
# Random tests cases
([10, 15, 3, 7], 17, True),
([10, 15, 3, 7], 18, True),
([10, 15, 3, 7], 19, False),
# Edge case
(, 0, False),
(, 1, False),
# Same value
([5, 5], 10, True),
([5], 10, True),
# Zero
([5, 0], 0, True),
([5, 0], 5, True),
([5, 0], 2, False),
]
for (lst, n, expected_res) in TESTS:
res = anyequalto(lst, n)
if res != expected_res:
print("Error with ", lst, n, "got", res, "expected", expected_res)
Data structure
At the moment, you can iterate on the list (via the in
check) for each element of the list. This leads to an O(n²)
behavior.
You can makes t hings more efficient by building a set to perform the in
test in constant time and have an overall O(n)
behavior.
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
for i in num_set:
if n - i in num_set:
return True
return False
Using the Python toolbox
The solution you are using could be written in a more concise and efficient way using the all
or any
builtin.
You have something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
return any(n - i in num_set for i in num_set)
The any builtin is something I hadn't seen yet. Since someone pointed out that if the value of n is the exact double of an item in num_lst. How would that work out with any? If possible.
– SimonJ
Dec 3 at 18:37
anyequalto([5], 10)
should beFalse
, so you cannot simply createnum_set
like this. You need to iterate over the numbers, check ifn-i
is in the set and only then addi
to the set.
– Eric Duminil
Dec 4 at 10:30
And[5, 0], 0
should beFalse
.
– Eric Duminil
Dec 4 at 10:41
I relied on the original code behavior and made sure I didn't change it while making it more explicit via the corresponding test cases. Maybe I should have hilighted it as a bug (I do not quite remember how the problem was expressed initially).
– Josay
Dec 4 at 10:46
Also, if I were to re-implement it based on that new behavior, I'd probably just use a Counter...
– Josay
Dec 4 at 10:51
|
show 1 more comment
Return value
Adding a few tests, we have:
print(anyequalto([10, 15, 3, 7], 17))
print(anyequalto([10, 15, 3, 7], 18))
print(anyequalto([10, 15, 3, 7], 19))
giving
True
True
None
The None
value seems a bit unexpected to me. We'd probably want False
to be returned in that particular case.
Also, even if you expect None
to be returned in that case, the Python Style Guide recommends being explicit for that (emphasis is mine):
Be consistent in return statements. Either all return statements in a
function should return an expression, or none of them should. If any
return statement returns an expression, any return statements where no
value is returned should explicitly state this as return None, and an
explicit return statement should be present at the end of the function
(if reachable).
Names, documentation and tests
The variables names could be clearer.
Also, the function behavior can be described in a docstring.
Finally, it could be worth writing tests for it.
You'd get something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
for i in num_lst:
if n - i in num_lst:
return True
return False
TESTS = [
# Random tests cases
([10, 15, 3, 7], 17, True),
([10, 15, 3, 7], 18, True),
([10, 15, 3, 7], 19, False),
# Edge case
(, 0, False),
(, 1, False),
# Same value
([5, 5], 10, True),
([5], 10, True),
# Zero
([5, 0], 0, True),
([5, 0], 5, True),
([5, 0], 2, False),
]
for (lst, n, expected_res) in TESTS:
res = anyequalto(lst, n)
if res != expected_res:
print("Error with ", lst, n, "got", res, "expected", expected_res)
Data structure
At the moment, you can iterate on the list (via the in
check) for each element of the list. This leads to an O(n²)
behavior.
You can makes t hings more efficient by building a set to perform the in
test in constant time and have an overall O(n)
behavior.
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
for i in num_set:
if n - i in num_set:
return True
return False
Using the Python toolbox
The solution you are using could be written in a more concise and efficient way using the all
or any
builtin.
You have something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
return any(n - i in num_set for i in num_set)
The any builtin is something I hadn't seen yet. Since someone pointed out that if the value of n is the exact double of an item in num_lst. How would that work out with any? If possible.
