LeetCode #3470 — HARD

Permutations IV

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

4
6

Example 2

3
2

Example 3

2
3

Core Insight

What unlocks the optimal approach

If <code>n</code> is odd, the first number must be odd.
If <code>n</code> is even, the first number can be either odd or even.
From smallest to largest, place each number and subtract the number of permutations from <code>k</code>.
The number of permutations can be calculated using factorials.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3470: Permutations IV
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3470: Permutations IV
// package main
// 
// import "slices"
// 
// // https://space.bilibili.com/206214
// // 预处理交替排列的方案数
// var f = []int{1}
// 
// func init() {
// 	for i := 1; f[len(f)-1] < 1e15; i++ {
// 		f = append(f, f[len(f)-1]*i)
// 		f = append(f, f[len(f)-1]*i)
// 	}
// }
// 
// func permute(n int, K int64) []int {
// 	// k 改成从 0 开始，方便计算
// 	k := int(K - 1)
// 	if n < len(f) && k >= f[n]*(2-n%2) { // n 是偶数的时候，方案数乘以 2
// 		return nil
// 	}
// 
// 	// cand 表示剩余未填入 ans 的数字
// 	// cand[0] 保存偶数，cand[1] 保存奇数
// 	cand := [2][]int{}
// 	for i := 2; i <= n; i += 2 {
// 		cand[0] = append(cand[0], i)
// 	}
// 	for i := 1; i <= n; i += 2 {
// 		cand[1] = append(cand[1], i)
// 	}
// 
// 	ans := make([]int, n)
// 	parity := 1 // 当前要填入 ans[i] 的数的奇偶性
// 	for i := range n {
// 		j := 0
// 		if n-1-i < len(f) {
// 			// 比如示例 1，按照第一个数分组，每一组的大小都是 size=2
// 			// 知道 k 和 size 就知道我们要去哪一组
// 			size := f[n-1-i]
// 			j = k / size // 去第 j 组
// 			k %= size
// 			// n 是偶数的情况，第一个数既可以填奇数又可以填偶数，要特殊处理
// 			if n%2 == 0 && i == 0 {
// 				parity = 1 - j%2
// 				j /= 2
// 			}
// 		} // else n 很大的情况下，只能按照 1,2,3,... 的顺序填
// 		ans[i] = cand[parity][j]
// 		cand[parity] = slices.Delete(cand[parity], j, j+1)
// 		parity ^= 1 // 下一个数的奇偶性
// 	}
// 	return ans
// }

// Accepted solution for LeetCode #3470: Permutations IV
package main

import "slices"

// https://space.bilibili.com/206214
// 预处理交替排列的方案数
var f = []int{1}

func init() {
	for i := 1; f[len(f)-1] < 1e15; i++ {
		f = append(f, f[len(f)-1]*i)
		f = append(f, f[len(f)-1]*i)
	}
}

func permute(n int, K int64) []int {
	// k 改成从 0 开始，方便计算
	k := int(K - 1)
	if n < len(f) && k >= f[n]*(2-n%2) { // n 是偶数的时候，方案数乘以 2
		return nil
	}

	// cand 表示剩余未填入 ans 的数字
	// cand[0] 保存偶数，cand[1] 保存奇数
	cand := [2][]int{}
	for i := 2; i <= n; i += 2 {
		cand[0] = append(cand[0], i)
	}
	for i := 1; i <= n; i += 2 {
		cand[1] = append(cand[1], i)
	}

	ans := make([]int, n)
	parity := 1 // 当前要填入 ans[i] 的数的奇偶性
	for i := range n {
		j := 0
		if n-1-i < len(f) {
			// 比如示例 1，按照第一个数分组，每一组的大小都是 size=2
			// 知道 k 和 size 就知道我们要去哪一组
			size := f[n-1-i]
			j = k / size // 去第 j 组
			k %= size
			// n 是偶数的情况，第一个数既可以填奇数又可以填偶数，要特殊处理
			if n%2 == 0 && i == 0 {
				parity = 1 - j%2
				j /= 2
			}
		} // else n 很大的情况下，只能按照 1,2,3,... 的顺序填
		ans[i] = cand[parity][j]
		cand[parity] = slices.Delete(cand[parity], j, j+1)
		parity ^= 1 // 下一个数的奇偶性
	}
	return ans
}

