Introduction

In this blog post, we will explore how to solve the LeetCode problem "1954" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. The value of |x| is defined as: x if x >= 0 -x if x < 0 Example 1: Input: neededApples = 1 Output: 8 Explanation: A square plot of side length 1 does not contain any apples. However, a square plot of side length 2 has 12 apples inside (as depicted in the image above). The perimeter is 2 * 4 = 8. Example 2: Input: neededApples = 13 Output: 16 Example 3: Input: neededApples = 1000000000 Output: 5040 Constraints: 1 <= neededApples <= 1015

Explanation

Here's a breakdown of the solution:

• Binary Search: The key insight is that the number of apples within a square plot is monotonically increasing with the side length of the square. This allows us to efficiently search for the minimum side length using binary search.
• Apple Calculation: The core of the solution lies in calculating the total number of apples within a square plot of a given side length. We can derive a formula to do this efficiently, avoiding iteration over individual coordinates.

• Formula Derivation: The total apples within the square with side 2*size can be calculated by summing the apples in each layer from the center. Each layer i where 1 <= i <= size has 8*i positions with apple counts of i. Also sum up positions (i, i), (-i, -i), (i, -i), (-i, i) each has apple count = 2 * i.

• Time & Space Complexity: O(log N) time, where N is the search space for the side length, and O(1) space.

Code

    def calculate_apples(size):
    """Calculates the total number of apples within a square plot of side length 2*size."""
    return 2 * size * (size + 1) * (2 * size + 1)

def minimumPerimeter(neededApples):
    """
    Finds the minimum perimeter of a square plot containing at least neededApples apples.
    """
    left = 0
    right = 100000  # Set upper bound based on constraint 10^15
    ans = 0

    while left <= right:
        mid = (left + right) // 2
        apples = calculate_apples(mid)

        if apples >= neededApples:
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return 8 * ans