Problem Description

Given an integer n, 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 all such alternating permutations sorted in lexicographical order.

Intuition

The problem revolves around generating all permutations of the first n positive integers, with the added constraint that adjacent numbers must alternate in parity (odd/even). An effective approach to tackle this is through backtracking, which allows us to explore all potential permutations while adhering to the rules.

Solution Approach

1. Backtracking Design: We define a function dfs(i) to fill the i-th position in a permutation. The index i starts from 0 up to n-1.

2. Base Condition: If i equals n, it signifies that all positions in the permutation have been correctly filled. At this point, the current permutation is added to the list of valid permutations (ans).

3. Recursive Exploration: We then consider each integer j from 1 to n. A number j can be placed in the current position if:

   • j has not been used yet (vis[j] is False).
   • j is of different parity compared to the last inserted number in the permutation.

4. Backtracking Step: If the conditions are met, we include j and proceed to fill the next position by making a recursive call to dfs(i + 1). After exploring that path, we backtrack by marking j as not used and removing it from the current permutation (t).

Overall, this approach efficiently generates all possible alternating permutations using depth-first search, ensuring the permutations are produced in lexicographical order. This is facilitated by iterating sequentially from 1 to n during exploration.

Learn more about Backtracking patterns.

Solution Approach

The solution utilizes a backtracking approach to generate all valid alternating permutations. Here's how the implementation works:

1. Data Structures:

   • t: A list that temporarily stores the current partial permutation being constructed.
   • vis: A list of boolean values where vis[j] indicates whether the number j has been used in the current permutation.

2. Recursive Function dfs(i):

   • Base Case: If i is greater than or equal to n, the permutation in t is complete, and it is added to the results ans.
   • Recursive Case: Iterate over each number j from 1 to n.
     • Constraints:
       • Check if j is available (i.e., it hasn't been used, so vis[j] is False).
       • Ensure j can be placed next to the last number in t by checking their parity: t[-1] % 2 != j % 2.
     • Action:
       • Append j to t, marking j as used (vis[j] = True).
       • Call dfs(i + 1) to fill the next position.
       • After returning from recursion, backtrack by removing j from t and marking it as unused (vis[j] = False).

3. Initialization and Invocation:

   • Start with an empty permutation (t = []) and all numbers available (vis = [False] * (n + 1)).
   • Call dfs(0) to begin filling permutations from the first position.

4. Return: The list ans contains all lexicographically sorted alternating permutations.

This step-by-step permutation building provides a robust method to explore all valid combinations while adhering to the alternating parity rule, efficiently traversing the decision tree using backtracking.

Example Walkthrough

Let's illustrate the solution approach using a small example where n = 3. We want to find all alternating permutations of the integers 1, 2, 3.

Initialization:

• Start with an empty permutation list t = [].
• Initialize a vis list to keep track of used numbers, vis = [False, False, False, False] (since n = 3, the length is n + 1).

Process using Backtracking:

1. First Level (i = 0):

   • We begin at the 0th position of the permutation.
   • Consider placing each integer j (1 to 3):
     • For j = 1:
       • Not used (vis[1] = False).
       • Place 1 into t, so t = [1], mark 1 as used, vis = [False, True, False, False].
       • Recursively call dfs(1) to fill the next position.

2. Second Level (i = 1):

   • We are at the 1st position, and t = [1].
   • Consider each j:
     • For j = 1: Already used, skip.
     • For j = 2:
       • Not used (vis[2] = False).
       • Check parity: Previous number is 1 (odd), 2 is even, satisfies condition.
       • Place 2 into t, t = [1, 2], mark 2 as used, vis = [False, True, True, False].
       • Recursively call dfs(2) to fill the final position.

3. Third Level (i = 2):

   • We are at the 2nd position, and t = [1, 2].
   • Consider each j:
     • For j = 1: Already used, skip.
     • For j = 2: Already used, skip.
     • For j = 3:
       • Not used (vis[3] = False).
       • Check parity: Previous number is 2 (even), 3 is odd, satisfies condition.
       • Place 3 into t, t = [1, 2, 3], mark 3 as used, vis = [False, True, True, True].
       • Reached base condition i = 3 (all positions filled), add [1, 2, 3] to ans.
       • Backtrack: Remove 3, reset vis[3] = False, return to previous level i = 2.

4. Backtracking:

   • At i = 2 with t = [1, 2], try next potential values (j) once 3 is backtracked.
   • Continue exploring by changing the first choice (t = [1]) at i = 1, and eventually restarting from different j values at i = 0.

Repeat these steps to explore all valid alternating permutations ans = [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]].

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n * n!) because the algorithm generates all permutations of numbers from 1 to n while considering additional constraints. During the recursive depth-first search, there are n! permutations, and for each permutation, operations proportional to n are carried out.

The space complexity of the code is O(n). This is primarily due to the space used by the recursion stack, which can go as deep as n, and additional space to store the current permutation t, which can also have at most n elements.

Learn more about how to find time and space complexity quickly.