<p>Given an array&nbsp;of <code>intervals</code>&nbsp;where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>, merge all overlapping intervals, and return <em>an array of the non-overlapping intervals that cover all the intervals in the input</em>.</p>

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

<pre>
<strong>Input:</strong> intervals = [[1,3],[2,6],[8,10],[15,18]]
<strong>Output:</strong> [[1,6],[8,10],[15,18]]
<strong>Explanation:</strong> Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
</pre>

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

<pre>
<strong>Input:</strong> intervals = [[1,4],[4,5]]
<strong>Output:</strong> [[1,5]]
<strong>Explanation:</strong> Intervals [1,4] and [4,5] are considered overlapping.
</pre>

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

<ul>
	<li><code>1 &lt;= intervals.length &lt;= 10<sup>4</sup></code></li>
	<li><code>intervals[i].length == 2</code></li>
	<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
</ul>


bool merge_into(const std::vector<int>& source, std::vector<int>& target) {
    if (source[0] > target[1]) return false;
    if (source[1] < target[0]) return false;

    target[0] = std::min(target[0], source[0]);
    target[1] = std::max(target[1], source[1]);
    return true;
}

class Solution {
public:
    std::vector<std::vector<int>> merge(std::vector<std::vector<int>>& intervals) {
        for (int i = intervals.size() - 1; i >= 0; --i) {
            for (int j = 0; j < intervals.size(); ++j) {
                if (i == j) continue;

                if (merge_into(intervals[i], intervals[j])) {
                    std::swap(intervals.back(), intervals[i]);
                    intervals.pop_back();
                    break;
                }
            }
        }

        return std::move(intervals);
    }
};
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