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.
Build confidence with an intuition-first walkthrough focused on math fundamentals.
You are given integers n, m, and k.
There are two logs of lengths n and m units, which need to be transported in three trucks where each truck can carry one log with length at most k units.
You may cut the logs into smaller pieces, where the cost of cutting a log of length x into logs of length len1 and len2 is cost = len1 * len2 such that len1 + len2 = x.
Return the minimum total cost to distribute the logs onto the trucks. If the logs don't need to be cut, the total cost is 0.
Example 1:
Input: n = 6, m = 5, k = 5
Output: 5
Explanation:
Cut the log with length 6 into logs with length 1 and 5, at a cost equal to 1 * 5 == 5. Now the three logs of length 1, 5, and 5 can fit in one truck each.
Example 2:
Input: n = 4, m = 4, k = 6
Output: 0
Explanation:
The two logs can fit in the trucks already, hence we don't need to cut the logs.
Constraints:
2 <= k <= 1051 <= n, m <= 2 * kProblem summary: You are given integers n, m, and k. There are two logs of lengths n and m units, which need to be transported in three trucks where each truck can carry one log with length at most k units. You may cut the logs into smaller pieces, where the cost of cutting a log of length x into logs of length len1 and len2 is cost = len1 * len2 such that len1 + len2 = x. Return the minimum total cost to distribute the logs onto the trucks. If the logs don't need to be cut, the total cost is 0.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math
6 5 5
4 4 6
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3560: Find Minimum Log Transportation Cost
class Solution {
public long minCuttingCost(int n, int m, int k) {
int x = Math.max(n, m);
return x <= k ? 0 : 1L * k * (x - k);
}
}
// Accepted solution for LeetCode #3560: Find Minimum Log Transportation Cost
func minCuttingCost(n int, m int, k int) int64 {
x := max(n, m)
if x <= k {
return 0
}
return int64(k * (x - k))
}
# Accepted solution for LeetCode #3560: Find Minimum Log Transportation Cost
class Solution:
def minCuttingCost(self, n: int, m: int, k: int) -> int:
x = max(n, m)
return 0 if x <= k else k * (x - k)
// Accepted solution for LeetCode #3560: Find Minimum Log Transportation Cost
fn min_cutting_cost(n: i32, m: i32, k: i32) -> i64 {
fn f(n: i64, k: i64) -> i64 {
if n <= k {
0
} else {
let limit = n / 2;
let mut ret = i64::MAX;
for i in 1..=limit {
if i <= k && (n - i) <= k {
ret = std::cmp::min(ret, i * (n - i));
}
}
ret
}
}
let n = n as i64;
let m = m as i64;
let k = k as i64;
f(n, k) + f(m, k)
}
fn main() {
let ret = min_cutting_cost(6, 5, 5);
println!("ret={ret}");
}
#[test]
fn test() {
assert_eq!(min_cutting_cost(6, 5, 5), 5);
assert_eq!(min_cutting_cost(4, 4, 6), 0);
}
// Accepted solution for LeetCode #3560: Find Minimum Log Transportation Cost
function minCuttingCost(n: number, m: number, k: number): number {
const x = Math.max(n, m);
return x <= k ? 0 : k * (x - k);
}
Use this to step through a reusable interview workflow for this problem.
Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.
Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.
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.