Introduction

In this blog post, we will explore how to solve the LeetCode problem "3149" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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].

Explanation

Here's the breakdown of the approach, complexity, and the Python code solution:

• High-Level Approach:

  • The problem requires finding the permutation with the minimum score, prioritizing the lexicographically smallest one in case of ties. Since n is small (<=14), we can exhaustively search all possible permutations.
  • Iterate through all permutations of range(n) using itertools.permutations.
  • For each permutation, calculate the score and compare it with the current minimum score. Update the best permutation if a lower score is found, or if the score is equal but the current permutation is lexicographically smaller.

• Complexity:

  • Runtime Complexity: O(n! * n), where n! is for generating all permutations and n is for calculating the score of each permutation.
  • Storage Complexity: O(n), primarily for storing the best permutation found so far.

• Python Code:

Code

    import itertools
def min_score_permutation(nums):
    """
    Finds the permutation of [0, 1, 2, ..., n-1] with the minimum possible score.

    Args:
        nums: An array which is a permutation of [0, 1, 2, ..., n-1].

    Returns:
        The permutation with the minimum score and lexicographically smallest if multiple exist.
    """

    n = len(nums)
    best_perm = None
    min_score = float('inf')

    for perm in itertools.permutations(range(n)):
        score = 0
        for i in range(n):
            score += abs(perm[i] - nums[perm[(i + 1) % n]])

        if score < min_score:
            min_score = score
            best_perm = list(perm)
        elif score == min_score:
            if list(perm) < best_perm:
                best_perm = list(perm)

    return best_perm