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This approach centers on calculating the time taken to move from one point to another based solely on the maximum of the differences between x and y coordinates. This works because moving diagonally allows us to cover one unit in both x and y directions simultaneously. Therefore, the time taken to move from one point to another is always determined by the greater of the horizontal or vertical distances needed to cover.
Time Complexity: O(n), where n is the number of points. We process each pair of points once.
Space Complexity: O(1), as we use a constant amount of extra space.
1public class MinimumTime {
2 public static int minTimeToVisitAllPoints(int[][] points) {
3 int totalTime = 0;
4 for (int i = 0; i < points.length - 1; i++) {
5 int xDiff = Math.abs(points[i + 1][0] - points[i][0]);
6 int yDiff = Math.abs(points[i + 1][1] - points[i][1]);
7 totalTime += Math.max(xDiff, yDiff);
8 }
9 return totalTime;
10 }
11
12 public static void main(String[] args) {
13 int[][] points = {{1, 1}, {3, 4}, {-1, 0}};
14 System.out.println(minTimeToVisitAllPoints(points));
15 }
16}The Java solution leverages Math.abs to compute the magnitudes of differences and Math.max to find the larger. This value is added to the running total time required to travel through all pairs of points.
This approach involves calculating the total distance for each x and y direction separately, but also accounting for the diagonal moves that can reduce total movement time. Here, the diagonal path is favored when both x and y movements can be made simultaneously, which is reflected in the use of the maximum function across the differences.
Time Complexity: O(n), where n is the number of points. We process each pair once.
Space Complexity: O(1), as we are using a constant amount of space.
The C solution involves computing the minimum time based on adjusting x and y coordinates according to maximum differences while iterating through each pair of points.