Efficient Pathfinding: Solving the Minimum Time Visiting All Points Problem in Rust

When navigating a 2D grid, the shortest path depends entirely on how you are allowed to move. In many competitive programming problems, you are restricted to horizontal and vertical movements (Manhattan Distance). However, what happens when you can move diagonally?

In this post, we’ll explore how to calculate the minimum time to visit a sequence of points using Chebyshev Distance and how to implement it cleanly in Rust.

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

The Problem

Given an array of points, we need to find the minimum time to visit them in the given order. We can move:

• Vertically (1 unit = 1s)

• Horizontally (1 unit = 1s)

• Diagonally (1 unit diagonally = 1s)

The ability to move diagonally is the cheat code here. Moving diagonally allows us to cover one unit of both and distance simultaneously in a single second.

The Intuition: Chebyshev Distance

If you want to move from to , you have two gaps to close:

1. X-axis gap: units.

2. Y-axis gap: units.

By moving diagonally for 2 seconds, you cover both and distances. After 2 seconds, you’ve closed the gap entirely and are now at . You then spend 1 more second moving vertically to reach .

Total time? 3 seconds. Mathematically, the time taken between two points and is always the maximum of the two differences:

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The Rust Implementation

Rust provides powerful iterator methods like .windows() and .map() that make this calculation expressive and concise.

impl Solution {
    pub fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 {
        points
            .windows(2) // 1. Look at points in pairs: [p1, p2], [p2, p3]...
            .map(|p| {
                // 2. Calculate absolute difference for X and Y
                let dx = (p[1][0] - p[0][0]).abs();
                let dy = (p[1][1] - p[0][1]).abs();

                // 3. The time taken is the maximum of the two
                dx.max(dy)
            })
            .sum() // 4. Add all the times together
    }
}

Why this approach?

1. windows(2): This is a safe way to iterate through a slice in overlapping chunks. It avoids manual indexing (i and i+1), which prevents index out of bounds errors.

2. Immutability: We aren't managing a mutable total_time variable; we are transforming a stream of data into a final result.

3. Performance: This runs in time and extra space, which is optimal for this problem.

Conclusion

The key to geometry-based coding challenges is identifying which distance metric applies. Once you realize that diagonal movement translates to the Chebyshev Distance, the code becomes a simple matter of summing up the max differences.