Minimum Garden Perimeter to Collect Enough Apples

Med

#1781Time: O(N^(1/3)), where `N` is `neededApples`. The number of apples `C(k)` grows approximately as `4k^3`. Thus, the value of `k` we are looking for is proportional to the cube root of `N`. The loop runs `k` times.Space: O(1), as we only use a few variables to store the current `k` and apple count.

Patterns

Math

Algorithms

Binary Search

Prompt

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

Approaches

3 approaches with complexity analysis and trade-offs.

This is a straightforward brute-force approach. We test values of k starting from 1, where 2k is the side length of the square plot. For each k, we calculate the total number of apples within the plot and stop as soon as this number meets or exceeds neededApples.

Algorithm

Start a loop with k = 1, where 2k represents the side length of the square plot.
In each iteration, calculate the total apples for the square defined by [-k, k] on both axes using the formula: currentApples = 2L * k * (k + 1) * (2L * k + 1). Note the use of L to ensure calculations are done using 64-bit longs to prevent overflow.
Check if currentApples >= neededApples.
If the condition is met, we have found the smallest k. The perimeter is 8 * k. Return this value and terminate.
If not, increment k and continue to the next iteration.

Walkthrough

The total number of apples C(k) for a square plot extending from -k to k on both axes is given by the formula C(k) = 2 * k * (k+1) * (2k+1). The algorithm iterates k from 1 upwards. In each step, it computes C(k) and compares it with neededApples. The first k for which C(k) >= neededApples is the one we need. The perimeter is then 8 * k. We must use 64-bit integers (long in Java) for the apple count to avoid overflow, as neededApples can be up to 10^15.

1class Solution {2    public long minimumPerimeter(long neededApples) {3        long k = 0;4        while (true) {5            k++;6            // Use long for all parts of the calculation to prevent overflow7            long currentApples = 2 * k * (k + 1) * (2 * k + 1);8            if (currentApples >= neededApples) {9                break;10            }11        }12        return 8 * k;13    }14}

Complexity

Time

O(N^(1/3)), where `N` is `neededApples`. The number of apples `C(k)` grows approximately as `4k^3`. Thus, the value of `k` we are looking for is proportional to the cube root of `N`. The loop runs `k` times.

Space

O(1), as we only use a few variables to store the current `k` and apple count.

Trade-offs

Pros

Simple to understand and implement.
Requires minimal mathematical analysis beyond deriving the apple count formula.

Cons

While it passes for the given constraints, it can be inefficient if neededApples were significantly larger, as the number of iterations is proportional to the cube root of neededApples.

The number of apples, C(k) = 2k(k+1)(2k+1), is a monotonically increasing function of k. This property allows us to use binary search to find the smallest k that satisfies C(k) >= neededApples much more efficiently than a linear search.

Algorithm

Establish a search range for k. The lower bound low can be 1. The upper bound high can be a safe, sufficiently large number like 100000 (since k for 10^15 apples is ~63000).
Initialize a variable ans to hold the result.
Loop while low <= high:
- Calculate the midpoint mid = low + (high - low) / 2.
- Compute the number of apples for k = mid: apples = 2L * mid * (mid + 1) * (2L * mid + 1).
- If apples >= neededApples, it means mid is a possible answer. We store it (ans = mid) and try to find a smaller k by searching in the left half (high = mid - 1).
- If apples < neededApples, mid is too small. We must search for a larger k in the right half (low = mid + 1).
After the loop terminates, ans holds the smallest k found. Return the perimeter 8 * ans.

Walkthrough

Instead of checking every k from 1, binary search intelligently narrows down the possibilities. We define a search space for k and repeatedly divide it in half. For each midpoint mid, we calculate the apple count C(mid). If C(mid) is sufficient, we know mid is a possible answer, and we try to find an even smaller k that also works by searching in the lower half of the range. If C(mid) is insufficient, we know mid and all smaller values are not the answer, so we must search for a larger k in the upper half. This process quickly converges to the smallest k that satisfies the condition.

