Introduction

In this blog post, we will explore how to solve the LeetCode problem "3470" 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

Given two integers, n and k, an alternating permutation is a permutation of the first n positive integers such that no two adjacent elements are both odd or both even. Return the k-th alternating permutation sorted in lexicographical order. If there are fewer than k valid alternating permutations, return an empty list. Example 1: Input: n = 4, k = 6 Output: [3,4,1,2] Explanation: The lexicographically-sorted alternating permutations of [1, 2, 3, 4] are: [1, 2, 3, 4] [1, 4, 3, 2] [2, 1, 4, 3] [2, 3, 4, 1] [3, 2, 1, 4] [3, 4, 1, 2] ← 6th permutation [4, 1, 2, 3] [4, 3, 2, 1] Since k = 6, we return [3, 4, 1, 2]. Example 2: Input: n = 3, k = 2 Output: [3,2,1] Explanation: The lexicographically-sorted alternating permutations of [1, 2, 3] are: [1, 2, 3] [3, 2, 1] ← 2nd permutation Since k = 2, we return [3, 2, 1]. Example 3: Input: n = 2, k = 3 Output: [] Explanation: The lexicographically-sorted alternating permutations of [1, 2] are: [1, 2] [2, 1] There are only 2 alternating permutations, but k = 3, which is out of range. Thus, we return an empty list []. Constraints: 1 <= n <= 100 1 <= k <= 1015

Explanation

Here's a breakdown of the solution:

• Understanding Alternating Permutations: The core idea is to generate alternating permutations efficiently. We can categorize numbers into odd and even sets. A valid alternating permutation must have elements from these sets alternatingly.

• Counting Valid Permutations: We need to calculate the total number of possible alternating permutations for a given n. If the k-th permutation exceeds this count, we return an empty list.

• Generating the k-th Permutation: The algorithm constructs the permutation digit by digit. At each step, it calculates how many permutations start with a specific number (either odd or even, based on the desired alternating pattern). If k falls within that count, the number is added to the result; otherwise, k is adjusted, and the process continues with the next candidate.

• Runtime Complexity: O(n^2), Storage Complexity: O(n)

Code

    def alternating_permutations(n, k):
    """
    Calculates the k-th alternating permutation of the first n positive integers.

    Args:
        n: The number of integers to permute.
        k: The index of the desired permutation (1-based).

    Returns:
        A list representing the k-th alternating permutation, or an empty list if
        there are fewer than k valid alternating permutations.
    """

    if n == 0:
        return []

    # Calculate the number of alternating permutations.
    if n <= 2:
        num_alternating = factorial(n)  # Simple case for n = 1 or 2
    else:
        num_alternating = count_alternating_permutations(n)


    if k > num_alternating:
        return []

    k -= 1  # Convert to 0-based index for calculations.
    nums = list(range(1, n + 1))
    result = []

    # Determine if the permutation should start with an odd or even number.
    start_with_odd = (n + 1) // 2 >= (n // 2)

    for i in range(n):
        # Determine whether we're looking for odd or even for this index
        if i % 2 == 0:
            eligible_nums = [num for num in nums if (num % 2 != 0) == start_with_odd]
        else:
            eligible_nums = [num for num in nums if (num % 2 != 0) != start_with_odd]

        num_eligible = len(eligible_nums)

        for j in range(num_eligible):
            num = eligible_nums[j]

            # Calculate how many permutations start with num
            remaining_nums = nums[:]
            remaining_nums.remove(num)
            num_permutations_starting_with_num = count_alternating_permutations_starting(n - i, remaining_nums, num, start_with_odd if i%2 == 0 else not start_with_odd)

            if k < num_permutations_starting_with_num:
                result.append(num)
                nums.remove(num)
                break
            else:
                k -= num_permutations_starting_with_num
    return result

def count_alternating_permutations(n):
    """
    Counts the number of alternating permutations for a given n.
    """
    dp = {}

    def solve(index, last_parity, remaining):
        if index == n:
            return 1

        key = (index, last_parity, tuple(sorted(remaining)))
        if key in dp:
            return dp[key]

        count = 0
        for num in remaining:
            parity = num % 2
            if index == 0 or parity != last_parity:
                new_remaining = remaining[:]
                new_remaining.remove(num)
                count += solve(index + 1, parity, new_remaining)

        dp[key] = count
        return count

    all_nums = list(range(1, n + 1))
    return solve(0, -1, all_nums) # -1 to indicate no previous parity at start.

def count_alternating_permutations_starting(remaining_length, remaining_nums, first_num, start_with_odd):
  """
  Calculates the number of alternating permutations of the remaining numbers,
  given that the permutation must start with `first_num` and alternate.
  """
  if remaining_length <= 1:
    return 1

  dp = {}

  def solve(index, last_parity, remaining):
    if index == remaining_length:
      return 1

    key = (index, last_parity, tuple(sorted(remaining)))
    if key in dp:
      return dp[key]

    count = 0
    for num in remaining:
      parity = num % 2
      if parity != last_parity:
        new_remaining = remaining[:]
        new_remaining.remove(num)
        count += solve(index + 1, parity, new_remaining)

    dp[key] = count
    return count

  initial_parity = first_num % 2
  return solve(1, initial_parity, remaining_nums)


def factorial(n):
    if n == 0:
        return 1
    else:
        return n * factorial(n-1)