LeetCode in Kotlin
3560. Find Minimum Log Transportation Cost
Easy
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 * k- The input is generated such that it is always possible to transport the logs.
Solution
class Solution {
fun minCuttingCost(n: Int, m: Int, k: Int): Long {
if (n == 0 || m == 0 || k == 0) {
return 0
}
var ans: Long = 0
if (m <= k && n <= k) {
return 0
}
if (m > k && n <= k) {
ans += (m - k).toLong() * k
}
if (n > k && m <= k) {
ans += (n - k).toLong() * k
}
return ans
}
}