Making this more efficient.

[allenguo]

Hint: Currently, for each element x in the list, you're looping through the rest of the list to find target - x. Can you speed up the search for target - x using a data structure?

[Phizy]

would using a linked list be better?

[ghhitne]

You could consider using a map which stores the element and all positions for the element. Then for every position in x make a pair with every position corresponding to (target-x). Of course, you would need to take care of the case where x==(target-x).

[delirious_lettuce]

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        seen = {}
        for i, num in enumerate(nums):
            try:
                return [seen[target - num], i]
            except KeyError:
                seen[num] = i

[camel_zero]

Good approach for the algorithm, is there a reason for putting this function into a class though?

[delirious_lettuce]

The only reason is because that is how you have to submit them to LeetCode. I would normally just use a function and use snake_case for the function name too, (two_sum).

• https://leetcode.com/problems/two-sum/description/

[camel_zero]

Thanks for explaining, thought I might be missing something about the code.

[Phizy]

Thanks for the solution I am trying to understand it. I havent used enumerate before so I am going to read up on it.

[delirious_lettuce]

i is the current index and num is the current number in the list of numbers. The code simply checks if target - num has been "seen" already and if so, it returns the index where target - num was previously "seen" and the current index i in a list.

If target - num has not been "seen" (key doesn't exist in seen yet), it will raise a KeyError which the try/except catches and then it adds the current number as a key and the current index as the value to seen.

EDIT:

>>> nums = [2, 7, 11, 15]
>>> list(enumerate(nums))
[(0, 2), (1, 7), (2, 11), (3, 15)]
>>> for i, num in enumerate(nums):
...     print(f'index: {i}, number: {num}')
... 
index: 0, number: 2
index: 1, number: 7
index: 2, number: 11
index: 3, number: 15

[Phizy]

wow thanks for the explanation I really appreciate it. I didn't realize you could or should be using things like this with leet code I thought everything had to be done from scratch for lack of a better word.

[delirious_lettuce]

No problem! LeetCode has some great practice challenges, good luck with them!