1266. Minimum Time Visiting All Points

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https://leetcode.com/problems/minimum-time-visiting-all-points/

On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit all points.

You can move according to the next rules:

  • In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
  • You have to visit the points in the same order as they appear in the array.

 

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
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Intuition
Min dist is covered by moving diagonally to align with min diff in X, Y, and cover remaining along the longer axes

Equivalently, min dist = max(abs diff X, abs diff Y)
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Time - O(N)
Space - O(1)
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class Solution {
public int minTimeToVisitAllPoints(int[][] points) {
int ans = 0;
for (int i = 1; i < points.length; i++) {
int x = Math.abs(points[i][0] - points[i - 1][0]);
int y = Math.abs(points[i][1] - points[i - 1][1]);
ans += Math.max(x, y);
}
return ans;
}
}
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