Smallest lexicographical permutation such that a[i] % i is equal to 0

Given the integers N and X, the task is to find the smallest lexicographical permutation a[], using all integers [1, N] such that a[i]%i is equal to 0 given that the a[1] = X (the first element of permutation) and a[N] = 1 (the last element of permutation) which means a[1], a[N] are constants and the elements from indices [2, N - 1] should follow a[i] % i = 0.

Note: Follow 1-based indexing

Examples:

Input: N = 4, X = 2
Output: 2 4 3 1
Explanation: The a[1] = 2, a[4] = 1 and a[2]%2, a[3]%3 is equal to 0 so the given condition as per question is satisfied and it is the smallest lexicographical permutation.

Input: N = 5, X = 4
Output: -1
Explanation: There is no such possible permutation that satisfies the given condition.

Approach: To solve the problem follow the below approach:

The approach is based on the logic that out of integers of permutation from 1 to N, 1 and X are placed and the X^th indexed position should be replaced by smallest multiple of X. Hashing is used to check whether any number to be placed at i^th position is already used or not.

Follow the steps mentioned below to solve the problem:

• Initialize the check[] array to check whether the element to be placed at the i^th index is visited or not.
• Initialize vector permutation to store the final result permutation.
• Initialize a flag variable to check if there is a possibility to form a permutation.
• As given in the question, mark 1 and x as visited as they are placed.
• Traverse from 2 till N
• Now, check for the number to be placed at i^th position to satisfy the condition. Iterate from j = i to all multiples of j to place the lowest possible multiple.
• If the element itself is not visited already then place it.
• Else, check for a multiple that is not visited and divides N.
• If the ith indexed element is still -1 we are unable to find an element to place so mark the flag and break.
• Print -1 if not possible to form a permutation
• Print the permutation if it is possible to form a permutation.

Below is the implementation of the above approach:

2 4 3 1 
-1

Time Complexity: O(nlogn), where n is the size of the array.
Auxiliary Space: O(n)