Fill two instances of all numbers from 1 to n in a specific way

3.8

Given a number n, create an array of size 2n such that the array contains 2 instances of every number from 1 to n, and the number of elements between two instances of a number i is equal to i. If such a configuration is not possible, then print the same.

Examples:

Input: n = 3
Output: res[] = {3, 1, 2, 1, 3, 2}

Input: n = 2
Output: Not Possible

Input: n = 4
Output: res[] = {4, 1, 3, 1, 2, 4, 3, 2}

We strongly recommend to minimize the browser and try this yourself first.

One solution is to Backtracking. The idea is simple, we place two instances of n at a place, then recur for n-1. If recurrence is successful, we return true, else we backtrack and try placing n at different location. Following is C implementation of the idea.

// A backtracking based C Program to fill two instances of all numbers 
// from 1 to n in a specific way
#include <stdio.h>
#include <stdbool.h>

// A recursive utility function to fill two instances of numbers from 
// 1 to n in res[0..2n-1].  'curr' is current value of n.
bool fillUtil(int res[], int curr, int n)
{
     // If current number becomes 0, then all numbers are filled
     if (curr == 0) return true;

     // Try placing two instances of 'curr' at all possible locations
     // till solution is found
     int i;
     for (i=0; i<2*n-curr-1; i++)
     {
        // Two 'curr' should be placed at 'curr+1' distance
        if (res[i] == 0 && res[i + curr + 1] == 0)
        {
           // Plave two instances of 'curr'
           res[i] = res[i + curr + 1] = curr;

           // Recur to check if the above placement leads to a solution
           if (fillUtil(res, curr-1, n))
               return true;

           // If solution is not possible, then backtrack
           res[i] = res[i + curr + 1] = 0;
        }
     }
     return false;
}

// This function prints the result for input number 'n' using fillUtil()
void fill(int n)
{
    // Create an array of size 2n and initialize all elements in it as 0
    int res[2*n], i;
    for (i=0; i<2*n; i++)
       res[i] = 0;

    // If solution is possible, then print it.
    if (fillUtil(res, n, n))
    {
        for (i=0; i<2*n; i++)
           printf("%d ", res[i]);
    }
    else
        puts("Not Possible");
}

// Driver program
int main()
{
  fill(7);
  return 0;
}

Output:

7 3 6 2 5 3 2 4 7 6 5 1 4 1

The above solution may not be the best possible solution. There seems to be a pattern in the output. I an Looking for a better solution from other geeks.

This article is contributed by Asif. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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