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
This problem asks for the minimum perimeter of a square plot centered at (0,0) containing at least neededApples
. The number of apples in a square plot with side length k
is given by the formula: 2 * k * (k + 1) * (2 * k + 1)
. This formula is derived by summing the apples in each concentric square layer of the plot.
We can solve this problem using two main approaches:
Approach 1: Iterative Solution
This approach directly utilizes the formula for the number of apples. It iterates through possible side lengths (x
), calculating the number of apples for each until the required number of apples is met or exceeded.
Code (Python):
class Solution:
def minimumPerimeter(self, neededApples: int) -> int:
x = 1
while 2 * x * (x + 1) * (2 * x + 1) < neededApples:
x += 1
return x * 8
Explanation:
x = 1
(side length).while
condition checks if the number of apples for the current x
is less than neededApples
.x
is incremented.x
represents the minimum side length.x * 8
(4 sides * side length * 2) and returned.Time Complexity: O(√N), where N is neededApples
. The loop iterates approximately up to the square root of neededApples
.
Space Complexity: O(1) - Constant extra space is used.
Approach 2: Binary Search
This approach employs binary search to efficiently find the minimum side length. Binary search is suitable because the number of apples is a monotonically increasing function of the side length.
Code (Python):
class Solution:
def minimumPerimeter(self, neededApples: int) -> int:
l, r = 1, 100000 # reasonable upper bound
while l < r:
mid = (l + r) // 2
if 2 * mid * (mid + 1) * (2 * mid + 1) >= neededApples:
r = mid
else:
l = mid + 1
return l * 8
Explanation:
l
and r
define the search space (initially 1 to a large upper bound).while
loop performs binary search.mid
calculates the middle point.if
condition checks if the number of apples for mid
is greater than or equal to neededApples
.mid
, so r
is updated to mid
.mid
, so l
is updated to mid + 1
.l
will hold the minimum side length, and the perimeter is calculated.Time Complexity: O(log N), where N is neededApples
. Binary search has logarithmic time complexity.
Space Complexity: O(1) - Constant extra space is used.
Conclusion:
Both approaches correctly solve the problem. The binary search approach is more efficient for large values of neededApples
due to its logarithmic time complexity. The iterative solution is simpler to understand and implement but less efficient for extremely large inputs. The choice of approach depends on the trade-off between code simplicity and performance requirements. The provided code snippets in other languages (Java, C++, Go, TypeScript) follow the same logic as the Python examples.