Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Move from brute-force thinking to an efficient approach using math strategy.
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 < 0Example 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 <= 1015Problem summary: 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
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Binary Search
1
13
1000000000
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1954: Minimum Garden Perimeter to Collect Enough Apples
class Solution {
public long minimumPerimeter(long neededApples) {
long x = 1;
while (2 * x * (x + 1) * (2 * x + 1) < neededApples) {
++x;
}
return 8 * x;
}
}
// Accepted solution for LeetCode #1954: Minimum Garden Perimeter to Collect Enough Apples
func minimumPerimeter(neededApples int64) int64 {
var x int64 = 1
for 2*x*(x+1)*(2*x+1) < neededApples {
x++
}
return 8 * x
}
# Accepted solution for LeetCode #1954: Minimum Garden Perimeter to Collect Enough Apples
class Solution:
def minimumPerimeter(self, neededApples: int) -> int:
x = 1
while 2 * x * (x + 1) * (2 * x + 1) < neededApples:
x += 1
return x * 8
// Accepted solution for LeetCode #1954: Minimum Garden Perimeter to Collect Enough Apples
/**
* [1954] Minimum Garden Perimeter to Collect Enough Apples
*
* 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:
* <img alt="" src="https://assets.leetcode.com/uploads/2019/08/30/1527_example_1_2.png" style="width: 442px; height: 449px;" />
* 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 <= 10^15
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/minimum-garden-perimeter-to-collect-enough-apples/
// discuss: https://leetcode.com/problems/minimum-garden-perimeter-to-collect-enough-apples/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn minimum_perimeter(needed_apples: i64) -> i64 {
let mut cubrt = 3.0f64.powf((needed_apples as f64 / 4.0).log(3.0) / 3.0) as i64;
while needed_apples <= (cubrt * (cubrt + 1) * ((cubrt << 1) + 1)) << 1 {
cubrt -= 1;
}
8 * (cubrt + 1)
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_1954_example_1() {
let needed_apples = 1;
let result = 8;
assert_eq!(Solution::minimum_perimeter(needed_apples), result);
}
#[test]
fn test_1954_example_2() {
let needed_apples = 13;
let result = 16;
assert_eq!(Solution::minimum_perimeter(needed_apples), result);
}
#[test]
fn test_1954_example_3() {
let needed_apples = 1000000000;
let result = 5040;
assert_eq!(Solution::minimum_perimeter(needed_apples), result);
}
}
// Accepted solution for LeetCode #1954: Minimum Garden Perimeter to Collect Enough Apples
function minimumPerimeter(neededApples: number): number {
let x = 1;
while (2 * x * (x + 1) * (2 * x + 1) < neededApples) {
++x;
}
return 8 * x;
}
Use this to step through a reusable interview workflow for this problem.
Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.
Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.
Usually fails on: Two-element ranges never converge.
Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.