Problem Description

You have an infinite 2D grid where an apple tree is planted at every integer coordinate point. Each tree at coordinate (i, j) has exactly |i| + |j| apples growing on it, where |x| represents the absolute value of x.

You need to buy a square plot of land that:

• Is centered at the origin (0, 0)
• Has sides parallel to the axes (axis-aligned)
• Contains at least neededApples apples within or on its perimeter

Your task is to find the minimum perimeter of such a square plot.

For example:

• A tree at (0, 0) has |0| + |0| = 0 apples
• A tree at (2, 3) has |2| + |3| = 5 apples
• A tree at (-1, 2) has |-1| + |2| = 1 + 2 = 3 apples

The square plot is centered at the origin, so if it extends to distance x from the center, it covers all points (i, j) where -x ≤ i ≤ x and -x ≤ j ≤ x. The perimeter of such a square would be 8x.

Given an integer neededApples, return the minimum perimeter required to enclose at least that many apples.

Intuition

Since the square is centered at the origin and axis-aligned, if it extends to distance x from the center, it forms a square with corners at (-x, -x), (-x, x), (x, -x), and (x, x). The key insight is recognizing the pattern of how apples are distributed based on their Manhattan distance from the origin.

For any point (i, j), the number of apples is |i| + |j|. This creates a diamond-like pattern where all points with the same Manhattan distance have the same number of apples. For instance, all points at Manhattan distance 1 from the origin (like (1, 0), (0, 1), (-1, 0), (0, -1)) have exactly 1 apple each.

To count the total apples in a square extending to distance x, we need to sum up all apples at each point within the square. Due to symmetry, we can focus on one quadrant and multiply by 4. In the first quadrant (including axes), for each row i from 0 to x, we have columns j from 0 to x, giving us apples count of i + j at each point.

The total for one quadrant (including positive axes) is the sum of (i + j) for all 0 ≤ i ≤ x and 0 ≤ j ≤ x. This can be simplified using the formula for sum of integers: the sum becomes 2 * x * (x + 1) * (2 * x + 1) / 2 for the entire square after accounting for all four quadrants and avoiding double-counting the axes.

The final formula for total apples in a square of half-width x is 2 * x * (x + 1) * (2 * x + 1). We simply need to find the smallest x where this value is at least neededApples, and return the perimeter which is 8 * x.

Learn more about Math and Binary Search patterns.

Solution Approach

The solution uses a simple linear search approach to find the minimum value of x (half-width of the square) that satisfies our apple requirement.

Starting with x = 1, we iteratively check if a square with half-width x contains enough apples using the formula 2 * x * (x + 1) * (2 * x + 1). This formula represents the total number of apples within a square that extends x units from the origin in all four directions.

The algorithm works as follows:

1. Initialize x = 1 (starting with the smallest possible square)

2. While the total apples in the current square is less than neededApples:

   • Check if 2 * x * (x + 1) * (2 * x + 1) < neededApples
   • If true, increment x by 1 to try a larger square
   • If false, we've found our minimum x

3. Once we find the minimum x that satisfies the condition, calculate the perimeter as 8 * x

The perimeter calculation is straightforward: since the square extends x units in all four directions from the origin, each side has length 2 * x, giving us a total perimeter of 4 * (2 * x) = 8 * x.

This brute force approach is efficient enough because the number of apples grows cubically with x (proportional to x³), meaning we don't need to iterate through many values before finding our answer. For example, when x = 10, we already have over 4,000 apples, and when x = 100, we have over 4 million apples.

Example Walkthrough

Let's walk through finding the minimum perimeter for neededApples = 12.

Step 1: Start with x = 1

• Square extends from (-1, -1) to (1, 1)
• Trees included: all points where -1 ≤ i ≤ 1 and -1 ≤ j ≤ 1
• Let's count the apples:
  • Corner trees: (±1, ±1) each have |±1| + |±1| = 2 apples → 4 trees × 2 = 8 apples
  • Edge trees: (±1, 0) and (0, ±1) each have 1 apple → 4 trees × 1 = 4 apples
  • Center tree: (0, 0) has 0 apples
• Total: 8 + 4 + 0 = 12 apples
• Using formula: 2 × 1 × (1 + 1) × (2 × 1 + 1) = 2 × 1 × 2 × 3 = 12 apples ✓

Since 12 ≥ 12, we found our answer! The minimum x = 1.

Step 2: Calculate perimeter

• Perimeter = 8 × x = 8 × 1 = 8

Let's verify with a larger example where we need to iterate. For neededApples = 50:

x = 1: 2 × 1 × 2 × 3 = 12 apples < 50 ❌ x = 2: 2 × 2 × 3 × 5 = 60 apples ≥ 50 ✓

The square with x = 2 extends from (-2, -2) to (2, 2), containing trees like:

• (0, 0): 0 apples
• (1, 1): 2 apples
• (2, 0): 2 apples
• (2, 2): 4 apples
• And many more...

Total: 60 apples (verified by formula) Perimeter = 8 × 2 = 16

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^(1/3)) where n is the value of neededApples.

The algorithm uses a linear search starting from x = 1 and incrementing until the condition 2 * x * (x + 1) * (2 * x + 1) >= neededApples is satisfied. The formula 2 * x * (x + 1) * (2 * x + 1) represents the total number of apples within a square garden of side length 2x. This formula grows as O(x^3), so when it equals neededApples, we have x^3 ≈ neededApples, which means x ≈ neededApples^(1/3). Therefore, the loop runs approximately O(neededApples^(1/3)) times.

Space Complexity: O(1)

The algorithm only uses a constant amount of extra space for the variable x. No additional data structures are used that scale with the input size.

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

Common Pitfalls

1. Integer Overflow in Formula Calculation

The formula 2 * x * (x + 1) * (2 * x + 1) involves multiplying large numbers together. For large values of x, intermediate calculations might exceed integer limits in some programming languages. While Python handles arbitrary precision integers automatically, this could be problematic in languages like Java or C++.

Solution:

• In Python, this isn't an issue due to automatic handling of large integers
• In other languages, use long/long long data types or check for overflow conditions
• Consider breaking the multiplication into steps and checking for overflow at each stage

2. Inefficient Search for Very Large Values

While the linear search works well for most cases due to cubic growth, for extremely large neededApples values (e.g., 10^15), incrementing by 1 could require millions of iterations.

Solution: Use binary search instead of linear search for better time complexity:

class Solution:
    def minimumPerimeter(self, neededApples: int) -> int:
        left, right = 1, 100000  # Set reasonable bounds
      
        while left < right:
            mid = (left + right) // 2
            apples = 2 * mid * (mid + 1) * (2 * mid + 1)
          
            if apples < neededApples:
                left = mid + 1
            else:
                right = mid
      
        return left * 8

3. Misunderstanding the Apple Count Formula

Developers might incorrectly derive or implement the formula for counting apples. Common mistakes include:

• Forgetting to account for all four quadrants
• Double-counting the axes or origin
• Using an incorrect summation formula

Solution: Verify the formula by manually calculating small cases:

• For x=1: The square contains points from (-1,-1) to (1,1), total apples = 12
• For x=2: The square contains points from (-2,-2) to (2,2), total apples = 60
• Confirm these match 2 * x * (x + 1) * (2 * x + 1)

4. Off-by-One Errors in Perimeter Calculation

Some might mistakenly calculate the perimeter as 4 * x (thinking of radius) or 4 * (2x + 1) (thinking of discrete points).

Solution: Remember that:

• The square extends x units in each direction from origin
• Each side has length 2x (from -x to x)
• Total perimeter = 4 sides × 2x = 8x