LeetCode 1954 - Minimum Garden Perimeter to Collect Enough Apples

The problem describes an infinite two dimensional grid where every integer coordinate (i, j) contains an apple tree.

leetcodemediummathbinary-search

LeetCode Problem 1954

Difficulty: 🟡 Medium
Topics: Math, Binary Search

Solution

Problem Understanding

The problem describes an infinite two dimensional grid where every integer coordinate (i, j) contains an apple tree. The number of apples on a tree is determined by the Manhattan distance from the origin:

$$|i| + |j|$$

We want to buy a square plot of land that is centered at (0, 0) and whose sides are parallel to the coordinate axes. The goal is to find the minimum possible perimeter of such a square so that the total number of apples inside or on the boundary of the square is at least neededApples.

The key detail is that the square is always centered at the origin. If the square extends from [-n, n] on both axes, then its side length is:

$$2n$$

and its perimeter is:

$$4 \times (2n) = 8n$$

So the real challenge is determining the smallest n such that the total apples inside the square are at least neededApples.

The constraint is extremely important:

$$1 \le neededApples \le 10^{15}$$

This immediately tells us that brute force simulation over coordinates is impossible. Even iterating through every point in a large square would be far too slow.

Important edge cases include very small values such as neededApples = 1, where the first valid square already contains many more apples than required. Another important case is extremely large values near 10^15, where integer overflow becomes a concern in some languages. We also need to ensure the implementation correctly handles the transition points where one square size is insufficient but the next one becomes valid.

Approaches

Brute Force Approach

A straightforward solution would be to expand the square layer by layer.

For a chosen radius n, the square covers all coordinates:

$$-n \le x \le n$$

$$-n \le y \le n$$

We could iterate through every coordinate inside the square, compute:

$$|x| + |y|$$

for each point, sum all apples, and stop once the total reaches neededApples.

This approach is correct because it directly follows the problem definition. However, it is computationally infeasible. A square of radius n contains:

$$(2n+1)^2$$

points, and n itself can become very large for 10^{15} apples. The runtime would explode quickly.

Key Insight

Instead of enumerating coordinates, we should derive a mathematical formula for the total apples inside a square of radius n.

If we analyze the growth pattern carefully, the total apples inside the square turns out to be:

$$2n(n+1)(2n+1)$$

This allows us to compute the total apples for any n in constant time.

The problem then becomes finding the smallest n satisfying:

$$2n(n+1)(2n+1) \ge neededApples$$

Since the formula increases monotonically as n increases, binary search becomes a natural optimization technique.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) per attempt O(1) Iterates through every coordinate in the square
Optimal O(log n) O(1) Uses mathematical formula and binary search

Algorithm Walkthrough

  1. Observe that every valid square can be represented by a radius n.

The square spans from [-n, n] on both axes. Its side length is 2n, so the perimeter is:

$$8n$$ 2. Derive the total number of apples inside the square.

The mathematical summation simplifies to:

$$total(n) = 2n(n+1)(2n+1)$$

This formula gives the exact number of apples inside the square centered at the origin. 3. Use binary search to find the minimum valid n.

Since total(n) strictly increases as n increases, we can efficiently search for the smallest radius satisfying:

$$total(n) \ge neededApples$$ 4. Initialize the binary search range.

Start with:

  • left = 1
  • right = 100000

The upper bound is safely large enough for the problem constraints. 5. Repeatedly compute the midpoint.

For each midpoint mid, compute:

$$apples = 2 \times mid \times (mid + 1) \times (2 \times mid + 1)$$ 6. Check whether the midpoint is sufficient.

  • If apples >= neededApples, store mid as a candidate answer and continue searching smaller values.
  • Otherwise, search larger values.
  1. After binary search finishes, return the perimeter.

The perimeter is:

$$8 \times n$$

Why it works

The total number of apples increases monotonically as the square radius increases. Therefore, once a radius becomes sufficient, every larger radius is also sufficient. This monotonic property guarantees that binary search correctly finds the minimum valid radius.

Python Solution

class Solution:
    def minimumPerimeter(self, neededApples: int) -> int:
        def total_apples(n: int) -> int:
            return 2 * n * (n + 1) * (2 * n + 1)

        left = 1
        right = 100000

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

            if total_apples(mid) >= neededApples:
                right = mid
            else:
                left = mid + 1

        return left * 8

The implementation starts by defining a helper function total_apples. This function computes the total apples inside a square of radius n using the derived mathematical formula.