– SimonJ
Dec 3 at 18:37
anyequalto([5], 10)
should beFalse
, so you cannot simply createnum_set
like this. You need to iterate over the numbers, check ifn-i
is in the set and only then addi
to the set.
– Eric Duminil
Dec 4 at 10:30
And[5, 0], 0
should beFalse
.
– Eric Duminil
Dec 4 at 10:41
I relied on the original code behavior and made sure I didn't change it while making it more explicit via the corresponding test cases. Maybe I should have hilighted it as a bug (I do not quite remember how the problem was expressed initially).
– Josay
Dec 4 at 10:46
Also, if I were to re-implement it based on that new behavior, I'd probably just use a Counter...
– Josay
Dec 4 at 10:51
|
show 1 more comment
Return value
Adding a few tests, we have:
print(anyequalto([10, 15, 3, 7], 17))
print(anyequalto([10, 15, 3, 7], 18))
print(anyequalto([10, 15, 3, 7], 19))
giving
True
True
None
The None
value seems a bit unexpected to me. We'd probably want False
to be returned in that particular case.
Also, even if you expect None
to be returned in that case, the Python Style Guide recommends being explicit for that (emphasis is mine):
Be consistent in return statements. Either all return statements in a
function should return an expression, or none of them should. If any
return statement returns an expression, any return statements where no
value is returned should explicitly state this as return None, and an
explicit return statement should be present at the end of the function
(if reachable).
Names, documentation and tests
The variables names could be clearer.
Also, the function behavior can be described in a docstring.
Finally, it could be worth writing tests for it.
You'd get something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
for i in num_lst:
if n - i in num_lst:
return True
return False
TESTS = [
# Random tests cases
([10, 15, 3, 7], 17, True),
([10, 15, 3, 7], 18, True),
([10, 15, 3, 7], 19, False),
# Edge case
(, 0, False),
(, 1, False),
# Same value
([5, 5], 10, True),
([5], 10, True),
# Zero
([5, 0], 0, True),
([5, 0], 5, True),
([5, 0], 2, False),
]
for (lst, n, expected_res) in TESTS:
res = anyequalto(lst, n)
if res != expected_res:
print("Error with ", lst, n, "got", res, "expected", expected_res)
Data structure
At the moment, you can iterate on the list (via the in
check) for each element of the list. This leads to an O(n²)
behavior.
You can makes t hings more efficient by building a set to perform the in
test in constant time and have an overall O(n)
behavior.
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
for i in num_set:
if n - i in num_set:
return True
return False
Using the Python toolbox
The solution you are using could be written in a more concise and efficient way using the all
or any
builtin.
You have something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
return any(n - i in num_set for i in num_set)
Return value
Adding a few tests, we have:
print(anyequalto([10, 15, 3, 7], 17))
print(anyequalto([10, 15, 3, 7], 18))
print(anyequalto([10, 15, 3, 7], 19))
giving
True
True
None
The None
value seems a bit unexpected to me. We'd probably want False
to be returned in that particular case.
Also, even if you expect None
to be returned in that case, the Python Style Guide recommends being explicit for that (emphasis is mine):
Be consistent in return statements. Either all return statements in a
function should return an expression, or none of them should. If any
return statement returns an expression, any return statements where no
value is returned should explicitly state this as return None, and an
explicit return statement should be present at the end of the function
(if reachable).
Names, documentation and tests
The variables names could be clearer.
Also, the function behavior can be described in a docstring.
Finally, it could be worth writing tests for it.
You'd get something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
for i in num_lst:
if n - i in num_lst:
return True
return False
TESTS = [
# Random tests cases
([10, 15, 3, 7], 17, True),
([10, 15, 3, 7], 18, True),
([10, 15, 3, 7], 19, False),
# Edge case
(, 0, False),
(, 1, False),
# Same value
([5, 5], 10, True),
([5], 10, True),
# Zero
([5, 0], 0, True),
([5, 0], 5, True),
([5, 0], 2, False),
]
for (lst, n, expected_res) in TESTS:
res = anyequalto(lst, n)
if res != expected_res:
print("Error with ", lst, n, "got", res, "expected", expected_res)
Data structure
At the moment, you can iterate on the list (via the in
check) for each element of the list. This leads to an O(n²)
behavior.
