# Arrange first N natural numbers such that absolute difference between all adjacent elements > 1?

We have the first N natural numbers. Our task is to get one permutation of them where the absolute difference between every two consecutive elements is > 1. If no such permutation is present, return -1.

The approach is simple. We will use the greedy approach. We will arrange all odd numbers in increasing or decreasing order, then arrange all even numbers in decreasing or increasing order

## Algorithm

### arrangeN(n)

Begin
if N is 1, then return 1
if N is 2 or 3, then return -1 as no such permutation is not present
even_max and odd_max is set as max even and odd number less or equal to n
arrange all odd numbers in descending order
arrange all even numbers in descending order
End

## Example

#include <iostream>
using namespace std;
void arrangeN(int N) {
if (N == 1) { //if N is 1, only that will be placed
cout << "1";
return;
}
if (N == 2 || N == 3) { //for N = 2 and 3, no such permutation is available
cout << "-1";
return;
}
int even_max = -1, odd_max = -1;
//find max even and odd which are less than or equal to N
if (N % 2 == 0) {
even_max = N;
odd_max = N - 1;
} else {
odd_max = N;
even_max = N - 1;
}
while (odd_max >= 1) { //print all odd numbers in decreasing order
cout << odd_max << " ";
odd_max -= 2;
}
while (even_max >= 2) { //print all even numbers in decreasing order
cout << even_max << " ";
even_max -= 2;
}
}
int main() {
int N = 8;
arrangeN(N);
}

## Output

7 5 3 1 8 6 4 2