# [Leetcode]1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums + nums = 2 + 7 = 9, return [0, 1].

This is the first question in Leetcode, and it’s more inspiring than I thought when I saw its echo in two recent contest questions: [Leetcode]1498. Number of Subsequences That Satisfy the Given Sum Condition and [Leetcode]1497. Check If Array Pairs Are Divisible by k. Therefore I’d like to share it here and let’s see how to use hashmap to improve the time complexity.

#### Logic

The brute force approach would be scanning nums looking for the partner for each number, which results in O(n^2). To achieve O(n) time complexity, we can use a hashmap to store the number that we have scanned, and when we find target-current has been scanned, we can return it.

#### Code

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
dic = {}
for i,val in enumerate(nums):
if target-val in dic:
return [dic[target-val],i] 