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Problem solution · Java

Find the Minimum Cost Array Permutation

Find the Minimum Cost Array Permutation: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
48 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Find the Minimum Cost Array Permutation, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 48 lines of Java from the credited upstream file 3149.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind the Minimum Cost Array Permutation · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int[] findPermutation(int[] nums) {    final int n = nums.length;    int[][] mem = new int[n][1 << n];    // bestPick[last][mask] := the best pick, where `last` is the last chosen    // number and `mask` is the bitmask of the chosen numbers    int[][] bestPick = new int[n][1 << n];     // Choose 0 as perm[0] since the score function is cyclic.    getScore(nums, /*last=*/0, /*mask=*/1, bestPick, mem);    return construct(bestPick);  }   // Returns the minimum score, where `last` is the last chosen number and  // `mask` is the bitmask of the chosen numbers.  private int getScore(int[] nums, int last, int mask, int[][] bestPick, int[][] mem) {    if (Integer.bitCount(mask) == nums.length)      return Math.abs(last - nums[0]); // |perm[n - 1] - nums[perm[0]]|    if (mem[last][mask] > 0)      return mem[last][mask];     int minScore = Integer.MAX_VALUE;    for (int i = 1; i < nums.length; ++i) {      if ((mask >> i & 1) == 1)        continue;      int nextMinScore = Math.abs(last - nums[i]) + getScore(nums, i, mask | 1 << i, bestPick, mem);      if (nextMinScore < minScore) {        minScore = nextMinScore;        bestPick[last][mask] = i;      }    }     return mem[last][mask] = minScore;  }   private int[] construct(int[][] bestPick) {    int[] ans = new int[bestPick.length];    int last = 0;    int mask = 1;    for (int i = 0; i < bestPick.length; ++i) {      ans[i] = last;      last = bestPick[last][mask];      mask |= 1 << last;    }    return ans;  }}

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