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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