LeetCode in Kotlin
3444. Minimum Increments for Target Multiples in an Array
Hard
You are given two arrays, nums and target.
In a single operation, you may increment any element of nums by 1.
Return the minimum number of operations required so that each element in target has at least one multiple in nums.
Example 1:
Input: nums = [1,2,3], target = [4]
Output: 1
Explanation:
The minimum number of operations required to satisfy the condition is 1.
- Increment 3 to 4 with just one operation, making 4 a multiple of itself.
Example 2:
Input: nums = [8,4], target = [10,5]
Output: 2
Explanation:
The minimum number of operations required to satisfy the condition is 2.
- Increment 8 to 10 with 2 operations, making 10 a multiple of both 5 and 10.
Example 3:
Input: nums = [7,9,10], target = [7]
Output: 0
Explanation:
Target 7 already has a multiple in nums, so no additional operations are needed.
Constraints:
1 <= nums.length <= 5 * 1041 <= target.length <= 4target.length <= nums.length1 <= nums[i], target[i] <= 104
Solution
import kotlin.math.min
class Solution {
fun minimumIncrements(nums: IntArray, target: IntArray): Int {
val m = target.size
val fullMask = (1 shl m) - 1
val lcmArr = LongArray(1 shl m)
for (mask in 1..<(1 shl m)) {
var l: Long = 1
for (j in 0..<m) {
if ((mask and (1 shl j)) != 0) {
l = lcm(l, target[j].toLong())
}
}
lcmArr[mask] = l
}
val inf = Long.Companion.MAX_VALUE / 2
var dp = LongArray(1 shl m)
for (i in 0..<(1 shl m)) {
dp[i] = inf
}
dp[0] = 0
for (x in nums) {
val newdp = dp.clone()
for (mask in 1..<(1 shl m)) {
val l = lcmArr[mask]
val r = x % l
val cost = (if (r == 0L) 0 else l - r)
for (old in 0..<(1 shl m)) {
val newMask = old or mask
newdp[newMask] = min(newdp[newMask], (dp[old] + cost))
}
}
dp = newdp
}
return dp[fullMask].toInt()
}
private fun gcd(a: Long, b: Long): Long {
return if (b == 0L) a else gcd(b, a % b)
}
private fun lcm(a: Long, b: Long): Long {
return a / gcd(a, b) * b
}
}