Today we will be looking at one of the interviewers’ favorite question for the technical coding interview. Two Sum. 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 ...

Today we will be looking at one of the interviewers’ favorite question for the technical coding interview.

**1****. Two Sum**

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

You can return the answer in any order.

**Example 1:**

```
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: 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 <= 105`

`-109 <= nums[i] <= 109`

`-109 <= target <= 109`

**Only one valid answer exists.**

One of the first solutions that came to my mind after reading the question is going through each element and check if the addition of any 2 elements results in the answer. If yes, we return the index of both the elements else we continue the process until we find the answer.

The code would look like the following.

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

Complexity

**Time Complexity**

Considering the worst-case scenario, wherein our output is at the end of the list. In that case, we would take up O(N*(N-1)) time which is O(N²).

**Space Complexity**

We are not using any extra space and hence the space complexity will be O(1).

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