1266. Minimum Time Visiting All Points
On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return the minimum time in seconds to visit all the points in the order given by points.
You can move according to these rules:
- In 1 second, you can either: move vertically by one unit, move horizontally by one unit, or move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second).
- You have to visit the points in the same order as they appear in the array.
- You are allowed to pass through points that appear later in the order, but these do not count as visits.
Example 1:
Input: points = [[1,1],[3,4],[-1,0]]
Output: 7
Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0]
Time from [1,1] to [3,4] = 3 seconds
Time from [3,4] to [-1,0] = 4 seconds
Total time = 7 seconds
Example 2:
Input: points = [[3,2],[-2,2]]
Output: 5
Constraints:
- points.length == n
- 1 <= n <= 100
- points[i].length == 2
- -1000 <= points[i][0], points[i][1] <= 1000
Solution
class Solution {
public int minTimeToVisitAllPoints(int[][] points) {
int count = 0;
int i = 0;
while (i < points.length - 1) {
count += chebishevDistance(points[i], points[i + 1]);
i++;
}
return count;
}
int chebishevDistance(int[] a, int[] b) {
return Math.max(Math.abs(a[0] - b[0]), Math.abs(a[1] - b[1]));
}
int[][] DIRECTIONS = new int[][] {
new int[] {0,1}, // top
new int[] {1,1}, // top r
new int[] {1,0}, // r
new int[] {1,-1}, // down r
new int[] {0,-1}, // down
new int[] {-1,-1}, // down l
new int[] {-1,0}, // l
new int[] {-1,1}, // top l
};
public int minTimeToVisitAllPoints2(int[][] points) {
int i = 0;
int[] current = points[i];
int count = 0;
while (i < points.length - 1) {
int[] next = points[i + 1];
if (Arrays.equals(current, next)) {
i++;
continue;
}
double minDist = Double.MAX_VALUE;
int[] bestStep = null;
for (int[] dir: DIRECTIONS) {
int[] nextStep = new int[] {current[0] + dir[0], current[1] + dir[1]};
double distance = distance(nextStep, next);
if (distance < minDist) {
minDist = distance;
bestStep = nextStep;
}
}
current = bestStep;
count++;
}
return count;
}
double distance(int[] a, int[] b) {
double x2 = (a[0] - b[0]) * (a[0] - b[0]);
double y2 = (a[1] - b[1]) * (a[1] - b[1]);
return Math.sqrt(x2 + y2);
}
}