1class Solution {2    public long minimumPerimeter(long neededApples) {3        long low = 1, high = 100000; // A safe upper bound for k4        long ans = high;5        while (low <= high) {6            long mid = low + (high - low) / 2;7            long currentApples = 2 * mid * (mid + 1) * (2 * mid + 1);8            9            if (currentApples >= neededApples) {10                ans = mid;11                high = mid - 1;12            } else {13                low = mid + 1;14            }15        }16        return ans * 8;17    }18}

Complexity

Time

O(log U), where `U` is the upper bound of the search space for `k`. This is extremely fast.

Space

O(1), as it only requires a few variables for the search bounds and answer.

Trade-offs

Pros

Significantly faster than linear search, with logarithmic time complexity.
Guaranteed to find the solution quickly and is a robust method for monotonic functions.

Cons

Slightly more complex to implement than a simple loop.
Requires choosing a safe upper bound for the search space, which might need some preliminary analysis.

This is the most efficient approach, leveraging a mathematical approximation to find the required value of k almost directly. This nearly eliminates the need for a search, resulting in a constant-time solution.

Algorithm

We need to find the smallest integer k such that 2k(k+1)(2k+1) >= neededApples.
Approximate the formula 2k(k+1)(2k+1) as 4k^3 for large k.
Solve the approximation for k: 4k^3 ≈ neededApples => k ≈ (neededApples / 4)^(1/3).
Calculate an initial guess for k using the cube root function: k_guess = (long) Math.cbrt(neededApples / 4.0).
The actual integer k will be very close to k_guess. Start a search from k_guess (or a slightly smaller value like k_guess - 2 to be safe) and increment until the exact formula 2k(k+1)(2k+1) >= neededApples is satisfied.
The number of checks required will be very small, making the search part negligible.

Walkthrough

The formula for the number of apples, C(k) = 2k(k+1)(2k+1), is dominated by the k^3 term for large values of k. We can approximate C(k) as 4k^3. By setting 4k^3 ≈ neededApples, we can solve for k to get an excellent estimate: k ≈ (neededApples / 4)^(1/3). We can compute this value using a cube root function (Math.cbrt in Java). The true integer solution will be extremely close to this floating-point estimate. A simple search starting from the floor of this estimate will find the correct k in just a few iterations (usually 1 or 2). This makes the overall time complexity effectively constant.

1class Solution {2    public long minimumPerimeter(long neededApples) {3        // Start with a guess based on the approximation 4k^3 ≈ neededApples.4        // Subtract a small constant to be safe from floating point and approximation errors.5        long k = (long) Math.cbrt(neededApples / 4.0) - 2;6        if (k < 1) {7            k = 1;8        }9 10        // This loop will start very close to the answer and run only a few times.11        while (2 * k * (k + 1) * (2 * k + 1) < neededApples) {12            k++;13        }14        15        return 8 * k;16    }17}

Complexity

Time

O(1). The `cbrt` function is a constant time operation, and the subsequent loop runs for a small, constant number of iterations because the initial guess is very accurate.

Space

O(1).

Trade-offs

Pros

The fastest possible approach with O(1) time complexity.
Demonstrates a deep understanding of the problem's mathematical properties.

Cons

Relies on floating-point arithmetic (cbrt), which can have precision issues if not handled carefully.
The implementation can be slightly more complex than binary search if aiming for perfect robustness against edge cases.

Solutions

Solution

1class Solution { public long minimumPerimeter ( long neededApples ) { long x = 1 ; while ( 2 * x * ( x + 1 ) * ( 2 * x + 1 ) < neededApples ) { ++ x ; } return 8 * x ; } }

1class Solution { public: long long minimumPerimeter ( long long neededApples ) { long long x = 1 ; while ( 2 * x * ( x + 1 ) * ( 2 * x + 1 ) < neededApples ) { ++ x ; } return 8 * x ; } };

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

Video walkthrough

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1class Solution { public long minimumPerimeter ( long neededApples ) { long x = 1 ; while ( 2 * x * ( x + 1 ) * ( 2 * x + 1 ) < neededApples ) { ++ x ; } return 8 * x ; } }

1class Solution { public: long long minimumPerimeter ( long long neededApples ) { long long x = 1 ; while ( 2 * x * ( x + 1 ) * ( 2 * x + 1 ) < neededApples ) { ++ x ; } return 8 * x ; } };

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

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