### Problem description

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.

### Constraints

Constraints:

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

### Testcases

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]
```

### My solution

It is easy to come up with an $O(n^2)$ solution, just iterate through every index, and for every index, iterate again, then one can get the answer

However, can we solve this problem using less than $O(n^2)$ time complexity?

The answer is yes.

First we will provide a $O(n \log n)$ time complexity solution

to be continue

### Code

$O(n^2)$ solution:

```
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> sol;
bool flag = false;
for(int i = 0 ; i < nums.size() ; i++){
if(flag == true) break;
for(int j = i + 1 ; j < nums.size() ; j++){
if(nums[i] + nums[j] == target){
sol.push_back(i), sol.push_back(j);
flag = true;
break;
}
}
}
return sol;
}
};
```

$O(n\log n)$ solution:

to be continue