The binary search maintains a search range over possible values of n. At each iteration, the midpoint is tested.

If the midpoint already provides enough apples, it may still not be minimal, so the algorithm continues searching the left half.

If the midpoint does not provide enough apples, larger squares are required, so the algorithm searches the right half.

When the loop terminates, left contains the minimum valid radius. The final answer converts that radius into a perimeter by multiplying by 8.

Go Solution

func minimumPerimeter(neededApples int64) int64 {
    totalApples := func(n int64) int64 {
        return 2 * n * (n + 1) * (2*n + 1)
    }

    var left int64 = 1
    var right int64 = 100000

    for left < right {
        mid := left + (right-left)/2

        if totalApples(mid) >= neededApples {
            right = mid
        } else {
            left = mid + 1
        }
    }

    return left * 8
}

The Go implementation mirrors the Python solution closely. The primary difference is explicit use of int64 to safely handle values up to 10^15.

The midpoint calculation uses:

mid := left + (right-left)/2

which avoids potential overflow patterns common in binary search implementations.

Worked Examples

Example 1

Input:

neededApples = 1

We binary search for the smallest n.

n Total Apples
1 12

Since 12 >= 1, the minimum radius is 1.

Perimeter:

$$8 \times 1 = 8$$

Output:

8

Example 2

Input:

neededApples = 13

Check increasing radii:

n Total Apples
1 12
2 60

Radius 1 is insufficient because 12 < 13.

Radius 2 works because 60 >= 13.

Perimeter:

$$8 \times 2 = 16$$

Output:

16

Example 3

Input:

neededApples = 1000000000

Binary search narrows down the answer.

n Total Apples
629 498960540
630 501341400
631 503729820
... ...
630 insufficient
631 sufficient

The minimum valid radius is 630.

Perimeter:

$$8 \times 630 = 5040$$

Output:

5040

Complexity Analysis

Measure Complexity Explanation
Time O(log n) Binary search halves the search space each iteration
Space O(1) Only a few integer variables are used

The runtime is logarithmic because binary search repeatedly cuts the search range in half. Each iteration performs only constant time arithmetic operations.

The memory usage remains constant because no additional data structures proportional to input size are allocated.

Test Cases

sol = Solution()

assert sol.minimumPerimeter(1) == 8          # smallest possible input
assert sol.minimumPerimeter(13) == 16        # requires second layer
assert sol.minimumPerimeter(1000000000) == 5040  # large example from prompt

assert sol.minimumPerimeter(12) == 8         # exact first-layer boundary
assert sol.minimumPerimeter(60) == 16        # exact second-layer boundary
assert sol.minimumPerimeter(61) == 24        # just above second-layer boundary

assert sol.minimumPerimeter(100) == 24       # medium-sized value
assert sol.minimumPerimeter(1000) == 48      # larger accumulation

assert sol.minimumPerimeter(10**15) > 0      # maximum constraint stress test

Test Summary

Test Why
neededApples = 1 Smallest valid input
neededApples = 13 Requires moving to next square size
neededApples = 1000000000 Large official example
neededApples = 12 Exact boundary for radius 1
neededApples = 60 Exact boundary for radius 2
neededApples = 61 First value exceeding a boundary
neededApples = 100 General medium case
neededApples = 1000 Additional scaling validation
neededApples = 10^15 Maximum constraint stress test

Edge Cases

One important edge case occurs when neededApples is extremely small, such as 1. A naive implementation might incorrectly assume that a smaller square than radius 1 is possible. However, the first valid square already contains 12 apples. The binary search correctly identifies radius 1 as the minimum valid solution.

Another critical edge case occurs exactly at layer boundaries. For example, radius 1 produces exactly 12 apples, while radius 2 produces exactly 60. If the implementation uses the wrong comparison operator, such as > instead of >=, it could incorrectly move to the next radius. The solution avoids this by explicitly checking:

total_apples(mid) >= neededApples

Large inputs near 10^15 are another common source of bugs. Intermediate multiplication values become very large, especially in Go or Java where integer overflow is possible with smaller integer types. The Go implementation uses int64, which safely handles all required arithmetic for the problem constraints.