# Two sum | leetcode solution

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.

Follow-up: Can you come up with an algorithm that is less than `O(n2) `time complexity?

Code(C++):-

class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
int n=nums.size();
int s,e;
vector<int> ans;
int flag=0;
for(int i=0;i<n;i++)
{
s=i;
for(int j=i+1;j<n;j++)
{
if(nums[i]+nums[j]==target)
{
e=j;
flag=1;
break;
}
}
if(flag==1)
break;
}
ans.push_back(s);
ans.push_back(e);
return ans;
}

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