<p>You are given an array <code>nums</code> which is a <span data-keyword="permutation">permutation</span> of <code>[0, 1, 2, ..., n - 1]</code>. The <strong>score</strong> of any permutation of <code>[0, 1, 2, ..., n - 1]</code> named <code>perm</code> is defined as:</p>

<p><code>score(perm) = |perm[0] - nums[perm[1]]| + |perm[1] - nums[perm[2]]| + ... + |perm[n - 1] - nums[perm[0]]|</code></p>

<p>Return the permutation <code>perm</code> which has the <strong>minimum</strong> possible score. If <em>multiple</em> permutations exist with this score, return the one that is <span data-keyword="lexicographically-smaller-array">lexicographically smallest</span> among them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,0,2]</span></p>

<p><strong>Output:</strong> <span class="example-io">[0,1,2]</span></p>

<p><strong>Explanation:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2024/04/04/example0gif.gif" style="width: 235px; height: 235px;" /></strong></p>

<p>The lexicographically smallest permutation with minimum cost is <code>[0,1,2]</code>. The cost of this permutation is <code>|0 - 0| + |1 - 2| + |2 - 1| = 2</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [0,2,1]</span></p>

<p><strong>Output:</strong> <span class="example-io">[0,2,1]</span></p>

<p><strong>Explanation:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2024/04/04/example1gif.gif" style="width: 235px; height: 235px;" /></strong></p>

<p>The lexicographically smallest permutation with minimum cost is <code>[0,2,1]</code>. The cost of this permutation is <code>|0 - 1| + |2 - 2| + |1 - 0| = 2</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n == nums.length &lt;= 14</code></li>
	<li><code>nums</code> is a permutation of <code>[0, 1, 2, ..., n - 1]</code>.</li>
</ul>

class Solution:
    def findPermutation(self, nums: List[int]) -> List[int]:
        n = len(nums)
        dp = [[inf]*n for _ in range(1<<n)]
        jump = [[0]*n for _ in range(1<<n)]
        for m in range((1<<n)-1, 0, -1): 
            i = m.bit_count()
            for p in range(n): 
                if i == n: dp[m][p] = abs(p - nums[0])
                else: 
                    for k in range(n): 
                        if not m & 1<<k: 
                            cand = abs(p - nums[k]) + dp[m^1<<k][k]
                            if cand < dp[m][p]: 
                                dp[m][p] = cand
                                jump[m][p] = k 
        ans = [0]
        m = 1
        for _ in range(n-1): 
            k = jump[m][ans[-1]]
            ans.append(k)
            m ^= 1<<k
        return ans 
22300