LeetCode 1266. Minimum Time Visiting All Points Solution in Java, C++, Python & More | Explanation + Code

Description

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:

  <ul>
  	<li>move vertically by one&nbsp;unit,</li>
  	<li>move horizontally by one unit, or</li>
  	<li>move diagonally <code>sqrt(2)</code> units (in other words, move one unit vertically then one unit horizontally in <code>1</code> second).</li>
  </ul>
  </li>
  <li>You have to visit the points in the same order as they appear in the array.</li>
  <li>You are allowed to pass through points that appear later in the order, but these do not count as visits.</li>
  

 

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

Solutions

Solution 1: Simulation

For two points p₁=(x₁, y₁) and p₂=(x₂, y₂), the distances moved in the horizontal and vertical directions are d_x = |x₁ - x₂| and d_y = |y₁ - y₂| respectively.

If d_x \ge d_y, we move diagonally for d_y steps, then move horizontally for d_x - d_y steps; if d_x < d_y, we move diagonally for d_x steps, then move vertically for d_y - d_x steps. Therefore, the shortest distance between two points is max(d_x, d_y).

We can iterate through all pairs of points, calculate the shortest distance between each pair, and sum them up.

The time complexity is O(n), where n is the number of points. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustC

class Solution: def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int: return sum( max(abs(p1[0] - p2[0]), abs(p1[1] - p2[1])) for p1, p2 in pairwise(points) )(code-box)

class Solution { public int minTimeToVisitAllPoints(int[][] points) { int ans = 0; for (int i = 1; i < points.length; ++i) { int dx = Math.abs(points[i][0] - points[i - 1][0]); int dy = Math.abs(points[i][1] - points[i - 1][1]); ans += Math.max(dx, dy); } return ans; } }(code-box)

class Solution { public: int minTimeToVisitAllPoints(vector<vector<int>>& points) { int ans = 0; for (int i = 1; i < points.size(); ++i) { int dx = abs(points[i][0] - points[i - 1][0]); int dy = abs(points[i][1] - points[i - 1][1]); ans += max(dx, dy); } return ans; } };(code-box)

func minTimeToVisitAllPoints(points [][]int) (ans int) { for i, p := range points[1:] { dx := abs(p[0] - points[i][0]) dy := abs(p[1] - points[i][1]) ans += max(dx, dy) } return } func abs(x int) int { if x < 0 { return -x } return x }(code-box)

function minTimeToVisitAllPoints(points: number[][]): number { let ans = 0; for (let i = 1; i < points.length; i++) { const dx = Math.abs(points[i][0] - points[i - 1][0]); const dy = Math.abs(points[i][1] - points[i - 1][1]); ans += Math.max(dx, dy); } return ans; }(code-box)

impl Solution { pub fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 { let mut ans = 0; for i in 1..points.len() { let dx = (points[i][0] - points[i - 1][0]).abs(); let dy = (points[i][1] - points[i - 1][1]).abs(); ans += dx.max(dy); } ans } }(code-box)

#define max(a, b) ((a) > (b) ? (a) : (b)) int minTimeToVisitAllPoints(int** points, int pointsSize, int* pointsColSize) { int ans = 0; for (int i = 1; i < pointsSize; ++i) { int dx = abs(points[i][0] - points[i - 1][0]); int dy = abs(points[i][1] - points[i - 1][1]); ans += max(dx, dy); } return ans; }(code-box)