You can makes t hings more efficient by building a set to perform the in
test in constant time and have an overall O(n)
behavior.
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
for i in num_set:
if n - i in num_set:
return True
return False
Using the Python toolbox
The solution you are using could be written in a more concise and efficient way using the all
or any
builtin.
You have something like:
def anyequalto(num_lst, n):
"""Return True if 2 numbers from `num_lst` add up to n, False otherwise."""
num_set = set(num_lst)
return any(n - i in num_set for i in num_set)
answered Dec 3 at 18:11
Josay
24.9k13784
24.9k13784
The any builtin is something I hadn't seen yet. Since someone pointed out that if the value of n is the exact double of an item in num_lst. How would that work out with any? If possible.
– SimonJ
Dec 3 at 18:37
anyequalto([5], 10)
should beFalse
, so you cannot simply createnum_set
like this. You need to iterate over the numbers, check ifn-i
is in the set and only then addi
to the set.
– Eric Duminil
Dec 4 at 10:30
And[5, 0], 0
should beFalse
.
– Eric Duminil
Dec 4 at 10:41
I relied on the original code behavior and made sure I didn't change it while making it more explicit via the corresponding test cases. Maybe I should have hilighted it as a bug (I do not quite remember how the problem was expressed initially).
– Josay
Dec 4 at 10:46
Also, if I were to re-implement it based on that new behavior, I'd probably just use a Counter...
– Josay
Dec 4 at 10:51
|
show 1 more comment
The any builtin is something I hadn't seen yet. Since someone pointed out that if the value of n is the exact double of an item in num_lst. How would that work out with any? If possible.
– SimonJ
Dec 3 at 18:37
anyequalto([5], 10)
should beFalse
, so you cannot simply createnum_set
like this. You need to iterate over the numbers, check ifn-i
is in the set and only then addi
to the set.
– Eric Duminil
Dec 4 at 10:30
And[5, 0], 0
should beFalse
.
– Eric Duminil
Dec 4 at 10:41
I relied on the original code behavior and made sure I didn't change it while making it more explicit via the corresponding test cases. Maybe I should have hilighted it as a bug (I do not quite remember how the problem was expressed initially).
– Josay
Dec 4 at 10:46
Also, if I were to re-implement it based on that new behavior, I'd probably just use a Counter...
– Josay
Dec 4 at 10:51
The any builtin is something I hadn't seen yet. Since someone pointed out that if the value of n is the exact double of an item in num_lst. How would that work out with any? If possible.
– SimonJ
Dec 3 at 18:37
The any builtin is something I hadn't seen yet. Since someone pointed out that if the value of n is the exact double of an item in num_lst. How would that work out with any? If possible.
– SimonJ
Dec 3 at 18:37
anyequalto([5], 10)
should be False
, so you cannot simply create num_set
like this. You need to iterate over the numbers, check if n-i
is in the set and only then add i
to the set.– Eric Duminil
Dec 4 at 10:30
anyequalto([5], 10)
should be False
, so you cannot simply create num_set
like this. You need to iterate over the numbers, check if n-i
is in the set and only then add i
to the set.– Eric Duminil
Dec 4 at 10:30
And
[5, 0], 0
should be False
.– Eric Duminil
Dec 4 at 10:41
And
[5, 0], 0
should be False
.– Eric Duminil
Dec 4 at 10:41
I relied on the original code behavior and made sure I didn't change it while making it more explicit via the corresponding test cases. Maybe I should have hilighted it as a bug (I do not quite remember how the problem was expressed initially).
– Josay
Dec 4 at 10:46
I relied on the original code behavior and made sure I didn't change it while making it more explicit via the corresponding test cases. Maybe I should have hilighted it as a bug (I do not quite remember how the problem was expressed initially).
