# 1. Two Sum

<https://leetcode.com/problems/two-sum/description/>

## Solution

* Brute Force
  * Use every element as an anchor and search its right element

```python
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for i in range(len(nums)):
            for j in range(i+1, len(nums)):
                if nums[i] + nums[j] == target:
                    return [i,j]
```

时间复杂度：O(n^2)\
空间复杂度：O(1)

* Hash Map
  * Leverage the power of hashmap and store the number and its index in the map
  * When the map contain target - curNum, we know the search is complete

```python
class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        residual_dict = {}

        for i, num in enumerate(nums):
            if num in residual_dict:
                return [residual_dict[num], i]
            else:
                residual_dict[target - num] = i
```

时间复杂度：O(n)\
空间复杂度：O(n)