<p>A city&#39;s <strong>skyline</strong> is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Given the locations and heights of all the buildings, return <em>the <strong>skyline</strong> formed by these buildings collectively</em>.</p>

<p>The geometric information of each building is given in the array <code>buildings</code> where <code>buildings[i] = [left<sub>i</sub>, right<sub>i</sub>, height<sub>i</sub>]</code>:</p>

<ul>
	<li><code>left<sub>i</sub></code> is the x coordinate of the left edge of the <code>i<sup>th</sup></code> building.</li>
	<li><code>right<sub>i</sub></code> is the x coordinate of the right edge of the <code>i<sup>th</sup></code> building.</li>
	<li><code>height<sub>i</sub></code> is the height of the <code>i<sup>th</sup></code> building.</li>
</ul>

<p>You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height <code>0</code>.</p>

<p>The <strong>skyline</strong> should be represented as a list of &quot;key points&quot; <strong>sorted by their x-coordinate</strong> in the form <code>[[x<sub>1</sub>,y<sub>1</sub>],[x<sub>2</sub>,y<sub>2</sub>],...]</code>. Each key point is the left endpoint of some horizontal segment in the skyline except the last point in the list, which always has a y-coordinate <code>0</code> and is used to mark the skyline&#39;s termination where the rightmost building ends. Any ground between the leftmost and rightmost buildings should be part of the skyline&#39;s contour.</p>

<p><b>Note:</b> There must be no consecutive horizontal lines of equal height in the output skyline. For instance, <code>[...,[2 3],[4 5],[7 5],[11 5],[12 7],...]</code> is not acceptable; the three lines of height 5 should be merged into one in the final output as such: <code>[...,[2 3],[4 5],[12 7],...]</code></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/01/merged.jpg" style="width: 800px; height: 331px;" />
<pre>
<strong>Input:</strong> buildings = [[2,9,10],[3,7,15],[5,12,12],[15,20,10],[19,24,8]]
<strong>Output:</strong> [[2,10],[3,15],[7,12],[12,0],[15,10],[20,8],[24,0]]
<strong>Explanation:</strong>
Figure A shows the buildings of the input.
Figure B shows the skyline formed by those buildings. The red points in figure B represent the key points in the output list.
</pre>

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

<pre>
<strong>Input:</strong> buildings = [[0,2,3],[2,5,3]]
<strong>Output:</strong> [[0,3],[5,0]]
</pre>

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

<ul>
	<li><code>1 &lt;= buildings.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= left<sub>i</sub> &lt; right<sub>i</sub> &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>1 &lt;= height<sub>i</sub> &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>buildings</code> is sorted by <code>left<sub>i</sub></code> in&nbsp;non-decreasing order.</li>
</ul>

class Solution {
public:
    vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
        // the intuition is to start from left-most point, then search all the
        // coordinates of the buildings to determine the next left-most point;
        // for each point, determine the height of the tallest building whose
        // span covers this point - this seems simple enough
        
        vector<vector<int>> res;
        // determine the right-most point so we know when we're done
        int end = rightMostPoint(buildings);
        int prev = -1, prev_high = 0, cur = 0;
        while (cur != end) {
            cur = nextPoint(buildings, prev);
            int high = tallestHeightAt(buildings, cur);

            // decision for this point
            if (high != prev_high) {
                res.push_back({cur, high});
                prev_high = high;
            }
            prev = cur;
        }
        
        return res;
    }
    int nextPoint(vector<vector<int>>& b, int prev) {
        int x = INT_MAX;
        for (int i= 0; i<b.size(); i++) {
            if (b[i][0] <= prev && b[i][1] > prev) {
                x = min(x, b[i][1]);
            }
            else if (b[i][0] > prev) {
                x = min(x, b[i][0]);
                break;
            }
        }
        return x;
    }
    int tallestHeightAt(vector<vector<int>>& b, int cur) {
        int h = 0;
        for (int i= 0; i<b.size(); i++) {
            if (b[i][0] <= cur && b[i][1] > cur) {
                h = max(h, b[i][2]);
            }
            else if (b[i][0] > cur)
                break;
        }
        return h;
    }
    int rightMostPoint(vector<vector<int>>& b) {
        int x = b.back()[1];
        for (int i= 0; i<b.size()-1; i++) {
            if (b[i][1] > x)
                x = b[i][1];
        }
        return x;
    }
};
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