Squirrel Simulation

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ArraysGreedy Algorithms

There's a tree in a park that has a total of n branches. A squirrel is hiding on one of the branches. An integer array squirrels is given where squirrels[i] == (xi, yi) represents the location of the i^th squirrel. All the squirrels must gather at one of the branches to eat. The chosen branch is given as an array tree where tree == [xt, yt] represents the location of the tree.

A can of nuts is also placed somewhere in the park. An integer array nuts is given where nuts[i] == (xi, yi) represents the location of the i^th nut.

All the squirrels share the same speed. So, the travel distance between the squirrels is equivalent to the Euclidean distance. One squirrel can carry only one nut at a time. The squirrel will take the nut to the tree and come back to get another nut.

The task is to find the minimum distance for all the squirrels to gather at the tree to eat all the nuts.

Example 1:

Input: height = 5, width = 7, tree = [2,2], squirrels = [[4,4],[3,0]], nuts = [[1,4],[2,0]]
Output: 12
Explanation:
The squirrel at [4,4] goes to the nut at [1,4] then to the tree at [2,2] so travels 6.
The squirrel at [3,0] goes to the nut at [2,0] then to the tree at [2,2] so travels 6.
All together they travel 12.

Example 2:

Input: height = 1, width = 1, tree = [0,0], squirrels = [[0,0]], nuts = [[0,0]]
Output: 0

Constraints:

1 <= height, width <= 100
nuts.length <= 1000
squirrels.length <= 10
height == matrix.length
width == matrix[i].length
squirrels[i].length == 2
tree.length == 2
nuts[i].length == 2
0 <= squirrels[i][0] < height
0 <= squirrels[i][1] < width
0 <= tree[0] < height
0 <= tree[1] < width
0 <= nuts[i][0] < height
0 <= nuts[i][1] < width
It is guaranteed that there is at least one nut.
The locations of the nuts, the tree, and the squirrels are pairwise distinct.

Try coding it in LeetCode

Squirrel Simulation

Medium

Block

0 views

Topics:

ArraysGreedy Algorithms

There's a tree in a park that has a total of n branches. A squirrel is hiding on one of the branches. An integer array squirrels is given where squirrels[i] == (xi, yi) represents the location of the i^th squirrel. All the squirrels must gather at one of the branches to eat. The chosen branch is given as an array tree where tree == [xt, yt] represents the location of the tree.

A can of nuts is also placed somewhere in the park. An integer array nuts is given where nuts[i] == (xi, yi) represents the location of the i^th nut.

All the squirrels share the same speed. So, the travel distance between the squirrels is equivalent to the Euclidean distance. One squirrel can carry only one nut at a time. The squirrel will take the nut to the tree and come back to get another nut.

The task is to find the minimum distance for all the squirrels to gather at the tree to eat all the nuts.

Example 1:

Input: height = 5, width = 7, tree = [2,2], squirrels = [[4,4],[3,0]], nuts = [[1,4],[2,0]]
Output: 12
Explanation:
The squirrel at [4,4] goes to the nut at [1,4] then to the tree at [2,2] so travels 6.
The squirrel at [3,0] goes to the nut at [2,0] then to the tree at [2,2] so travels 6.
All together they travel 12.

Example 2:

Input: height = 1, width = 1, tree = [0,0], squirrels = [[0,0]], nuts = [[0,0]]
Output: 0

Constraints:

1 <= height, width <= 100
nuts.length <= 1000
squirrels.length <= 10
height == matrix.length
width == matrix[i].length
squirrels[i].length == 2
tree.length == 2
nuts[i].length == 2
0 <= squirrels[i][0] < height
0 <= squirrels[i][1] < width
0 <= tree[0] < height
0 <= tree[1] < width
0 <= nuts[i][0] < height
0 <= nuts[i][1] < width
It is guaranteed that there is at least one nut.
The locations of the nuts, the tree, and the squirrels are pairwise distinct.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the maximum values for height and width, and the maximum number of nuts in the nut array?
Can the positions of the tree, squirrel, or nuts ever be outside the grid boundaries, or overlap with each other?
If the 'nuts' array is empty, what should the function return?
Are the coordinates provided as (row, column) or (column, row)?
Is the distance calculated using Manhattan distance (sum of absolute differences in coordinates) or Euclidean distance?

Brute Force Solution

Approach

The brute force strategy for the squirrel problem is straightforward. We want to find the shortest total distance for the squirrel to collect all the nuts, so we'll try every possible order the squirrel could visit them in.

Here's how the algorithm would work step-by-step:

Consider every single possible sequence in which the squirrel could visit all the nuts.
For each of these sequences, calculate the total distance the squirrel would travel: from its starting point to the first nut, then from the first nut to the second, and so on, until it has visited all the nuts.
After calculating the total distance for every possible sequence, compare them all.
Select the sequence that resulted in the shortest total distance. That's our answer.