– Josay
Dec 4 at 10:46
Also, if I were to re-implement it based on that new behavior, I'd probably just use a Counter...
– Josay
Dec 4 at 10:51
Also, if I were to re-implement it based on that new behavior, I'd probably just use a Counter...
– Josay
Dec 4 at 10:51
|
show 1 more comment
Here's a fix to the edge case. We want to check explicitly that we have a valid solution, which means that we want something akin to any2
instead of any
(since if we have just a single match then we have the problem case).
def anyequalto(numbers, k):
number_set = set(numbers)
solutions = [num for num in numbers
if k - num in number_set]
return len(solutions) > 1
This fixes the edge case and still retains O(n)
runtime since it uses a set
.
>>> anyequalto([5], 10)
False
Aside
Right now this produces a list with the list comprehension used to generate solutions
which one might argue is needlessly inefficient. I couldn't think of a stylistically good way to assert that a generator has more than one element aside from checking for StopIteration
which I think is kinda clunky.
next(iterator, sentinel)
will returnsentinel
if the iterator is exhausted.
– spectras
Dec 4 at 3:16
Nice, it also works withanyequalto([5, 5], 10) # True
.
– Eric Duminil
Dec 4 at 19:25
add a comment |
Here's a fix to the edge case. We want to check explicitly that we have a valid solution, which means that we want something akin to any2
instead of any
(since if we have just a single match then we have the problem case).
def anyequalto(numbers, k):
number_set = set(numbers)
solutions = [num for num in numbers
if k - num in number_set]
return len(solutions) > 1
This fixes the edge case and still retains O(n)
runtime since it uses a set
.
>>> anyequalto([5], 10)
False
Aside
Right now this produces a list with the list comprehension used to generate solutions
which one might argue is needlessly inefficient. I couldn't think of a stylistically good way to assert that a generator has more than one element aside from checking for StopIteration
which I think is kinda clunky.
next(iterator, sentinel)
will returnsentinel
if the iterator is exhausted.
– spectras
Dec 4 at 3:16
Nice, it also works withanyequalto([5, 5], 10) # True
.
– Eric Duminil
Dec 4 at 19:25
add a comment |
Here's a fix to the edge case. We want to check explicitly that we have a valid solution, which means that we want something akin to any2
instead of any
(since if we have just a single match then we have the problem case).
def anyequalto(numbers, k):
number_set = set(numbers)
solutions = [num for num in numbers
if k - num in number_set]
return len(solutions) > 1
This fixes the edge case and still retains O(n)
runtime since it uses a set
.
>>> anyequalto([5], 10)
False
Aside
Right now this produces a list with the list comprehension used to generate solutions
which one might argue is needlessly inefficient. I couldn't think of a stylistically good way to assert that a generator has more than one element aside from checking for StopIteration
which I think is kinda clunky.
Here's a fix to the edge case. We want to check explicitly that we have a valid solution, which means that we want something akin to any2
instead of any
(since if we have just a single match then we have the problem case).
def anyequalto(numbers, k):
number_set = set(numbers)
solutions = [num for num in numbers
if k - num in number_set]
return len(solutions) > 1
This fixes the edge case and still retains O(n)
runtime since it uses a set
.
>>> anyequalto([5], 10)
False
Aside
Right now this produces a list with the list comprehension used to generate solutions
which one might argue is needlessly inefficient. I couldn't think of a stylistically good way to assert that a generator has more than one element aside from checking for StopIteration
which I think is kinda clunky.
answered Dec 3 at 19:53
cole
2614
2614
next(iterator, sentinel)
will returnsentinel
if the iterator is exhausted.
– spectras
Dec 4 at 3:16
Nice, it also works withanyequalto([5, 5], 10) # True
.
– Eric Duminil
Dec 4 at 19:25
add a comment |
next(iterator, sentinel)
will returnsentinel
if the iterator is exhausted.
