LeetCode in Kotlin
3149. Find the Minimum Cost Array Permutation
Hard
You are given an array nums which is a permutation of [0, 1, 2, ..., n - 1]. The score of any permutation of [0, 1, 2, ..., n - 1] named perm is defined as:
score(perm) = |perm[0] - nums[perm[1]]| + |perm[1] - nums[perm[2]]| + ... + |perm[n - 1] - nums[perm[0]]|
Return the permutation perm which has the minimum possible score. If multiple permutations exist with this score, return the one that is lexicographically smallest among them.
Example 1:
Input: nums = [1,0,2]
Output: [0,1,2]
Explanation:
The lexicographically smallest permutation with minimum cost is [0,1,2]. The cost of this permutation is |0 - 0| + |1 - 2| + |2 - 1| = 2.
Example 2:
Input: nums = [0,2,1]
Output: [0,2,1]
Explanation:
The lexicographically smallest permutation with minimum cost is [0,2,1]. The cost of this permutation is |0 - 1| + |2 - 2| + |1 - 0| = 2.
Constraints:
2 <= n == nums.length <= 14numsis a permutation of[0, 1, 2, ..., n - 1].
Solution
import kotlin.math.abs
import kotlin.math.min
class Solution {
private fun findMinScore(mask: Int, prevNum: Int, nums: IntArray, dp: Array<IntArray>): Int {
val n = nums.size
if (Integer.bitCount(mask) == n) {
dp[mask][prevNum] = abs((prevNum - nums[0]))
return dp[mask][prevNum]
}
if (dp[mask][prevNum] != -1) {
return dp[mask][prevNum]
}
var minScore = Int.MAX_VALUE
for (currNum in 0 until n) {
if ((mask shr currNum and 1 xor 1) == 1) {
val currScore = (
abs((prevNum - nums[currNum])) + findMinScore(
mask or (1 shl currNum),
currNum,
nums,
dp,
)
)
minScore = min(minScore, currScore)
}
}
return minScore.also { dp[mask][prevNum] = it }
}
private fun constructMinScorePermutation(n: Int, nums: IntArray, dp: Array<IntArray>): IntArray {
val permutation = IntArray(n)
var i = 0
permutation[i++] = 0
var prevNum = 0
var mask = 1
while (i < n) {
for (currNum in 0 until n) {
if ((mask shr currNum and 1 xor 1) == 1) {
val currScore =
(abs((prevNum - nums[currNum])) + dp[mask or (1 shl currNum)][currNum])
val minScore = dp[mask][prevNum]
if (currScore == minScore) {
permutation[i++] = currNum
prevNum = currNum
break
}
}
}
mask = mask or (1 shl prevNum)
}
return permutation
}
fun findPermutation(nums: IntArray): IntArray {
val n = nums.size
val dp = Array(1 shl n) { IntArray(n) }
for (row in dp) {
row.fill(-1)
}
findMinScore(1, 0, nums, dp)
return constructMinScorePermutation(n, nums, dp)
}
}