Is Find the Minimum Cost Array Permutation easy or hard?

Find the Minimum Cost Array Permutation is categorized as a Hard problem. The difficulty comes from recognizing that brute force permutations are too slow and that the problem can be reframed as a subset DP using bitmasks. Understanding state compression and transition design is essential to solving it efficiently.

Find the Minimum Cost Array Permutation Python or Java solution

Python implementations usually rely on recursion with memoization or iterative DP using bitmasks. Java and C++ versions often implement the same DP idea with arrays or hash maps for memoization. The core logic remains identical: transition from (mask, last) to a new state by adding an unused element.

How to solve Find the Minimum Cost Array Permutation in O(n log n)?

An O(n log n) strategy relies on sorting the array and arranging elements so adjacent transitions stay minimal. After sorting, construct the permutation greedily by selecting elements that minimize the next transition cost relative to the previous element. This works when the cost depends on numeric distance or ordering properties.

What is the best approach for Find the Minimum Cost Array Permutation?

The most reliable approach models the problem using bitmask dynamic programming. Each state represents a subset of chosen elements and the last element used. Transitions add an unused element and update the cost based on the previous value. This reduces redundant permutation exploration and runs in O(n^2 * 2^n) time.

Is Find the Minimum Cost Array Permutation asked at Google/Amazon/Meta?

Permutation optimization and bitmask dynamic programming problems are common in interviews at companies like Google, Amazon, and Meta. Variants appear in problems similar to Traveling Salesman or optimal ordering tasks. The key skill tested is recognizing when to use state compression with bitmasks.

What data structure is used in Find the Minimum Cost Array Permutation?

Typical solutions use arrays for storing the input and a DP table indexed by bitmask states. The mask tracks which elements are already used in the permutation, while another dimension stores the last chosen index. This combination allows efficient exploration of all subsets.

What is the time complexity of Find the Minimum Cost Array Permutation?

The naive permutation approach runs in O(n! * n) time because every ordering is evaluated. A more efficient solution uses bitmask dynamic programming with O(n^2 * 2^n) time and O(n * 2^n) space. Greedy heuristics can sometimes reduce the complexity to O(n log n) when the cost structure allows ordering by value differences.

Description Hints 3 Solution Video 34:32

3149. Find the Minimum Cost Array Permutation

Hard26.1% acceptance1 company

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 <= 14
nums is a permutation of [0, 1, 2, ..., n - 1].

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

The score function is cyclic, so we can always set perm[0] = 0 for the smallest lexical order.

Hint 2

It’s similar to the Traveling Salesman Problem. Use Dynamic Programming.

Hint 3

Use a bitmask to track which elements have been assigned to perm.

34:32

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