查找给定元组的列表的索引号

Question

The input in the first parameter which is a list should add up to the second parameter which is an integer. using only two numbers inside the list only once and returning the index of only one pair of numbers. So the approach i took here using the itertools module is to make possible unique combinations of ints which i can use to add up the numbers to the second parameter. but the combination is giving me a tuple inside a loop. which i can't use inside the list. Is there a way to check if the values inside the tuple which is 'i' inside the last 'if' block exist inside the list. If so, then how can i return the indexes of the expected output from the list in the first parameter. At this point, i'm kinda stuck here and the answer might be obvious but what am i missing.

Input: ([1,2,3,5,6], 5)

from itertools import combinations
def twoSum(nums,target):

    a = list(nums)
    for i in combinations(a,2):
        if sum(i) == target and sum(i) in a:
            enumerate(a[i])

Expected output: [1, 2]

Answer 1

Using itertools.combinations is slow, you could do this instead:

def twoSum(nums, target):
    lookup = {target - e: i for i, e in enumerate(nums)}
    return next((lookup[e], i) for i, e in enumerate(nums) if e in lookup)


res = twoSum([1, 2, 3, 5, 6], 5)
print(res)

Output

(2, 1)

The time complexity of this approach is O(n). This returns only one pair.

The idea is to create a dictionary that maps the difference between the target and an element of nums to it's index. Then if an element of nums is in lookup it means that those values add up to target, and therefore you can return their indexes.

Answer 2

Not really clear what the OP is after, as there are multiple attempts to this question. My interpretation would give the same answer that Dani Mesejo posted . Here is another interpretation.

def twoSum(nums, target):
    return [i for i, x in enumerate(nums) if (target-x) in nums]

This returns all indices of elements in nums which sum up to target . So with you example, where this is only one valid pair

>>> nums = [1, 2, 3, 5, 6]
>>> twoSum(nums, 5)
[1, 2]

but if there are more pairs, you get all the indices in the same list

>>> nums = [1, 2, 3, 4, 5, 6]
>>> twoSum(nums, 5)
[0, 1, 2, 3]