<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) {
        
        // Create hashmap of (nlog(n))
        unordered_map<int, vector<int>> numIndices;

        for (int numIdx = 0; numIdx < nums.size(); ++numIdx) {
            int num = nums.at(numIdx);

            vector<int>& indices = numIndices[num];
            indices.push_back(numIdx);

            int toTarget = target - num;

            auto toTargetIt = numIndices.find(toTarget);

            if (toTargetIt == numIndices.end())
                continue;
            
            int firstIdx = indices.at(0);

            vector<int>& otherIndices = toTargetIt->second;

            if (num == toTarget) {
                if (otherIndices.size() <= 1)
                    continue;
                
                return {firstIdx, otherIndices.at(1)};
            }

            return {firstIdx, otherIndices.at(0)};
        }

        return {-1, -1};
    }
};
15000