<p>Given an array of integers <code>nums</code>&nbsp;and an integer <code>target</code>, return <em>indices of the two numbers such that they add up to <code>target</code></em>.</p>

<p>You may assume that each input would have <strong><em>exactly</em> one solution</strong>, and you may not use the <em>same</em> element twice.</p>

<p>You can return the answer in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,7,11,15], target = 9
<strong>Output:</strong> [0,1]
<strong>Explanation:</strong> Because nums[0] + nums[1] == 9, we return [0, 1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,4], target = 6
<strong>Output:</strong> [1,2]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,3], target = 6
<strong>Output:</strong> [0,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>
	<li><strong>Only one valid answer exists.</strong></li>
</ul>

<p>&nbsp;</p>
<strong>Follow-up:&nbsp;</strong>Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>)</code><font face="monospace">&nbsp;</font>time complexity?
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {

        int need;
        unordered_map<int,int>m;
        int n=nums.size();
        for(int i=0;i<n;i++)
        {
            need=target-nums[i];
            if(m.find(need)!=m.end())
            {
                return {m[need],i};
            }
            m[nums[i]]=i;
        }
    return {}; 


    // int n=nums.size();
    // unordered_map<int,int>m;
    // for(int i=0;i<n;i++)
    // {
    //     m[nums[i]]=i;
    // }
    // for(int i=0;i<n;i++)
    // {
    //     int complement=target-nums[i];
    //     if(m.count(complement) && m[complement]!=i)
    //     {
    //         return{i,m[complement]};
    //     }
    // }
    // return {};



    //  int len=nums.size(),flag=0;
    //  vector<int> final;
    //  for(int j=0;j<len-1;j++)
    //  {
    //  for(int i=j+1;i<len;i++)
    //  {
    //      if(nums[j]+nums[i]==target)
    //      {
    //          final.push_back(j);
    //          final.push_back(i);
    //          flag=1;
    //          break;
    //      } 
    //  }
    //  if(flag==1)
    //  {
    //      break;
    //  }
    //  }
    //  return final;
    }
};
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