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

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.

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.8 Average Difficulty : 3.8/5.0
Based on 11 vote(s)