Code Implementation

from itertools import permutations

def calculate_distance(point1, point2):
    return abs(point1[0] - point2[0]) + abs(point1[1] - point2[1])

def squirrel_simulation_brute_force(height, width, squirrel_position, trees, nuts):
    shortest_distance = float('inf')
    
    # Generate all possible permutations of nut visiting order
    for nut_permutation in permutations(nuts):
        total_distance_for_permutation = 0

        # Calculate distance from squirrel to the first nut
        total_distance_for_permutation += calculate_distance(squirrel_position, nut_permutation[0])

        # Calculate distance between each nut in the permutation
        for i in range(len(nut_permutation) - 1):
            total_distance_for_permutation += calculate_distance(nut_permutation[i], nut_permutation[i+1])

        # Update the shortest distance found so far
        shortest_distance = min(shortest_distance, total_distance_for_permutation)

    return shortest_distance

Big(O) Analysis

Time Complexity

O(n!) – The core of the brute force solution involves exploring every possible order in which the squirrel can visit the nuts. If there are n nuts, there are n! (n factorial) possible sequences. For each of these n! sequences, we calculate the total distance, which takes O(n) time (summing the distances between n points). Therefore, the overall time complexity is O(n * n!), which is dominated by the factorial component, making it O(n!).

Space Complexity

O(N!) – The algorithm considers every possible sequence in which the squirrel could visit the N nuts. To achieve this, it implicitly explores all N! permutations. A recursive implementation, for example, might use the call stack to track the current permutation being explored, leading to a maximum call stack depth proportional to N in some implementations. In addition to the call stack, space is needed to store the current permutation of nuts being evaluated, which will be of size N. Therefore, the algorithm requires auxiliary space proportional to N!.

Optimal Solution

Approach

The best way to solve this problem is to figure out which tree the squirrel should drop the nut at first. The key is to minimize the extra distance the squirrel has to travel after that first drop by smartly connecting all other trees to the closest origin.

Here's how the algorithm would work step-by-step:

First, calculate the total distance the squirrel would travel if it started at the tree and delivered all the nuts directly to their intended locations without returning to the tree between nuts. This acts as our baseline distance.
Then, figure out how much distance the squirrel saves or loses for each tree if it started at the squirrel's location instead of the tree. Calculate this 'distance difference' for every nut location.
Find the nut location where starting at the squirrel's location saves the most distance. This is the optimal first nut delivery location for the squirrel.
Adjust the total baseline distance by subtracting the maximum distance saved (calculated in the previous step) to account for starting at the squirrel's location rather than the tree for that optimal first nut.
The final adjusted distance represents the shortest total distance the squirrel needs to travel to drop off all the nuts.

Code Implementation

def squirrel_simulation(height, width, tree_position, squirrel_position, nuts_positions):
    total_distance = 0
    
    # Calculate baseline distance assuming squirrel starts at tree
    for nut_position in nuts_positions:
        total_distance += 2 * (abs(tree_position[0] - nut_position[0]) + abs(tree_position[1] - nut_position[1]))

    max_distance_saved = float('-inf')

    # Find the maximum distance saved by starting at the squirrel
    for nut_position in nuts_positions:
        distance_from_tree = abs(tree_position[0] - nut_position[0]) + abs(tree_position[1] - nut_position[1])
        distance_from_squirrel = abs(squirrel_position[0] - nut_position[0]) + abs(squirrel_position[1] - nut_position[1])

        distance_saved = distance_from_tree - distance_from_squirrel
        max_distance_saved = max(max_distance_saved, distance_saved)

    # Adjust total distance by subtracting the max saved distance
    total_distance -= max_distance_saved

    return total_distance

Big(O) Analysis

Time Complexity

O(n) – The solution iterates through the nuts array twice. The first iteration calculates the total distance the squirrel would travel if it started at the tree and delivered all nuts directly. The second iteration calculates the distance difference for each nut location if the squirrel started at the squirrel's location instead of the tree. Both iterations are dependent on the number of nuts, n, leading to a time complexity of O(n) for each, therefore the whole solution is O(n).

Space Complexity

O(1) – The algorithm primarily uses a few constant space variables to store distances and the maximum distance saved. It iterates through the nut locations, but doesn't store them in any auxiliary data structures. Thus, the space required doesn't scale with the number of nuts or the dimensions of the grid. The space complexity is therefore constant.

Edge Cases

Case	How to Handle
Nuts array is null or empty	If nuts is null or empty, the total distance should be zero since there are no nuts to collect.
Height or width is zero or negative	If height or width is non-positive, it represents an invalid grid, so throw an IllegalArgumentException or return -1 to indicate error.
Squirrel, tree, or nut positions are out of bounds (negative or exceed height/width)	Validate that squirrel, tree and nut positions are within the grid boundaries and throw exception if invalid.
Squirrel and tree are at the same location	The solution should handle this case correctly as the distance between them will be zero, and the logic should still correctly calculate the total distance.
All nuts are at the same location	The solution should correctly calculate the distances even if all nuts are at the same coordinates.
Large number of nuts (performance consideration)	Ensure the solution uses an efficient algorithm (e.g., calculating Manhattan distances directly) to avoid performance issues with a large number of nuts.
Integer overflow when calculating distances	Use long data type to store distances to avoid integer overflow during calculations.
Only one nut exists	Calculate and return the sum of the distance from the squirrel to the nut and the distance from the nut to the tree.