Introduction

In this blog post, we will explore how to solve the LeetCode problem "2717" 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 a 0-indexed permutation of n integers nums. A permutation is called semi-ordered if the first number equals 1 and the last number equals n. You can perform the below operation as many times as you want until you make nums a semi-ordered permutation: Pick two adjacent elements in nums, then swap them. Return the minimum number of operations to make nums a semi-ordered permutation. A permutation is a sequence of integers from 1 to n of length n containing each number exactly once. Example 1: Input: nums = [2,1,4,3] Output: 2 Explanation: We can make the permutation semi-ordered using these sequence of operations: 1 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3]. 2 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4]. It can be proved that there is no sequence of less than two operations that make nums a semi-ordered permutation. Example 2: Input: nums = [2,4,1,3] Output: 3 Explanation: We can make the permutation semi-ordered using these sequence of operations: 1 - swap i = 1 and j = 2. The permutation becomes [2,1,4,3]. 2 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3]. 3 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4]. It can be proved that there is no sequence of less than three operations that make nums a semi-ordered permutation. Example 3: Input: nums = [1,3,4,2,5] Output: 0 Explanation: The permutation is already a semi-ordered permutation. Constraints: 2 <= nums.length == n <= 50 1 <= nums[i] <= 50 nums is a permutation.

Explanation

• Find the index of 1 and the index of n in the input array nums.
  • Calculate the number of swaps needed to move 1 to the beginning and n to the end. Be careful to avoid overcounting if the index of n is less than the index of 1.
  • Return the total number of swaps.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def semiOrderedPermutation(nums):
    """
    Calculates the minimum number of swaps to make a permutation semi-ordered.

    Args:
        nums: A list of integers representing the permutation.

    Returns:
        The minimum number of swaps required.
    """
    n = len(nums)
    index_1 = nums.index(1)
    index_n = nums.index(n)

    swaps = index_1 + (n - 1 - index_n)

    if index_1 > index_n:
        swaps -= 1

    return swaps