– spectras
Dec 4 at 3:16
Nice, it also works withanyequalto([5, 5], 10) # True
.
– Eric Duminil
Dec 4 at 19:25
next(iterator, sentinel)
will return sentinel
if the iterator is exhausted.– spectras
Dec 4 at 3:16
next(iterator, sentinel)
will return sentinel
if the iterator is exhausted.– spectras
Dec 4 at 3:16
Nice, it also works with
anyequalto([5, 5], 10) # True
.– Eric Duminil
Dec 4 at 19:25
Nice, it also works with
anyequalto([5, 5], 10) # True
.– Eric Duminil
Dec 4 at 19:25
add a comment |
- As others have mentioned, your code could be more efficient with set lookup.
- It should return
False
with[5]
and10
as input parameters. - Python method names are written in
snake_case
. - Python is a dynamic language, which means method names and parameter names are very important. They make it easier to read your code and understand it again 6 months after having written it.
x
sounds like an unknown object or a float, not like a list of integers.
Here's a possible way to write the method:
def contains_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return True
unique_numbers.add(number)
return False
It outputs:
contains_pair_with_given_sum([5, 10, 7], 12)
# True
contains_pair_with_given_sum([5], 10)
# False
You could also use the fact that a pair of integers is truthy in Python and return the pair when found:
def find_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return (number, desired_sum - number)
unique_numbers.add(number)
It outputs:
find_pair_with_given_sum([5, 10, 7], 12)
# (7, 5)
find_pair_with_given_sum([5], 10)
# None
add a comment |
- As others have mentioned, your code could be more efficient with set lookup.
- It should return
False
with[5]
and10
as input parameters. - Python method names are written in
snake_case
. - Python is a dynamic language, which means method names and parameter names are very important. They make it easier to read your code and understand it again 6 months after having written it.
x
sounds like an unknown object or a float, not like a list of integers.
Here's a possible way to write the method:
def contains_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return True
unique_numbers.add(number)
return False
It outputs:
contains_pair_with_given_sum([5, 10, 7], 12)
# True
contains_pair_with_given_sum([5], 10)
# False
You could also use the fact that a pair of integers is truthy in Python and return the pair when found:
def find_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return (number, desired_sum - number)
unique_numbers.add(number)
It outputs:
find_pair_with_given_sum([5, 10, 7], 12)
# (7, 5)
find_pair_with_given_sum([5], 10)
# None
add a comment |
- As others have mentioned, your code could be more efficient with set lookup.
- It should return
False
with[5]
and10
as input parameters. - Python method names are written in
snake_case
. - Python is a dynamic language, which means method names and parameter names are very important. They make it easier to read your code and understand it again 6 months after having written it.
x
sounds like an unknown object or a float, not like a list of integers.
Here's a possible way to write the method:
def contains_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return True
unique_numbers.add(number)
return False
It outputs:
contains_pair_with_given_sum([5, 10, 7], 12)
# True
contains_pair_with_given_sum([5], 10)
# False
You could also use the fact that a pair of integers is truthy in Python and return the pair when found:
def find_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return (number, desired_sum - number)
unique_numbers.add(number)
It outputs:
find_pair_with_given_sum([5, 10, 7], 12)
# (7, 5)
find_pair_with_given_sum([5], 10)
# None
- As others have mentioned, your code could be more efficient with set lookup.
- It should return
False
with[5]
and10
as input parameters. - Python method names are written in
snake_case
. - Python is a dynamic language, which means method names and parameter names are very important. They make it easier to read your code and understand it again 6 months after having written it.
x
sounds like an unknown object or a float, not like a list of integers.