# Accepted solution for LeetCode #3470: Permutations IV
class Solution:
  def permute(self, n: int, k: int) -> list[int]:
    ans = []
    isLookingForEven = True
    remainingNumbers = list(range(1, n + 1))

    for turn in range(n):
      remainingPermutations = (math.factorial((n - 1 - turn) // 2) *
                               math.factorial((n - turn) // 2))
      found = False
      for index, number in enumerate(remainingNumbers):
        if number % 2 != isLookingForEven and (turn > 0 or n % 2 == 1):
          continue
        if k <= remainingPermutations:
          ans.append(remainingNumbers.pop(index))
          isLookingForEven = ans[-1] % 2 == 0
          found = True
          break
        k -= remainingPermutations
      if not found:
        return []

    return ans

// Accepted solution for LeetCode #3470: Permutations IV
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3470: Permutations IV
// package main
// 
// import "slices"
// 
// // https://space.bilibili.com/206214
// // 预处理交替排列的方案数
// var f = []int{1}
// 
// func init() {
// 	for i := 1; f[len(f)-1] < 1e15; i++ {
// 		f = append(f, f[len(f)-1]*i)
// 		f = append(f, f[len(f)-1]*i)
// 	}
// }
// 
// func permute(n int, K int64) []int {
// 	// k 改成从 0 开始，方便计算
// 	k := int(K - 1)
// 	if n < len(f) && k >= f[n]*(2-n%2) { // n 是偶数的时候，方案数乘以 2
// 		return nil
// 	}
// 
// 	// cand 表示剩余未填入 ans 的数字
// 	// cand[0] 保存偶数，cand[1] 保存奇数
// 	cand := [2][]int{}
// 	for i := 2; i <= n; i += 2 {
// 		cand[0] = append(cand[0], i)
// 	}
// 	for i := 1; i <= n; i += 2 {
// 		cand[1] = append(cand[1], i)
// 	}
// 
// 	ans := make([]int, n)
// 	parity := 1 // 当前要填入 ans[i] 的数的奇偶性
// 	for i := range n {
// 		j := 0
// 		if n-1-i < len(f) {
// 			// 比如示例 1，按照第一个数分组，每一组的大小都是 size=2
// 			// 知道 k 和 size 就知道我们要去哪一组
// 			size := f[n-1-i]
// 			j = k / size // 去第 j 组
// 			k %= size
// 			// n 是偶数的情况，第一个数既可以填奇数又可以填偶数，要特殊处理
// 			if n%2 == 0 && i == 0 {
// 				parity = 1 - j%2
// 				j /= 2
// 			}
// 		} // else n 很大的情况下，只能按照 1,2,3,... 的顺序填
// 		ans[i] = cand[parity][j]
// 		cand[parity] = slices.Delete(cand[parity], j, j+1)
// 		parity ^= 1 // 下一个数的奇偶性
// 	}
// 	return ans
// }

// Accepted solution for LeetCode #3470: Permutations IV
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3470: Permutations IV
// package main
// 
// import "slices"
// 
// // https://space.bilibili.com/206214
// // 预处理交替排列的方案数
// var f = []int{1}
// 
// func init() {
// 	for i := 1; f[len(f)-1] < 1e15; i++ {
// 		f = append(f, f[len(f)-1]*i)
// 		f = append(f, f[len(f)-1]*i)
// 	}
// }
// 
// func permute(n int, K int64) []int {
// 	// k 改成从 0 开始，方便计算
// 	k := int(K - 1)
// 	if n < len(f) && k >= f[n]*(2-n%2) { // n 是偶数的时候，方案数乘以 2
// 		return nil
// 	}
// 
// 	// cand 表示剩余未填入 ans 的数字
// 	// cand[0] 保存偶数，cand[1] 保存奇数
// 	cand := [2][]int{}
// 	for i := 2; i <= n; i += 2 {
// 		cand[0] = append(cand[0], i)
// 	}
// 	for i := 1; i <= n; i += 2 {
// 		cand[1] = append(cand[1], i)
// 	}
// 
// 	ans := make([]int, n)
// 	parity := 1 // 当前要填入 ans[i] 的数的奇偶性
// 	for i := range n {
// 		j := 0
// 		if n-1-i < len(f) {
// 			// 比如示例 1，按照第一个数分组，每一组的大小都是 size=2
// 			// 知道 k 和 size 就知道我们要去哪一组
// 			size := f[n-1-i]
// 			j = k / size // 去第 j 组
// 			k %= size
// 			// n 是偶数的情况，第一个数既可以填奇数又可以填偶数，要特殊处理
// 			if n%2 == 0 && i == 0 {
// 				parity = 1 - j%2
// 				j /= 2
// 			}
// 		} // else n 很大的情况下，只能按照 1,2,3,... 的顺序填
// 		ans[i] = cand[parity][j]
// 		cand[parity] = slices.Delete(cand[parity], j, j+1)
// 		parity ^= 1 // 下一个数的奇偶性
// 	}
// 	return ans
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Permutations IV

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

Overflow in intermediate arithmetic