LeetCode-in-Java
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 java.util.Arrays;
public class Solution {
private int findMinScore(int mask, int prevNum, int[] nums, int[][] dp) {
int n = nums.length;
if (Integer.bitCount(mask) == n) {
dp[mask][prevNum] = Math.abs(prevNum - nums[0]);
return dp[mask][prevNum];
}
if (dp[mask][prevNum] != -1) {
return dp[mask][prevNum];
}
int minScore = Integer.MAX_VALUE;
for (int currNum = 0; currNum < n; currNum++) {
if ((mask >> currNum & 1 ^ 1) == 1) {
int currScore =
Math.abs(prevNum - nums[currNum])
+ findMinScore(mask | 1 << currNum, currNum, nums, dp);
minScore = Math.min(minScore, currScore);
}
}
dp[mask][prevNum] = minScore;
return dp[mask][prevNum];
}
private int[] constructMinScorePermutation(int n, int[] nums, int[][] dp) {
int[] permutation = new int[n];
int i = 0;
permutation[i++] = 0;
int prevNum = 0;
for (int mask = 1; i < n; mask |= 1 << prevNum) {
for (int currNum = 0; currNum < n; currNum++) {
if ((mask >> currNum & 1 ^ 1) == 1) {
int currScore =
Math.abs(prevNum - nums[currNum]) + dp[mask | 1 << currNum][currNum];
int minScore = dp[mask][prevNum];
if (currScore == minScore) {
permutation[i++] = currNum;
prevNum = currNum;
break;
}
}
}
}
return permutation;
}
public int[] findPermutation(int[] nums) {
int n = nums.length;
int[][] dp = new int[1 << n][n];
for (int[] row : dp) {
Arrays.fill(row, -1);
}
findMinScore(1, 0, nums, dp);
return constructMinScorePermutation(n, nums, dp);
}
}