Problem Description

You have a garden represented as a grid with dimensions height x width. In this garden, there are three types of entities:

1. A tree at position [tree_r, tree_c]
2. A squirrel starting at position [squirrel_r, squirrel_c]
3. Multiple nuts at various positions given in the array nuts, where each nuts[i] = [nut_ir, nut_ic] represents the position of the i-th nut

The squirrel needs to collect all the nuts and bring them to the tree, following these rules:

• The squirrel can carry only one nut at a time
• The squirrel can move in four directions (up, down, left, right) to adjacent cells
• Each move to an adjacent cell counts as one unit of distance
• The distance between any two positions is calculated using Manhattan distance: |r1 - r2| + |c1 - c2|

The squirrel's task is to:

1. Start from its initial position
2. Move to pick up a nut
3. Carry that nut to the tree
4. From the tree, go pick up another nut
5. Return to the tree with that nut
6. Repeat steps 4-5 until all nuts are collected

Your goal is to find the minimum total distance the squirrel needs to travel to collect all nuts and place them under the tree.

The key insight is that after the squirrel picks up and delivers the first nut to the tree, for all subsequent nuts, the squirrel must travel from the tree to each nut and back. This means each subsequent nut contributes 2 * (distance from nut to tree) to the total distance. The only variable part is which nut the squirrel chooses first, since for that first nut, the squirrel travels from its starting position to the nut, then to the tree, rather than starting from the tree.

Intuition

Let's think about the squirrel's journey step by step. After collecting all nuts, what path did the squirrel actually take?

First, observe that except for the very first nut, every other nut follows the same pattern: the squirrel starts from the tree, goes to get the nut, and comes back to the tree. This means each of these nuts contributes exactly 2 * (distance from nut to tree) to the total path length.

For the first nut, however, the pattern is different. The squirrel starts from its initial position, goes to the nut, then goes to the tree. So the first nut contributes (distance from squirrel to nut) + (distance from nut to tree) to the total path.

Now here's the key insight: If we imagine that the squirrel started at the tree instead of its actual starting position, then every nut (including the first) would contribute 2 * (distance from nut to tree) to the total. Let's call this hypothetical total distance S.

The actual distance differs from S only because of the first nut. For whichever nut we choose as the first:

• We remove one round trip from tree to that nut: -2 * (distance from nut to tree)
• We add the actual path for the first nut: +(distance from squirrel to nut) + (distance from nut to tree)

Simplifying: the difference is (distance from squirrel to nut) - (distance from nut to tree).

Therefore, the total actual distance is: S + (distance from squirrel to first nut) - (distance from first nut to tree)

Since S is fixed (it's 2 * sum of all distances from nuts to tree), we want to minimize the adjustment term. We should choose the first nut that minimizes (distance from squirrel to nut) - (distance from nut to tree).

By iterating through all nuts and checking which one gives us the minimum total distance when chosen as the first nut, we can find the optimal solution.

Learn more about Math patterns.

Solution Approach

Based on our intuition, we can implement the solution by calculating the total distance assuming all nuts require a round trip from the tree, then finding the optimal first nut to minimize the total distance.

Step 1: Calculate the base distance

First, we calculate the sum of distances from all nuts to the tree, multiplied by 2. This represents the total distance if the squirrel started at the tree:

s = sum(abs(r - tr) + abs(c - tc) for r, c in nuts) * 2

Here, (tr, tc) is the tree's position, and for each nut at position (r, c), we calculate the Manhattan distance abs(r - tr) + abs(c - tc).

Step 2: Find the optimal first nut

For each nut, we calculate what the total distance would be if we chose it as the first nut. The formula for the total distance when choosing nut i as the first is:

Total = s - (distance from nut_i to tree) + (distance from squirrel to nut_i)

Let's denote:

• a = distance from nut_i to tree = abs(r - tr) + abs(c - tc)
• b = distance from squirrel to nut_i = abs(r - sr) + abs(c - sc)

Then the total distance becomes: s - a + b

Step 3: Select the minimum

We iterate through all nuts and keep track of the minimum total distance:

ans = inf
for r, c in nuts:
    a = abs(r - tr) + abs(c - tc)  # distance from nut to tree
    b = abs(r - sr) + abs(c - sc)  # distance from squirrel to nut
    ans = min(ans, s - a + b)

The algorithm has a time complexity of O(n) where n is the number of nuts, as we iterate through the nuts twice (once for calculating s, once for finding the minimum). The space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given:

• Tree position: [2, 2]
• Squirrel position: [4, 4]
• Nuts positions: [[3, 0], [2, 5]]

Step 1: Calculate base distance (all nuts as round trips from tree)

