Given a positive number n, we need to find all the combinations of 2*n elements such that every element from 1 to n appears exactly twice and distance between its appearances is exactly equal to value of the element.
Examples:
Input : n = 3 Output : 3 1 2 1 3 2 2 3 1 2 1 3 All elements from 1 to 3 appear twice and distance between two appearances is equal to value of the element. Input : n = 4 Output : 4 1 3 1 2 4 3 2 2 3 4 2 1 3 1 4
Explanation
We can use backtracking to solve this problem. The idea is to all possible combinations for the first element and recursively explore remaining element to check if they will lead to the solution or not. If current configuration doesn’t result in solution, we backtrack. Note that an element k can be placed at position i and (i+k+1) in the output array i >= 0 and (i+k+1) < 2*n.
Note that no combination of element is possible for some value of n like 2, 5, 6 etc.
// C++ program to find all combinations where every // element appears twice and distance between // appearances is equal to the value #include <bits/stdc++.h> using namespace std;
// Find all combinations that satisfies given constraints void allCombinationsRec(vector< int > &arr, int elem, int n)
{ // if all elements are filled, print the solution
if (elem > n)
{
for ( int i : arr)
cout << i << " " ;
cout << endl;
return ;
}
// try all possible combinations for element elem
for ( int i = 0; i < 2*n; i++)
{
// if position i and (i+elem+1) are not occupied
// in the vector
if (arr[i] == -1 && (i + elem + 1) < 2*n &&
arr[i + elem + 1] == -1)
{
// place elem at position i and (i+elem+1)
arr[i] = elem;
arr[i + elem + 1] = elem;
// recurse for next element
allCombinationsRec(arr, elem + 1, n);
// backtrack (remove elem from position i and (i+elem+1) )
arr[i] = -1;
arr[i + elem + 1] = -1;
}
}
} void allCombinations( int n)
{ // create a vector of double the size of given number with
vector< int > arr(2*n, -1);
// all its elements initialized by 1
int elem = 1;
// start from element 1
allCombinationsRec(arr, elem, n);
} // Driver code int main()
{ // given number
int n = 3;
allCombinations(n);
return 0;
} |
// Java program to find all combinations where every // element appears twice and distance between // appearances is equal to the value import java.util.Vector;
class Test
{ // Find all combinations that satisfies given constraints
static void allCombinationsRec(Vector<Integer> arr, int elem, int n)
{
// if all elements are filled, print the solution
if (elem > n)
{
for ( int i : arr)
System.out.print(i + " " );
System.out.println();
return ;
}
// try all possible combinations for element elem
for ( int i = 0 ; i < 2 *n; i++)
{
// if position i and (i+elem+1) are not occupied
// in the vector
if (arr.get(i) == - 1 && (i + elem + 1 ) < 2 *n &&
arr.get(i + elem + 1 ) == - 1 )
{
// place elem at position i and (i+elem+1)
arr.set(i, elem);
arr.set(i + elem + 1 , elem);
// recurse for next element
allCombinationsRec(arr, elem + 1 , n);
// backtrack (remove elem from position i and (i+elem+1) )
arr.set(i, - 1 );
arr.set(i + elem + 1 , - 1 );
}
}
}
static void allCombinations( int n)
{
// create a vector of double the size of given number with
Vector<Integer> arr = new Vector<>();
for ( int i = 0 ; i < 2 *n; i++) {
arr.add(- 1 );
}
// all its elements initialized by 1
int elem = 1 ;
// start from element 1
allCombinationsRec(arr, elem, n);
}
// Driver method
public static void main(String[] args)
{
// given number
int n = 3 ;
allCombinations(n);
}
} |
# Python3 program to find all combinations # where every element appears twice and distance # between appearances is equal to the value # Find all combinations that # satisfies given constraints def allCombinationsRec(arr, elem, n):
# if all elements are filled,
# print the solution
if (elem > n):
for i in (arr):
print (i, end = " " )
print ("")
return
# Try all possible combinations
# for element elem
for i in range ( 0 , 2 * n):
# if position i and (i+elem+1) are
# not occupied in the vector
if (arr[i] = = - 1 and
(i + elem + 1 ) < 2 * n and
arr[i + elem + 1 ] = = - 1 ):
# place elem at position
# i and (i+elem+1)
arr[i] = elem
arr[i + elem + 1 ] = elem
# recurse for next element
allCombinationsRec(arr, elem + 1 , n)
# backtrack (remove elem from
# position i and (i+elem+1) )
arr[i] = - 1
arr[i + elem + 1 ] = - 1
def allCombinations(n):
# create a vector of double
# the size of given number with
arr = [ - 1 ] * ( 2 * n)
# all its elements initialized by 1
elem = 1
# start from element 1
allCombinationsRec(arr, elem, n)
# Driver code n = 3
allCombinations(n) # This code is contributed by Smitha Dinesh Semwal. |
// C# program to find all combinations where every // element appears twice and distance between // appearances is equal to the value using System;
using System.Collections.Generic;
class Test{
// Find all combinations that satisfies given // constraints static void allCombinationsRec(List< int > arr, int elem,
int n)
{ // If all elements are filled, print the solution
if (elem > n)
{
foreach ( int i in arr)
Console.Write(i + " " );
Console.WriteLine();
return ;
}
// Try all possible combinations for element elem
for ( int i = 0; i < 2 * n; i++)
{
// If position i and (i+elem+1) are not
// occupied in the vector
if (arr[i] == -1 && (i + elem + 1) < 2 * n &&
arr[i + elem + 1] == -1)
{
// Place elem at position i and (i+elem+1)
arr[i] = elem;
arr[i + elem + 1] = elem;
// Recurse for next element
allCombinationsRec(arr, elem + 1, n);
// Backtrack (remove elem from position i
// and (i+elem+1) )
arr[i] = -1;
arr[i + elem + 1] = -1;
}
}
} static void allCombinations( int n)
{ // Create a vector of double the size of given
// number with
List< int > arr = new List< int >();
for ( int i = 0; i < 2 * n; i++)
{
arr.Add(-1);
}
// All its elements initialized by 1
int elem = 1;
// Start from element 1
allCombinationsRec(arr, elem, n);
} // Driver Code public static void Main( string [] args)
{ // Given number
int n = 3;
allCombinations(n);
} } // This code is contributed by ukasp |
<script> // Javascript program to find all combinations where every // element appears twice and distance between // appearances is equal to the value // Find all combinations that satisfies given constraints function allCombinationsRec(arr, elem, n)
{ // if all elements are filled, print the solution
if (elem > n) {
for (i of arr)
document.write(i + " " );
document.write( "<br>" );
return ;
}
// try all possible combinations for element elem
for (let i = 0; i < 2 * n; i++)
{
// if position i and (i+elem+1) are not occupied
// in the vector
if (arr[i] == -1 && (i + elem + 1) < 2 * n &&
arr[i + elem + 1] == -1)
{
// place elem at position i and (i+elem+1)
arr[i] = elem;
arr[i + elem + 1] = elem;
// recurse for next element
allCombinationsRec(arr, elem + 1, n);
// backtrack (remove elem from position i and (i+elem+1) )
arr[i] = -1;
arr[i + elem + 1] = -1;
}
}
} function allCombinations(n)
{ // create a vector of double the size of given number with
let arr = new Array(2 * n).fill(-1);
// all its elements initialized by 1
let elem = 1;
// start from element 1
allCombinationsRec(arr, elem, n);
} // Driver code // given number let n = 3; allCombinations(n); // This code is contributed by gfgking. </script> |
Output:
3 1 2 1 3 2 2 3 1 2 1 3