Code Implementation

class Solution:
    def minDistance(self, height: int, width: int, tree: List[int], squirrel: List[int], nuts: List[List[int]]) -> int:
        def dist(a, b):
            return abs(a[0] - b[0]) + abs(a[1] - b[1])
        
        total = 0
        max_diff = float('-inf')
        for nut in nuts:
            d_tree = dist(nut, tree)
            d_squirrel = dist(nut, squirrel)
            total += 2 * d_tree
            max_diff = max(max_diff, d_tree - d_squirrel)
        return total - max_diff
      

class Solution {
public:
    int minDistance(int height, int width, vector& tree, vector& squirrel, vector<vector<int>>& nuts) {
        auto dist = [](vector<int>& a, vector<int>& b) {
            return abs(a[0] - b[0]) + abs(a[1] - b[1]);
        };
        int total = 0;
        int maxDiff = INT_MIN;
        for (auto& nut : nuts) {
            int dTree = dist(nut, tree);
            int dSquirrel = dist(nut, squirrel);
            total += 2 * dTree;
            maxDiff = max(maxDiff, dTree - dSquirrel);
        }
        return total - maxDiff;
    }
};
      

class Solution {
    public int minDistance(int height, int width, int[] tree, int[] squirrel, int[][] nuts) {
        int total = 0;
        int maxDiff = Integer.MIN_VALUE;
        for (int[] nut : nuts) {
            int dTree = Math.abs(nut[0] - tree[0]) + Math.abs(nut[1] - tree[1]);
            int dSquirrel = Math.abs(nut[0] - squirrel[0]) + Math.abs(nut[1] - squirrel[1]);
            total += 2 * dTree;
            maxDiff = Math.max(maxDiff, dTree - dSquirrel);
        }
        return total - maxDiff;
    }
}
      

var minDistance = function(height, width, tree, squirrel, nuts) {
    function dist(a, b) {
        return Math.abs(a[0] - b[0]) + Math.abs(a[1] - b[1]);
    }
    let total = 0;
    let maxDiff = -Infinity;
    for (let nut of nuts) {
        let dTree = dist(nut, tree);
        let dSquirrel = dist(nut, squirrel);
        total += 2 * dTree;
        maxDiff = Math.max(maxDiff, dTree - dSquirrel);
    }
    return total - maxDiff;
};
      

Problem Description

In the Squirrel Simulation problem, you are given a 2D grid representing a park. There are several nuts on the grid, a single tree, and a squirrel at a starting position. The squirrel can move in four directions (up, down, left, right) and can carry only one nut at a time. The goal is for the squirrel to collect all the nuts and bring them to the tree, minimizing the total distance traveled.

• The squirrel can start by picking up any nut and must bring each nut to the tree, one by one.
• The squirrel always starts at its initial position, but after bringing the first nut to the tree, it always starts the next trip from the tree.
• You are given the grid size (height and width), the position of the tree (tree), the squirrel's starting position (squirrel), and a list of nut positions (nuts).
• The objective is to compute the minimum total distance the squirrel must travel to collect all the nuts and bring them to the tree.

Thought Process

At first glance, a brute-force approach might be to try all possible orders in which the squirrel can pick up the nuts. However, this quickly becomes infeasible as the number of nuts increases due to the factorial number of permutations.

A key observation is that after the first nut, the squirrel always starts from the tree. Only the first trip starts from the squirrel's initial position. Therefore, the only choice we need to make is: which nut should the squirrel pick up first? For each nut, we can compute the difference in distance between starting from the squirrel's position versus starting from the tree.

The optimal solution is to pick as the first nut the one for which this difference is maximized. This ensures the total distance is minimized.

Solution Approach

• Step 1: For each nut, calculate the distance from the nut to the tree (d_tree) and from the nut to the squirrel's starting position (d_squirrel).
• Step 2: If the squirrel always started at the tree, the total distance would be 2 × (nut to tree) for each nut (go to nut, back to tree).
• Step 3: However, for the first nut, the squirrel starts at its own position, so the actual distance is squirrel to nut plus nut to tree.
• Step 4: The difference for each nut is d_tree - d_squirrel. We want to maximize this difference for the first nut, as it will reduce the total distance.
• Step 5: The minimal total distance is:
  • Sum of 2 × (nut to tree) for all nuts
  • Subtract the maximal (nut to tree - squirrel to nut) value
• Step 6: Implement this logic efficiently with a single pass through the list of nuts.

Example Walkthrough

Let's say the grid is 5x7, the tree is at position [2,2], the squirrel starts at [4,4], and the nuts are at positions [[3,0],[2,5]].

1. Calculate distances:
   • For [3,0]: nut to tree = |3-2| + |0-2| = 1 + 2 = 3; squirrel to nut = |4-3| + |4-0| = 1 + 4 = 5
   • For [2,5]: nut to tree = |2-2| + |5-2| = 0 + 3 = 3; squirrel to nut = |4-2| + |4-5| = 2 + 1 = 3
2. Total distance if always starting from tree: 2 × 3 + 2 × 3 = 12
3. The difference for [3,0] is 3-5 = -2; for [2,5] is 3-3 = 0
4. The maximum difference is 0 (for [2,5]). So the minimal total distance is 12 - 0 = 12.
5. The squirrel should pick [2,5] first, then [3,0].

Time and Space Complexity

• Brute-force: Trying all permutations of nut orders is O(N!), where N is the number of nuts. This is infeasible for large N.
• Optimized Approach: We do a single pass through the list of nuts, so the time complexity is O(N).
• Space Complexity: We use a constant amount of extra space (a few variables), so the space complexity is O(1).

Summary

The Squirrel Simulation problem can be solved efficiently by recognizing that only the first trip is special, and all subsequent trips start at the tree. By picking the nut that maximizes the saving on the first trip, we minimize the overall distance. This insight reduces the problem from a complex permutation to a simple linear scan, making the solution both elegant and efficient.