Here's a possible way to write the method:
def contains_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return True
unique_numbers.add(number)
return False
It outputs:
contains_pair_with_given_sum([5, 10, 7], 12)
# True
contains_pair_with_given_sum([5], 10)
# False
You could also use the fact that a pair of integers is truthy in Python and return the pair when found:
def find_pair_with_given_sum(numbers, desired_sum):
unique_numbers = set()
for number in numbers:
if (desired_sum - number) in unique_numbers:
return (number, desired_sum - number)
unique_numbers.add(number)
It outputs:
find_pair_with_given_sum([5, 10, 7], 12)
# (7, 5)
find_pair_with_given_sum([5], 10)
# None
edited Dec 4 at 10:55
answered Dec 4 at 10:47
Eric Duminil
2,0261613
2,0261613
add a comment |
add a comment |
for a much faster way to check if a number is in some container, use a
set()
:
anyequalto({10, 15, 3, 7}, 17)
. even if you have to convert from a list first,
it will still be faster to do that once rather than linearly search the list each
iteration. This will make your function O(n) rather than O(n^2)choose more expressive variable names that
x
andy
. something like
numbers
andk
j
If you want to write this in a more functional style, you could do:
def anyequalto(numbers, k):
numbers = set(numbers)
return any((k-num) in numbers
for num in numbers)
Because any will terminate early if the generator produces a true
value, this will be very similar in speed to an iterative solution.
1
It should beany(k in numbers for num in numbers)
, notany(for num in numbers k in numbers)
.
– Graipher
Dec 3 at 18:10
anyequalto([5], 10)
should beFalse
.
– Eric Duminil
Dec 4 at 10:32
add a comment |
for a much faster way to check if a number is in some container, use a
set()
:
anyequalto({10, 15, 3, 7}, 17)
. even if you have to convert from a list first,
it will still be faster to do that once rather than linearly search the list each
iteration. This will make your function O(n) rather than O(n^2)choose more expressive variable names that
x
andy
. something like
numbers
andk
j
If you want to write this in a more functional style, you could do:
def anyequalto(numbers, k):
numbers = set(numbers)
return any((k-num) in numbers
for num in numbers)
Because any will terminate early if the generator produces a true
value, this will be very similar in speed to an iterative solution.
1
It should beany(k in numbers for num in numbers)
, notany(for num in numbers k in numbers)
.
– Graipher
Dec 3 at 18:10
anyequalto([5], 10)
should beFalse
.
– Eric Duminil
Dec 4 at 10:32
add a comment |
for a much faster way to check if a number is in some container, use a
set()
:
anyequalto({10, 15, 3, 7}, 17)
. even if you have to convert from a list first,
it will still be faster to do that once rather than linearly search the list each
iteration. This will make your function O(n) rather than O(n^2)choose more expressive variable names that
x
andy
. something like
numbers
andk
j
If you want to write this in a more functional style, you could do:
def anyequalto(numbers, k):
numbers = set(numbers)
return any((k-num) in numbers
for num in numbers)
Because any will terminate early if the generator produces a true
value, this will be very similar in speed to an iterative solution.
for a much faster way to check if a number is in some container, use a
set()
:
anyequalto({10, 15, 3, 7}, 17)
. even if you have to convert from a list first,
it will still be faster to do that once rather than linearly search the list each
iteration. This will make your function O(n) rather than O(n^2)choose more expressive variable names that
x
andy
. something like
numbers
andk
j
If you want to write this in a more functional style, you could do:
def anyequalto(numbers, k):
numbers = set(numbers)
return any((k-num) in numbers
for num in numbers)
Because any will terminate early if the generator produces a true
value, this will be very similar in speed to an iterative solution.
edited Dec 3 at 19:18
answered Dec 3 at 18:05
Peter
69812
69812
1
It should beany(k in numbers for num in numbers)
, notany(for num in numbers k in numbers)
.
– Graipher
Dec 3 at 18:10
anyequalto([5], 10)
should beFalse
.
– Eric Duminil
Dec 4 at 10:32
add a comment |
1
It should beany(k in numbers for num in numbers)
, notany(for num in numbers k in numbers)
.
– Graipher
Dec 3 at 18:10
anyequalto([5], 10)
should beFalse
.
