How do you solve Permutations IV on LeetCode?

Permutations IV (LeetCode #3470) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding permutations and their properties is crucial.

What is the brute force solution for Permutations IV?

The brute force approach for Permutations IV is Brute Force, with O(n²) time complexity and O(n!) space complexity. Generate all permutations and filter for alternating ones. This method is straightforward but inefficient for larger n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Permutations IV?

Permutations IV uses the following concepts: Array, Math, Combinatorics, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Permutations IV?

Explain your thought process clearly. Consider edge cases and constraints. Practice combinatorial problems.

What are common mistakes when solving Permutations IV?

Not considering the base case when k exceeds the number of valid permutations. Miscalculating combinations.

#3470

Permutations IV

Hard

ArrayMathCombinatoricsEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n!)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use combinatorial properties to generate the k-th permutation directly without generating all permutations.

⚙️

Algorithm

3 steps

1Step 1: Determine the number of odd and even integers based on n.
2Step 2: Use combinatorial counting to find the k-th valid permutation directly.
3Step 3: Construct the permutation by selecting elements based on the calculated positions.

solution.py25 lines

1from math import comb
2
3def kth_alternating_permutation(n, k):
4    odds = (n + 1) // 2
5    evens = n // 2
6    result = []
7    if k > comb(odds + evens, odds): return []
8    for i in range(n):
9        if i % 2 == 0:
10            for j in range(1, odds + 1):
11                count = comb(odds - j + evens, odds - j)
12                if k <= count:
13                    result.append(2 * j - 1)
14                    odds -= 1
15                    break
16                k -= count
17        else:
18            for j in range(1, evens + 1):
19                count = comb(odds + evens - j, odds)
20                if k <= count:
21                    result.append(2 * j)
22                    evens -= 1
23                    break
24                k -= count
25    return result

ℹ

Complexity note: The optimal solution uses combinatorial counting, leading to linear time complexity for generating the k-th permutation.

1Understanding permutations and their properties is crucial.
2Combinatorial counting can significantly reduce complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.