# 1 Leetcode Python: Two Sum – Easy

### Problem

Given an array of integers `nums` and an integer `target`, return *indices of the two numbers such that they add up to `target`*.

You may assume that each input would have ***exactly***** one solution**, and you may not use the *same* element twice.

You can return the answer in any order.

**Example 1:**

```
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
```

**Example 2:**

```
Input: nums = [3,2,4], target = 6
Output: [1,2]
```

**Example 3:**

```
Input: nums = [3,3], target = 6
Output: [0,1]
```

**Constraints:**

* `2 <= nums.length <= 104`
* `-109 <= nums[i] <= 109`
* `-109 <= target <= 109`
* **Only one valid answer exists.**

### Thoughts: <a href="#thoughts" id="thoughts"></a>

The most straight forward solution would be a nested loop. This would end up with running complicity of O(n2).

With the help of a hash-table, we could improve running complicity to O(n), but this requires space O(n). This is very common in coding problem. Sometimes, in order to improve the running time, you have to trade off on space.

### O(n2) solution: <a href="#on2-solution" id="on2-solution"></a>

```
class Solution:    
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        dict_num = {}
        
        for i in range(len(nums)):
            number = nums[i]
            
            if target - number in dict_num:
                return (dict_num[target-number], i)
            
            dict_num[number] = i
```


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