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

<p>You can move according to these rules:</p>

<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>

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

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

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

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

<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>

class Solution:
    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
        totalTime = 0

        for i in range(1, len(points)):

            previousPoint = points[i - 1]
            currentPoint = points[i]

            xDistance = abs(currentPoint[0] - previousPoint[0])
            yDistance = abs(currentPoint[1] - previousPoint[1])

            totalTime += max(xDistance, yDistance)

        return totalTime

        
16500