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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.
1function minTimeToVisitAllPoints(points) {
2 let totalTime = 0;
3 for (let i = 0; i < points.length - 1; i++) {
4 const xDiff = Math.abs(points[i+1][0] - points[i][0]);
5 const yDiff = Math.abs(points[i+1][1] - points[i][1]);
6 totalTime += Math.max(xDiff, yDiff);
7 }
8 return totalTime;
9}
10
11const points = [[1, 1], [3, 4], [-1, 0]];
12console.log(minTimeToVisitAllPoints(points));The JavaScript solution follows similar logic to other languages, using Math.abs for absolute differences and Math.max for selecting the larger of the two, iterating through each point pair to derive a final time estimate.
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.
class MinimumTime {
public static int MinTimeToVisitAllPoints(int[][] points) {
int totalTime = 0;
for (int i = 0; i < points.Length - 1; i++) {
int xDiff = Math.Abs(points[i + 1][0] - points[i][0]);
int yDiff = Math.Abs(points[i + 1][1] - points[i][1]);
totalTime += Math.Max(xDiff, yDiff);
}
return totalTime;
}
static void Main() {
int[][] points = new int[][] {
new int[] {3, 2},
new int[] {-2, 2}
};
Console.WriteLine(MinTimeToVisitAllPoints(points));
}
}C# computes this straightforward maximum arithmetic across point pairs efficiently.