# 1266. Minimum Time Visiting All Points

T: O(n)

S: O(1)

````java
```java
class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int result = 0;
        for (int i = 0; i < points.length-1; i++) {
            int XCurr = points[i][0];
            int XNext = points[i+1][0];

            int YCurr = points[i][1];
            int YNext = points[i+1][1];

            result += Math.max(Math.abs(XCurr - XNext), Math.abs(YCurr - YNext));
        }
        return result;
    }
}
```

```java
/**


why min dist is max(diffX, diffY)?


because when we move x -> actually we also move y (by diagnal way)

so if we move x, but still can't arrive, the left is diffY - diffX) (this is positive number)

so we only need to know which is bigger -> so it's max(diffX, diffY)

key point is it can move by "diagnal way"

ex: [1,1],[3,4]

4               3,4

3               x -> 2 units (x can diagnal walk 2 units), then just vertically move to 3,4 by 1 units

2         x
  
1    1,1
      1    2    3

      max(abs(3-1), abs(4-1)) = max(2, 3) = 3
 */
```
````


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