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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int minTimeToVisitAllPoints(int** points, int pointsSize, int* pointsColSize) {
5 int totalTime = 0;
6 for (int i = 0; i < pointsSize - 1; ++i) {
7 int xDiff = abs(points[i + 1][0] - points[i][0]);
8 int yDiff = abs(points[i + 1][1] - points[i][1]);
9 totalTime += xDiff > yDiff ? xDiff : yDiff;
10 }
11 return totalTime;
12}
13
14int main() {
15 int pointsSize = 3;
16 int pointsColSize[] = {2, 2, 2};
17 int* points[] = {(int[]){1, 1}, (int[]){3, 4}, (int[]){-1, 0}};
18 printf("%d\n", minTimeToVisitAllPoints(points, pointsSize, pointsColSize));
19 return 0;
20}The C implementation iterates over the list of points, computes the differences in x and y coordinates between consecutive points, and adds the maximum of those differences to total time. It uses the built-in abs function to ensure it is working with positive differences.
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.
1
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.