For each nut, calculate distance to tree and multiply by 2:

• Nut 1 at [3, 0]: Distance to tree = |3-2| + |0-2| = 1 + 2 = 3
  • Round trip = 3 × 2 = 6
• Nut 2 at [2, 5]: Distance to tree = |2-2| + |5-2| = 0 + 3 = 3
  • Round trip = 3 × 2 = 6

Base distance s = 6 + 6 = 12

Step 2: Try each nut as the first nut

For each nut as first choice, calculate: s - (nut to tree) + (squirrel to nut)

Option 1: Choose nut at [3, 0] first

• Distance from nut to tree: 3 (calculated above)
• Distance from squirrel to nut: |4-3| + |4-0| = 1 + 4 = 5
• Total distance: 12 - 3 + 5 = 14

Path visualization:

1. Squirrel [4,4] → Nut1 [3,0]: distance = 5
2. Nut1 [3,0] → Tree [2,2]: distance = 3
3. Tree [2,2] → Nut2 [2,5]: distance = 3
4. Nut2 [2,5] → Tree [2,2]: distance = 3 Total: 5 + 3 + 3 + 3 = 14 ✓

Option 2: Choose nut at [2, 5] first

• Distance from nut to tree: 3 (calculated above)
• Distance from squirrel to nut: |4-2| + |4-5| = 2 + 1 = 3
• Total distance: 12 - 3 + 3 = 12

Path visualization:

1. Squirrel [4,4] → Nut2 [2,5]: distance = 3
2. Nut2 [2,5] → Tree [2,2]: distance = 3
3. Tree [2,2] → Nut1 [3,0]: distance = 3
4. Nut1 [3,0] → Tree [2,2]: distance = 3 Total: 3 + 3 + 3 + 3 = 12 ✓

Step 3: Select minimum

The minimum total distance is 12, achieved by choosing the nut at [2, 5] first.

Notice how the formula correctly captures the optimization: we save distance when the first nut is close to the squirrel relative to its distance from the tree. In this case, the nut at [2, 5] has the same distance to both the squirrel and the tree (3 units each), making (squirrel to nut) - (nut to tree) = 0, which gives us the best adjustment to the base distance.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the number of nuts. The algorithm iterates through the nuts list twice - once to calculate the sum of Manhattan distances from all nuts to the tree (which takes O(n) time), and once more to find the minimum cost by trying each nut as the first nut the squirrel picks up (another O(n) iteration). Within each iteration of the second loop, the operations (calculating Manhattan distances and updating the minimum) take O(1) time. Therefore, the overall time complexity is O(n) + O(n) = O(n).

The space complexity is O(1). The algorithm only uses a constant amount of extra space to store variables like tr, tc, sr, sc, s, ans, a, and b. These variables don't depend on the input size. The input lists are not copied or modified, and no additional data structures that scale with input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Handle Empty Nuts Array

The most common pitfall is not considering the edge case where the nuts array is empty. If there are no nuts to collect, the squirrel doesn't need to move at all, and the minimum distance should be 0.

Problematic Code:

# This will fail if nuts is empty
total_tree_distance = sum(
    abs(nut_row - tree_row) + abs(nut_col - tree_col) 
    for nut_row, nut_col in nuts
) * 2  # sum() returns 0 for empty list, which is fine

min_distance = float('inf')
for nut_row, nut_col in nuts:
    # Never enters loop if nuts is empty
    ...
return min_distance  # Returns inf instead of 0!

Solution:

def minDistance(self, height, width, tree, squirrel, nuts):
    # Handle edge case: no nuts to collect
    if not nuts:
        return 0
  
    # Rest of the implementation...

2. Integer Overflow in Other Languages

While Python handles large integers automatically, implementing this in languages like C++ or Java could cause integer overflow when calculating the sum of distances for large grids or many nuts.

Solution: Use appropriate data types (e.g., long long in C++ or long in Java) to prevent overflow.

3. Misunderstanding the First Nut Logic

A conceptual pitfall is incorrectly calculating the adjustment for the first nut. Some might think:

• Wrong: total_distance = base_distance + squirrel_to_first_nut
• Correct: total_distance = base_distance - tree_to_first_nut + squirrel_to_first_nut

The key insight is that for the first nut, we save one leg of the round trip from the tree (hence subtracting tree_to_first_nut) and add the distance from squirrel's starting position to that nut.

4. Using Euclidean Distance Instead of Manhattan Distance

The problem specifically requires Manhattan distance, but it's easy to accidentally use Euclidean distance formula.

Wrong:

distance = ((r1 - r2) ** 2 + (c1 - c2) ** 2) ** 0.5

Correct:

distance = abs(r1 - r2) + abs(c1 - c2)