<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?
#pragma GCC optimize("Ofast")

static const int __ = []()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 0;
}();


int init = [] {
    
    ofstream out("user.out");
    cout.rdbuf(out.rdbuf());
    std::string first;
    std::string sec;
    int C{};
    for (string str; getline(cin, str);) {
        
        if(C == 0){ first = str; C++;}
        else
        {
            sec = str;

            int target = std::stoi(sec);
            std::unordered_map<int, int> mp;


            for(int i = 0,  indx{}; indx < first.size(); )
            {
                if(first[indx] == '[' || first[indx] == ']' || first[indx] == ',') {indx++; continue;}
                std::string temp = "";
                int j = indx;
                for(; first[j] != ',' && first[j] != ']'; j++) temp += first[j];
                indx = j;
                int cur = std::stoi(temp);

                

                if(mp.find(target - cur) != mp.end())
                {
                    int f = mp[target - cur];

                    std::cout << '[' << i << ',' << f << ']' << std::endl;
                    break;
                }
                else
                {
                    mp[cur] = i;
                    i++;
                }


            }

            C = 0;
        }
        

    }
    exit(0);
    return 1337;
}();



class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        return std::vector<int>();
    }

};
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