How do you solve Find the Minimum Cost Array Permutation on LeetCode?

Find the Minimum Cost Array Permutation (LeetCode #3149) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(n * 2^n) space complexity. Key insight: Fixing one element helps reduce the search space significantly.

What is the brute force solution for Find the Minimum Cost Array Permutation?

The brute force approach for Find the Minimum Cost Array Permutation is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach generates all possible permutations of the array and calculates the score for each permutation. This is straightforward but inefficient, especially as the number of permutations grows factorially with the size of the array. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Find the Minimum Cost Array Permutation?

Find the Minimum Cost Array Permutation uses the following concepts: Array, Dynamic Programming, Bit Manipulation, Bitmask. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Find the Minimum Cost Array Permutation?

Always clarify the problem constraints and requirements before jumping into coding. Discuss your thought process and potential approaches with the interviewer. Practice dynamic programming problems to get comfortable with state representation.

What are common mistakes when solving Find the Minimum Cost Array Permutation?

Not considering the cyclic nature of the score calculation. Overlooking the need to track lexicographically smallest permutations.

#3149

Find the Minimum Cost Array Permutation

Hard

ArrayDynamic ProgrammingBit ManipulationBitmaskDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n * 2^n)
Space	O(n)	O(n * 2^n)

💡

Intuition

Time O(n * 2^n)Space O(n * 2^n)

The optimal solution uses dynamic programming to efficiently calculate the minimum score by leveraging the cyclic nature of the problem. By fixing one element (e.g., perm[0] = 0), we can reduce the complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Fix perm[0] = 0 to ensure the lexicographically smallest permutation.
2Step 2: Use dynamic programming with bit masking to represent subsets of the remaining elements.
3Step 3: Recursively calculate the minimum score for each subset and store results to avoid redundant calculations.

solution.py19 lines

1def min_cost_permutation(nums):
2    n = len(nums)
3    dp = [[float('inf')] * n for _ in range(1 << n)]
4    dp[1][0] = 0
5    for mask in range(1 << n):
6        for i in range(n):
7            if mask & (1 << i):
8                for j in range(n):
9                    if not (mask & (1 << j)):
10                        new_mask = mask | (1 << j)
11                        dp[new_mask][j] = min(dp[new_mask][j], dp[mask][i] + abs(j - nums[i]))
12    min_score = float('inf')
13    best_perm = []
14    for i in range(1, n):
15        score = dp[(1 << n) - 1][i] + abs(i - nums[0])
16        if score < min_score:
17            min_score = score
18            best_perm = [0] + [j for j in range(1, n) if (1 << j) & (1 << i)]
19    return best_perm

ℹ

Complexity note: The time complexity is O(n * 2^n) due to the dynamic programming approach that explores all subsets of the elements, while the space complexity is O(n * 2^n) for storing the DP table.

1Fixing one element helps reduce the search space significantly.
2Using dynamic programming allows us to avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.