Code Implementation

class Solution:
    def minimumPerimeter(self, neededApples: int) -> int:
        def apples_in_layer(n):
            # Total apples in square from (-n,-n) to (n,n)
            return 2 * n * (n + 1) * (2 * n + 1)
        
        left, right = 1, 10**6
        answer = right
        while left <= right:
            mid = (left + right) // 2
            if apples_in_layer(mid) >= neededApples:
                answer = mid
                right = mid - 1
            else:
                left = mid + 1
        return answer * 8
      

class Solution {
public:
    long long applesInLayer(long long n) {
        return 2 * n * (n + 1) * (2 * n + 1);
    }
    long long minimumPerimeter(long long neededApples) {
        long long left = 1, right = 1e6, answer = right;
        while (left <= right) {
            long long mid = (left + right) / 2;
            if (applesInLayer(mid) >= neededApples) {
                answer = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return answer * 8;
    }
};
      

class Solution {
    private long applesInLayer(long n) {
        return 2 * n * (n + 1) * (2 * n + 1);
    }
    public long minimumPerimeter(long neededApples) {
        long left = 1, right = 1000000, answer = right;
        while (left <= right) {
            long mid = (left + right) / 2;
            if (applesInLayer(mid) >= neededApples) {
                answer = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return answer * 8;
    }
}
      

var minimumPerimeter = function(neededApples) {
    function applesInLayer(n) {
        return 2 * n * (n + 1) * (2 * n + 1);
    }
    let left = 1, right = 1e6, answer = right;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (applesInLayer(mid) >= neededApples) {
            answer = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    return answer * 8;
};
      

Problem Description

You are given an infinite 2D grid of integer coordinates, and at every point (i, j), there are |i| + |j| apples. You want to collect at least neededApples apples by picking all apples inside or on the boundary of a square region centered at the origin (0, 0). The square must have sides of even length, so its perimeter is always a multiple of 8.

Your task is to find the minimum perimeter of such a square region so that the total number of apples inside (and on the boundary) is at least neededApples.

• There is always one valid solution for any positive neededApples.
• You must not reuse any apples from outside the chosen region.

Thought Process

The naive approach would be to simulate the grid, sum apples for increasing square sizes, and stop once the total meets or exceeds neededApples. However, the grid is infinite and the numbers get large quickly, so brute-forcing would be slow and memory-intensive.

Instead, we look for a mathematical formula to calculate the total apples for a given square size. Since the apple count at each point is based on its Manhattan distance from the origin, and the region is always a square centered at the origin, we can derive a closed-form for the sum.

The main insight is that as the square grows, the number of apples increases rapidly, so we can use binary search to efficiently find the minimum square size needed.

Solution Approach

• 1. Model the Problem:
  • Let the square have side length 2n, so its perimeter is 8n.
  • We want to compute the total apples in the region from (-n, -n) to (n, n).
• 2. Derive the Formula:
  • For each point (i, j), apples = |i| + |j|.
  • Sum over all points in the square:
    • Total apples = 2 * n * (n + 1) * (2n + 1)
• 3. Use Binary Search:
  • Since apples increase rapidly with n, do binary search on n to find the smallest n such that the formula gives at least neededApples.
  • For each guess, calculate total apples with the formula and adjust the search range accordingly.
• 4. Return the Perimeter:
  • Once the minimum n is found, return 8 * n as the answer.

This approach is efficient, using mathematical insight and binary search to avoid unnecessary computation.

Example Walkthrough

Suppose neededApples = 100:

• Start with n = 1: Apples = 2 * 1 * 2 * 3 = 12
• n = 2: Apples = 2 * 2 * 3 * 5 = 60
• n = 3: Apples = 2 * 3 * 4 * 7 = 168

At n = 2, we have 60 apples, which is less than 100. At n = 3, we have 168 apples, which is enough. So, the minimum n is 3, and the minimum perimeter is 8 * 3 = 24.

Binary Search Steps:

• Set left = 1, right = large number (e.g., 1e6)
• Check mid = (left + right) // 2, compute apples, adjust left/right
• Continue until the smallest valid n is found

Time and Space Complexity

• Brute-force:
  • Would require O(n²) operations for each square, leading to extremely slow performance for large neededApples.
• Optimized Approach:
  • Binary search runs in O(log N), where N is the maximum search boundary (e.g., 1e6).
  • Each binary search step is O(1) due to the closed-form formula for apples.
  • Space complexity is O(1), as only a few variables are used.

Summary

The key to this problem is recognizing the mathematical pattern in the apple distribution and leveraging it to avoid brute-force computation. By deriving a closed-form formula and using binary search, we efficiently find the minimum perimeter required to collect enough apples. This approach is both elegant and highly performant, making it suitable for very large inputs.