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Permutations IV

Difficulty: Hard

Problem Description

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.

Key Insights

  • An alternating permutation must have odd and even numbers placed alternately.
  • The total number of odd and even integers in the set affects the possible configurations.
  • Lexicographical order requires careful placement of elements based on their values.
  • The maximum number of alternating permutations can be computed using combinatorial mathematics.
  • We need efficient ways to generate permutations without generating all of them, especially given the constraints on k.

Space and Time Complexity

Time Complexity: O(n) for generating the k-th permutation, with precomputation potentially taking O(n^2) for factorial values. Space Complexity: O(n) for storing the result and auxiliary data structures.

Solution

To solve the problem, we will use a combinatorial approach to count valid permutations without generating all permutations. We can maintain a balance between odd and even integers while constructing the permutation. A recursive or iterative approach can be employed to select the next integer based on the current permutation state and the value of k.

Code Solutions

def get_kth_permutation(n, k):
    if n < 1 or k < 1:
        return []
    
    odds = list(range(1, n + 1, 2))
    evens = list(range(2, n + 1, 2))
    total_odds = len(odds)
    total_evens = len(evens)

    # Function to compute factorial
    def factorial(num):
        if num == 0 or num == 1:
            return 1
        result = 1
        for i in range(2, num + 1):
            result *= i
        return result

    # Calculate total permutations
    total_permutations = factorial(total_odds + total_evens)
    if total_permutations < k:
        return []

    permutation = []
    is_odd_turn = True

    while total_odds > 0 or total_evens > 0:
        if is_odd_turn:
            # Count permutations starting with each odd number
            for i in range(total_odds):
                count = factorial(total_odds - 1) * factorial(total_evens)
                if k <= count:
                    permutation.append(odds[i])
                    odds.pop(i)
                    total_odds -= 1
                    break
                k -= count
        else:
            # Count permutations starting with each even number
            for i in range(total_evens):
                count = factorial(total_odds) * factorial(total_evens - 1)
                if k <= count:
                    permutation.append(evens[i])
                    evens.pop(i)
                    total_evens -= 1
                    break
                k -= count

        is_odd_turn = not is_odd_turn

    return permutation
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