#1954

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Minimum Garden Perimeter to Collect Enough Apples

Find the minimum perimeter of a square garden that holds enough apples based on a specific formula involving binary search.

math binary search

Problem Statement

In this problem, you are given an infinite 2D grid where each integer coordinate (i, j) contains |i| + |j| apples. Your goal is to buy a square plot of land centered at (0, 0), and you need to find the minimum perimeter of a square such that the total number of apples inside or on its boundary is at least equal to the given neededApples.

To solve this problem, you will use binary search to efficiently find the smallest possible side length of the square. The side length will determine the number of apples inside the square, and you will calculate the perimeter based on this side length.

Examples

Example 1

Input: neededApples = 1

Output: 8

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 details omitted.

Example 3

Input: neededApples = 1000000000

Output: 5040

Example details omitted.

Constraints

1 <= neededApples <= 1015

Solution Approach

Binary Search for Square Side Length

You can binary search over the possible side lengths of the square garden. Start with a reasonable range and check if a given side length can contain enough apples. Adjust the search bounds accordingly based on whether the number of apples inside is enough.

Formula for Apples Inside the Square

The number of apples inside a square garden can be calculated by summing the apples at all integer coordinates (i, j) within the square. You will derive a formula for the number of apples inside a square with side length L, which will be critical for efficiently checking each candidate side length.

Efficient Calculation of Perimeter

Once the side length L is determined via binary search, the perimeter of the square is calculated as 4 * L. This step is simple and directly follows the computation of the valid side length.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(log(N)) due to binary search over the side lengths, where N is the upper bound of possible side lengths. The space complexity is O(1), as we only need a few variables to track the search bounds and calculations.

What Interviewers Usually Probe

Candidate uses binary search effectively to minimize the number of checks needed to find the valid side length.
Candidate understands how to compute the apples inside the square using a formula.
Candidate knows how to calculate the perimeter after determining the side length of the square.

Common Pitfalls or Variants

Common pitfalls

Candidate fails to correctly derive or use the formula for the number of apples inside a square.
Candidate misuses binary search bounds, either searching too broadly or too narrowly.
Candidate fails to properly compute the perimeter once the side length is determined.

Follow-up variants

Add additional constraints on the grid size or allowed range for neededApples.
Allow for different shapes of land (rectangular instead of square) and adjust the perimeter calculation.
Use this problem as a stepping stone to more complex grid-based problems involving multiple plots or obstacles.

FAQ

What is the primary pattern used to solve 'Minimum Garden Perimeter to Collect Enough Apples'?

The main pattern is binary search over the valid answer space to find the minimum side length for the square garden.

How do you calculate the number of apples inside a square garden?

The number of apples inside a square garden is derived from a formula based on the sum of absolute values of the coordinates within the square.

What is the time complexity of solving the 'Minimum Garden Perimeter to Collect Enough Apples' problem?

The time complexity is O(log(N)), where N is the upper bound of possible side lengths of the square.

What happens if the side length of the square is too small to contain enough apples?

If the side length is too small, you adjust the binary search bounds to search for larger side lengths until the square contains enough apples.

Can this problem be solved without binary search?

While possible, it would be inefficient. Binary search significantly reduces the number of checks needed, making it the optimal approach.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minimumPerimeter(self, neededApples: int) -> int:
        x = 1
        while 2 * x * (x + 1) * (2 * x + 1) < neededApples:
            x += 1
        return x * 8

Solution 2

#### Python3

1

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class Solution:
    def minimumPerimeter(self, neededApples: int) -> int:
        x = 1
        while 2 * x * (x + 1) * (2 * x + 1) < neededApples:
            x += 1
        return x * 8

Continue Practicing

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