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Minimum Garden Perimeter to Collect Enough Apples

Difficulty: Medium

Problem Description

In a garden represented as an infinite 2D grid, there is an apple tree planted at every integer coordinate. The apple tree planted at an integer coordinate (i, j) has |i| + |j| apples growing on it. You will buy an axis-aligned square plot of land that is centered at (0, 0). Given an integer neededApples, return the minimum perimeter of a plot such that at least neededApples apples are inside or on the perimeter of that plot.

Key Insights

  • Each tree at coordinate (i, j) has |i| + |j| apples, which means the number of apples increases as you move away from the origin.
  • The square plot of side length L has corners at (-L/2, -L/2), (L/2, -L/2), (L/2, L/2), and (-L/2, L/2).
  • The total number of apples inside and on the perimeter of the square can be calculated based on the distance from the origin.
  • The perimeter of a square of side length L is 4 * L.
  • A binary search can be used to efficiently find the minimum side length L that yields at least neededApples.

Space and Time Complexity

Time Complexity: O(log(n)) where n is the neededApples, due to the binary search over possible lengths of the square. Space Complexity: O(1), as we only use a constant amount of space.

Solution

To solve the problem, we can use a binary search approach. We start with an initial range for the side length of the square, from 0 to a sufficiently large value. For each potential side length L, we calculate the number of apples contained within the square. If this number is greater than or equal to neededApples, we adjust our search range to potentially find a smaller valid L. Once we find the minimum L that meets the requirement, we calculate the perimeter as 4 * L.

Code Solutions

def minPerimeter(neededApples: int) -> int:
    # Function to calculate the number of apples inside the square of side length L
    def apples_in_square(L: int) -> int:
        # The total apples can be calculated based on the layers of squares
        total = 0
        for i in range(0, (L // 2) + 1):
            total += (L // 2 - i + 1) * (i + 1)
        return total

    # Binary search for the minimum side length L
    left, right = 0, 2 * int((neededApples**0.5)) + 1  # Initial guess for right
    while left < right:
        mid = (left + right) // 2
        if apples_in_square(mid) >= neededApples:
            right = mid  # Look for a smaller square
        else:
            left = mid + 1  # Look for a larger square

    return 4 * left  # Perimeter is 4 times the side length
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