573. Squirrel Simulation

Problem Description

Given a garden represented by its height and width, the position of a tree, the position of a squirrel, and the positions of multiple nuts, find the minimum distance for the squirrel to collect all nuts and place them under the tree. The squirrel can only move to adjacent cells (up, down, left, right), and it can only carry one nut at a time. Distance is measured by the number of moves.

Solution: Mathematical Optimization

The brute-force approach of trying all possible nut collection orders is computationally expensive. Instead, we can optimize using the following observation:

The total distance the squirrel travels is the sum of the distances from each nut to the tree (twice, since the squirrel has to go to the nut and back to the tree) plus the extra distance incurred by the squirrel's initial position. This means we can find the optimal order by minimizing the extra distance.

Let's define:

• tree: coordinates of the tree (tr, tc)
• squirrel: coordinates of the squirrel (sr, sc)
• nuts: coordinates of the nuts (list of [r, c])

The total distance dist_total can be expressed as:

dist_total = 2 * Σ(abs(nut_r - tr) + abs(nut_c - tc)) + extra_dist

Where the summation is over all nuts and extra_dist is the extra distance because the squirrel starts at squirrel instead of the tree. This extra_dist can be minimized by selecting the nut closest to the squirrel.

The algorithm consists of the following steps:

1. Calculate the total distance if the squirrel starts at the tree: This is 2 * Σ(abs(nut_r - tr) + abs(nut_c - tc)). This gives us a baseline.

2. Iterate through each nut: For each nut, calculate:

   • dist_nut_to_tree: The distance from the nut to the tree.
   • dist_squirrel_to_nut: The distance from the squirrel to the nut.
   • extra_dist: dist_squirrel_to_nut - dist_nut_to_tree. This is the extra distance incurred if we pick this nut first.

3. Find the minimum extra distance: Select the nut that minimizes extra_dist.

4. Calculate and return the total distance: dist_total = 2 * Σ(abs(nut_r - tr) + abs(nut_c - tc)) + min(extra_dist)

Code Implementation (Python)

import math
 
def minDistance(height, width, tree, squirrel, nuts):
    tr, tc = tree
    sr, sc = squirrel
    total_dist_from_tree = sum(abs(r - tr) + abs(c - tc) for r, c in nuts) * 2
    min_extra_dist = float('inf')
 
    for r, c in nuts:
        dist_nut_to_tree = abs(r - tr) + abs(c - tc)
        dist_squirrel_to_nut = abs(r - sr) + abs(c - sc)
        min_extra_dist = min(min_extra_dist, dist_squirrel_to_nut - dist_nut_to_tree)
 
    return total_dist_from_tree + min_extra_dist
 

Time and Space Complexity

• Time Complexity: O(N), where N is the number of nuts. We iterate through the nuts once.
• Space Complexity: O(1). We only use a few variables to store the distances.

This mathematical optimization drastically improves the efficiency compared to a brute-force approach, making the solution feasible even for a large number of nuts. The code implementations in other languages (Java, C++, Go, TypeScript, Rust, C#) will follow a similar structure and logic, with minor syntactic differences.