<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) {
        struct ElemWithIndex {
            int num;
            int index;
        };

        std::vector<ElemWithIndex> elems(nums.size());
        for(int i = 0; i < static_cast<int>(nums.size()); i++) {
            elems[i] = ElemWithIndex{nums[i], i};
        }

        std::sort(
            elems.begin(), elems.end(),
            [](const auto& a, const auto& b) {
                return a.num < b.num;
            }
        );
        
        int b_idx = static_cast<int>(
            std::upper_bound(
                elems.begin() + 1,
                elems.end(),
                target - elems[0].num,
                [](const int& a, const auto& b) {
                    return a < b.num;
                }
            ) - elems.begin() - 1
        );

        if(elems[0].num + elems[b_idx].num == target) {
            return {elems[0].index, elems[b_idx].index};
        }
        
        for(int a_idx = 1; a_idx < static_cast<int>(elems.size()); a_idx++) {
            while(elems[a_idx].num + elems[b_idx].num > target) {
                b_idx--;
            }

            if(elems[a_idx].num + elems[b_idx].num == target) {
                return {elems[a_idx].index, elems[b_idx].index};
            }
        }

        return {};
    }
};
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