– Eric Duminil
Dec 4 at 10:32
1
1
It should be
any(k in numbers for num in numbers)
, not any(for num in numbers k in numbers)
.– Graipher
Dec 3 at 18:10
It should be
any(k in numbers for num in numbers)
, not any(for num in numbers k in numbers)
.– Graipher
Dec 3 at 18:10
anyequalto([5], 10)
should be False
.– Eric Duminil
Dec 4 at 10:32
anyequalto([5], 10)
should be False
.– Eric Duminil
Dec 4 at 10:32
add a comment |
A simple solution would use any
and itertools.combinations
from itertools import combinations
def anytwoequalto(numbers, k):
return any(((i + j) == k) for i, j in combinations(numbers, 2))
itertools.combinations
iterates over the given object returning tuples with the given number of items in them with no repetitions. eg.
combinations('ABCD', 2) =>
(('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D'))
any
returns True
if any of the expressions ((i + j) == k)
evaluate to True
If you wish to consider that an item can be added to itself, use combinations_with_replacement
instead.
Note: depending on the version of Python you are using, you may need to add extra parentheses (()
) around the expression inside the any
call.
Hope this makes sense.
add a comment |
A simple solution would use any
and itertools.combinations
from itertools import combinations
def anytwoequalto(numbers, k):
return any(((i + j) == k) for i, j in combinations(numbers, 2))
itertools.combinations
iterates over the given object returning tuples with the given number of items in them with no repetitions. eg.
combinations('ABCD', 2) =>
(('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D'))
any
returns True
if any of the expressions ((i + j) == k)
evaluate to True
If you wish to consider that an item can be added to itself, use combinations_with_replacement
instead.
Note: depending on the version of Python you are using, you may need to add extra parentheses (()
) around the expression inside the any
call.
Hope this makes sense.
add a comment |
A simple solution would use any
and itertools.combinations
from itertools import combinations
def anytwoequalto(numbers, k):
return any(((i + j) == k) for i, j in combinations(numbers, 2))
itertools.combinations
iterates over the given object returning tuples with the given number of items in them with no repetitions. eg.
combinations('ABCD', 2) =>
(('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D'))
any
returns True
if any of the expressions ((i + j) == k)
evaluate to True
If you wish to consider that an item can be added to itself, use combinations_with_replacement
instead.
Note: depending on the version of Python you are using, you may need to add extra parentheses (()
) around the expression inside the any
call.
Hope this makes sense.
A simple solution would use any
and itertools.combinations
from itertools import combinations
def anytwoequalto(numbers, k):
return any(((i + j) == k) for i, j in combinations(numbers, 2))
itertools.combinations
iterates over the given object returning tuples with the given number of items in them with no repetitions. eg.
combinations('ABCD', 2) =>
(('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D'))
any
returns True
if any of the expressions ((i + j) == k)
evaluate to True
If you wish to consider that an item can be added to itself, use combinations_with_replacement
instead.
Note: depending on the version of Python you are using, you may need to add extra parentheses (()
) around the expression inside the any
call.
Hope this makes sense.
answered Dec 4 at 5:11
Ben Jaguar Marshall
1112
1112
add a comment |
add a comment |
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Your revised code is actually less efficient. There is no point in creating a new list each time.
– Solomon Ucko
Dec 3 at 20:11
This puzzle must be making the rounds, because someone posted it on this site a few days ago: codereview.stackexchange.com/questions/208138/…
– Eric Lippert
Dec 3 at 22:33
4
Please do not add new code to your question. Your new code has a few aspects that deserve to be reviewed. You can do so in a different follow-up question. Please refer to the rules for that.
– Josay
Dec 3 at 22:43
1
Your revised code doesn't work correctly with duplicate values. contains_pair_totalling([5, 5], 10) returns false. Sets don't contain duplicates, so your remove call breaks this case.
– jaxad0127
Dec 3 at 22:50
1
The original code didn’t work with the sum double any list value, either.
anyequalto([5], 10)
returnsTrue
!– AJNeufeld
Dec 3 at 22:59