minimum-time-visiting-all-points

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Algorithms

On a 2D plane, there are <code>n</code> points with integer coordinates <code>points[i] = [xi, yi]</code>. Return the minimum time in seconds to visit all the points in the order given by <code>points</code>.

You can move according to these rules:

<ul>
	<li>In <code>1</code> second, you can either:

	<ul>
		<li>move vertically by one&nbsp;unit,</li>
		<li>move horizontally by one unit, or</li>
		<li>move diagonally <code>sqrt(2)</code> units (in other words, move one unit vertically then one unit horizontally in <code>1</code> second).</li>
	</ul>
	</li>
	<li>You have to visit the points in the same order as they appear in the array.</li>
	<li>You are allowed to pass through points that appear later in the order, but these do not count as visits.</li>
</ul>

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2019/11/14/1626_example_1.PNG" style="width: 500px; height: 428px;" />
<pre>
Input: points = [[1,1],[3,4],[-1,0]]
Output: 7
Explanation: One optimal path is [1,1] -&gt; [2,2] -&gt; [3,3] -&gt; [3,4] -&gt; [2,3] -&gt; [1,2] -&gt; [0,1] -&gt; [-1,0] 
Time from [1,1] to [3,4] = 3 seconds 
Time from [3,4] to [-1,0] = 4 seconds
Total time = 7 seconds</pre>

Example 2:

<pre>
Input: points = [[3,2],[-2,2]]
Output: 5
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>points.length == n</code></li>
	<li><code>1 &lt;= n&nbsp;&lt;= 100</code></li>
	<li><code>points[i].length == 2</code></li>
	<li><code>-1000&nbsp;&lt;= points[i][0], points[i][1]&nbsp;&lt;= 1000</code></li>
</ul>