[Tutor] Trying to find all pairs of numbers from list whose sum is 10

[Oscar Benjamin]

On Sat, 19 Jun 2021 at 19:25, Manprit Singh <manpritsinghece at gmail.com> wrote:
>
> Dear sir,
>
> Let us consider a list
> lst = [2, 4, 7, 5, 9, 0, 8, 5, 3, 8]
> in this list there are 3 pairs (2, 8), (5, 5), (7, 3) whose sum is 10. To
> finding these pairs i have written the code as below :
>
> def pair_ten(x):
>     y, st  = x[:], set()
>     for i in x:
>         y.remove(i)
>         st.update((i, ele) for ele in y if i+ele == 10)
>     return st
>
> lst = [2, 4, 7, 5, 9, 0, 8, 5, 3, 8]
> ans = pair_ten(lst)
> print(ans)
>
> {(2, 8), (5, 5), (7, 3)}
>
> But i feel the code is still more complex, due to two for loops, in
> what way this ccan be dome more efficiently.

What does "more efficiently" mean? What is the scope of this problem?
Are the numbers always positive? Do you only want this for summing to
10 or is that just an example?

The task as stated finding pairs that sum to 10 can be done very
easily if all inputs are assumed to be positive:

def pair_ten(numbers):
    numbers_set = set(numbers)
    pairs10 = []
    for i in range(5+1):
        if i in numbers_set:
            i10 = 10 - i
            if i10 in numbers_set:
                pairs10.append((i, i10))
    return pairs10

This only has one for loop. Your version has two as well as a hidden
loop in the call to y.remove.

Alan's version is nice and simple and works fine for small inputs but
would be horribly inefficient if the size of the input list was large.
You haven't said whether it could be large though (if it's always
small then you are probably overthinking this). The version I showed
above would be inefficient if numbers was a small list and you were
looking for pairs that sum to say 1 million rather than 10. Also the
implementation I showed above converts the numbers list to a set which
is O(N) with no possibility of an early exit. If numbers was a huge
list that contained all the possible pairs many times then it could be
much quicker to loop through the first hundred or so items to see that
all 5 possible pairs are there rather than convert the entire list to
a set before doing anything else.

More information is needed to say what is actually efficient.

--
Oscar