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

We need to arrange the first N natural numbers such that the absolute difference between every two consecutive elements is greater than 1. If no such permutation exists, we return -1.

The approach uses a greedy strategy − arrange all odd numbers in descending order, followed by all even numbers in descending order. This ensures adjacent elements always differ by at least 2.

Syntax

void arrangeN(int N);

Algorithm

Begin
  if N is 1, then return 1
  if N is 2 or 3, then return -1 as no such permutation exists
  even_max and odd_max is set as max even and odd number ? n
  arrange all odd numbers in descending order
  arrange all even numbers in descending order
End

Example

Here's the complete C implementation of the arrangement algorithm −

#include <stdio.h>

void arrangeN(int N) {
    if (N == 1) { // if N is 1, only that will be placed
        printf("1");
        return;
    }
    if (N == 2 || N == 3) { // for N = 2 and 3, no such permutation is available
        printf("-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
        printf("%d ", odd_max);
        odd_max -= 2;
    }
    
    while (even_max >= 2) { // print all even numbers in decreasing order
        printf("%d ", even_max);
        even_max -= 2;
    }
    printf("<br>");
}

int main() {
    int N = 8;
    printf("Arrangement for N = %d: ", N);
    arrangeN(N);
    return 0;
}

Output

Arrangement for N = 8: 7 5 3 1 8 6 4 2 

How It Works

• For N = 1: Only one element, so condition is satisfied.
• For N = 2, 3: No valid arrangement exists since consecutive numbers differ by 1.
• For N ? 4: Place odds in descending order (7,5,3,1), then evens in descending order (8,6,4,2).
• Adjacent elements always differ by at least 2: |7-5|=2, |5-3|=2, |1-8|=7, etc.

Conclusion

This greedy approach efficiently solves the problem by separating odd and even numbers. The arrangement ensures all adjacent pairs have an absolute difference greater than 1, making it a valid solution for N ? 4.