Solution using Combinatorics and Policy-Based Data Structures in C++
IntuitionThe problem requires generating the k-th lexicographical alternating permutation of the first n positive integers. An alternating permutation ensures t
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires generating the k-th lexicographical alternating permutation of the first n positive integers. An alternating permutation ensures that no two adjacent elements are both odd or both even. To achieve this, we can:

Separate the numbers into two sets: odd and even.

Use combinatorial mathematics to determine the position of each element in the permutation.

Use an ordered set (from the Policy-Based Data Structures library) to efficiently find and remove elements based on their order.

The key idea is to treat the problem as a combinatorial one, where we calculate the number of valid permutations for each step and use this to determine the k-th permutation.
# Approach
<!-- Describe your approach to solving the problem. -->
Initialization:

Create two ordered sets: odd for odd numbers and even for even numbers.

Precompute factorials up to 18! (since 19! exceeds the range of long long).

Base Case:

If n is even, calculate the number of valid permutations as (n/2)! * (n/2 - 1)!.

If n is odd, calculate the number of valid permutations as (n/2)! * (n/2)!.

Determine the First Element:

If k is small, start with the smallest available element (1).

Otherwise, calculate the position of the first element in the permutation using combinatorial logic.

Build the Permutation:

Iteratively determine the next element by alternating between the odd and even sets.
Calculate the number of permutations for each position while excluding that position.

Use the precomputed factorials to determine the position of each element in the remaining set.

Remove the selected element from the corresponding set and add it to the result.

Edge Cases:

If k exceeds the total number of valid permutations, return an empty list.

Handle cases where n is large (e.g., n > 18) by returning early if the factorial calculation would overflow.

# Time Complexity
- O(n*log(n))
And as the value of n doesn't exceed 100 the time complexity is significantly low.


# Code
```cpp []
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 

#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> 

class Solution {
public:
    vector<int> permute(int n, long long k) {
        ordered_set even, odd;
        for (int i = 1; i <= n; i++) {
            if (i % 2) odd.insert(i);
            else even.insert(i);
        }
        vector<int> v;
        int i = 2;
        k--;
        vector<long long> f(19, 1);
        for (int j = 1; j <= 18; j++) {
            f[j] = j * f[j - 1];
        }
        int c = 0;
        if (n % 2 == 0) {
            if (n / 2 > 12 || f[n / 2] * f[n / 2 - 1] > k) {
                v.push_back(1);
                odd.erase(1);
            } else {
                if (k / (f[n / 2] * f[n / 2 - 1]) >= n) {
                    return v;
                }
                long long a = k / (f[n / 2] * f[n / 2 - 1]) + 1;
                if (a % 2 == 0) {
                    c = 1;
                    even.erase(a);
                }else{
                    odd.erase(a);
                }
                v.push_back(a);
                k %= (f[n / 2] * f[n / 2 - 1]);
            }
        } else {
            if (n / 2 > 12 || f[n / 2] * f[n / 2] > k) {
                v.push_back(1);
                odd.erase(1);
            } else {
                if (k / (f[n / 2] * f[n / 2]) > n / 2) {
                    return v;
                }
                v.push_back((k / (f[n / 2] * f[n / 2]))*2 + 1);
                odd.erase((k / (f[n / 2] * f[n / 2]))*2 + 1);
                k %= (f[n / 2] * f[n / 2]);
            }
        }
        while (i <= n) {
            for(auto it:odd) cout<<it<<" ";cout<<endl;
            for(auto it:even) cout<<it<<" ";cout<<endl;
            long long a = (n - i);
            int d;
            cout<<i<<" "<<k<<endl;
            if (a % 2) {
                if (a/2>12||f[a / 2] * f[a / 2 + 1] > k) {
                    d = 0;
                } else {
                    long long b = f[a / 2] * f[a / 2 + 1];
                    d = k / b;
                    k %= b;
                }
            } else {
                if (a/2>12||f[a / 2] * f[a / 2] > k) {
                    d = 0;
                } else {
                    long long b = f[a / 2] * f[a / 2];
                    d = k / b;
                    k %= b;
                }
            }
            // cout<<i<<k<<d<<endl;
            if (c == 0) {
                int x = *even.find_by_order(d);
                even.erase(x);
                v.push_back(x);
            } else {
                int x = *odd.find_by_order(d);
                odd.erase(x);
                v.push_back(x);
            }
            c = !c;
            i++;
        }
        for (auto it : v) {
            cout << it << " ";
        }
        cout << endl;
        return v;
    }
};
```
This solution is efficient and leverages combinatorial mathematics and ordered sets to generate the k-th alternating permutation. It is unique compared to other solutions because of its use of Policy-Based Data Structures and combinatorial logic.
![image.png](https://assets.leetcode.com/users/images/30603cf7-75c7-4852-8fbb-70d977419da5_1741121157.0948408.png)

['Math', 'Combinatorics', 'Ordered Set', 'C++']