Solution Approach

The solution uses binary search to find the minimum value of x (half-width of the square) that satisfies our apple requirement. Since the total apples in a square grows monotonically with x, we can use binary search to efficiently find the first x where the condition 2 * x * (x + 1) * (2 * x + 1) >= neededApples becomes true.

The algorithm follows the standard binary search template for finding the first true index:

1. Initialize search bounds:

   • left = 1 (minimum possible half-width)
   • right = 100000 (upper bound large enough for any input)
   • first_true_index = -1 (tracks the answer)

2. Binary search loop (while left <= right):

   • Calculate mid = (left + right) // 2
   • Check if the square with half-width mid has enough apples: 2 * mid * (mid + 1) * (2 * mid + 1) >= neededApples
   • If true (feasible): save first_true_index = mid and search for a smaller valid value with right = mid - 1
   • If false (not enough apples): need a larger square, so left = mid + 1

3. Return the perimeter: 8 * first_true_index

The perimeter calculation is straightforward: since the square extends x units in all four directions from the origin, each side has length 2 * x, giving us a total perimeter of 4 * (2 * x) = 8 * x.

This approach is more efficient than linear search, taking O(log n) iterations instead of O(n^(1/3)) where n is neededApples.

Example Walkthrough

Let's walk through finding the minimum perimeter for neededApples = 50 using binary search.

Setup:

• left = 1, right = 100000, first_true_index = -1
• We need to find the first x where 2 * x * (x + 1) * (2 * x + 1) >= 50

Iteration 1:

• mid = (1 + 100000) // 2 = 50000
• Check: 2 * 50000 * 50001 * 100001 is vastly greater than 50 ✓
• Since feasible: first_true_index = 50000, right = 50000 - 1 = 49999

Iterations 2-16: Binary search continues narrowing down...

Near the end:

• left = 1, right = 2, first_true_index = 2
• mid = (1 + 2) // 2 = 1
• Check: 2 * 1 * 2 * 3 = 12 < 50 ❌
• Not feasible: left = 1 + 1 = 2

Final iteration:

• left = 2, right = 2
• mid = 2
• Check: 2 * 2 * 3 * 5 = 60 >= 50 ✓
• Feasible: first_true_index = 2, right = 1
• Now left > right, loop ends

Result: first_true_index = 2, perimeter = 8 * 2 = 16

Let's verify the apple count for x = 2:

• Square extends from (-2, -2) to (2, 2)
• Formula: 2 * 2 * 3 * 5 = 60 apples
• This is ≥ 50, so it's valid
• For x = 1: 2 * 1 * 2 * 3 = 12 apples < 50, not enough

So x = 2 is indeed the minimum, giving us perimeter = 16.

Time and Space Complexity

Time Complexity: O(log n) where n is the upper bound of the search range (100000 in our implementation, or more precisely O(log(neededApples^(1/3)))).

The algorithm uses binary search to find the minimum x where the condition is satisfied. Since the search range is from 1 to 100000, the binary search takes at most log₂(100000) ≈ 17 iterations. Each iteration performs a constant-time comparison using the formula 2 * x * (x + 1) * (2 * x + 1).

Space Complexity: O(1)

The algorithm only uses a constant amount of extra space for the variables left, right, mid, and first_true_index. No additional data structures are used that scale with the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using Wrong Binary Search Template Variant

A common mistake is using the while left < right with right = mid variant instead of the standard template. This can lead to incorrect results or infinite loops.

Incorrect Implementation:

while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid  # Wrong pattern
    else:
        left = mid + 1
return left

Correct Implementation (using template):

left, right = 1, 100000
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index * 8

2. Integer Overflow in Formula Calculation

The formula 2 * x * (x + 1) * (2 * x + 1) involves multiplying large numbers together. For large values of x, intermediate calculations might exceed integer limits in some programming languages.

Solution:

• In Python, this isn't an issue due to automatic handling of large integers
• In Java/C++, use long/long long data types
• Consider the order of multiplication to minimize intermediate overflow

3. Misunderstanding the Apple Count Formula

Developers might incorrectly derive or implement the formula for counting apples. Common mistakes include:

• Forgetting to account for all four quadrants
• Double-counting the axes or origin
• Using an incorrect summation formula

Solution: Verify the formula by manually calculating small cases:

• For x=1: The square contains points from (-1,-1) to (1,1), total apples = 12
• For x=2: The square contains points from (-2,-2) to (2,2), total apples = 60
• Confirm these match 2 * x * (x + 1) * (2 * x + 1)

4. Off-by-One Errors in Perimeter Calculation

Some might mistakenly calculate the perimeter as 4 * x (thinking of radius) or 4 * (2x + 1) (thinking of discrete points).

Solution: Remember that:

• The square extends x units in each direction from origin
• Each side has length 2x (from -x to x)
• Total perimeter = 4 sides × 2x = 8x