Permutation of Array such that products of all adjacent elements are even

Given an array arr[] consisting of N positive integers, the task is to find any permutation of given array such that the product of adjacent elements is even. Print any such permutation or -1 if not possible.

Example:

Input: arr[] = {6,7,9,8,10,11}
Output: 8 9 10 7 6 11
Explanation: 
Product of adjacent elements =>
8 x 9 = 72 (even)
9 x 10 = 90 (even)
10 x 7 = 70 (even)
7 x 6 = 42 (even)
6 x 11 = 66 (even)

Input: arr[] = {3,2,5,7,1,4,9}
Output: -1
Explanation: There is no possible arrangements of elements such that product of adjacent elements is equal.

Naive Approach: The simplest approach to solve this problem is to try every possible arrangement of the elements and check the condition to be true.

Time Complexity: O(N*N!) where N is the number of elements in the array. O(N!) is the time taken to create all permutations of the given array and O(N) is the time required to check if the current permutation is the required one or not.
Auxiliary Space: O(N) to store the permutation each time.

Efficient Approach: The solution can be found using simple observations. If there are multiple odd and even elements in the array then an optimal arrangement of any adjacent elements can be either of the below cases for the product to be even:

{Odd, Even} 
{Even, Odd}
{Even, Even}

Please note that {Odd, Odd} arrangement of any adjacent element will give an Odd product. Hence, this arrangement is not possible. 

The above arrangements is only possible if 

number_of_odd_elements  <= number_of_even_elements + 1 in the array.

Follow the steps below to solve the problem.

1. Take two vectors even and odd to store the even and odd elements of the array separately.
2. If the size of odd vector is greater than size of even vector + 1, (as explained above) then solution is not possible. Therefore, print -1.
3. Else first print one element from the odd vector and then one element from even vector until both vectors are empty.

Below is the implementation of the above approach.

// C++ program to Permutation of Array 
// such that product of all 
// adjacent elements is even 

#include <bits/stdc++.h> 
using namespace std; 

// Function to print 
// the required permutation 

void printPermutation(int arr[], int n) 
{ 
    vector<int> odd, even; 

    // push odd elements in 'odd' 
    // and even elements in 'even' 
    for (int i = 0; i < n; i++) { 

        if (arr[i] % 2 == 0) 
            even.push_back(arr[i]); 
        else
            odd.push_back(arr[i]); 
    } 

    int size_odd = odd.size(); 
    int size_even = even.size(); 

    // Check if it possible to 
    // arrange the elements 
    if (size_odd > size_even + 1) 
        cout << -1 << endl; 

    // else print the permutation 
    else { 

        int i = 0; 
        int j = 0; 

        while (i < size_odd && j < size_even) { 

            cout << odd[i] << " "; 
            ++i; 
            cout << even[j] << " "; 
            ++j; 
        } 

        // Print remaining odds are even. 
        // and even elements 
        while (i < size_odd) { 
            cout << odd[i] << " "; 
            ++i; 
        } 

        while (j < size_even) { 
            cout << even[j] << " "; 
        } 
    } 
} 

// Driver code 
int main() 
{ 
    int arr[] = { 6, 7, 9, 8, 10, 11 }; 
    int N = sizeof(arr) / sizeof(arr[0]); 

    printPermutation(arr, N); 
    return 0; 
}

// Java program to permutation of array 
// such that product of all adjacent 
// elements is even 
import java.io.*;
import java.util.*;

class GFG{
    
// Function to print 
// the required permutation 
static void printPermutation(int arr[], int n) 
{ 
    ArrayList<Integer> odd = new ArrayList<Integer>();
    ArrayList<Integer> even = new ArrayList<Integer>();
    
    // push odd elements in 'odd' 
    // and even elements in 'even' 
    for(int i = 0; i < n; i++)
    { 
         if (arr[i] % 2 == 0) 
              even.add(arr[i]); 
           else
              odd.add(arr[i]); 
    } 
  
    int size_odd = odd.size(); 
    int size_even = even.size(); 
  
    // Check if it possible to 
    // arrange the elements 
    if (size_odd > size_even + 1) 
        System.out.println("-1"); 

    // Else print the permutation 
    else 
    { 
        int i = 0; 
        int j = 0; 

        while (i < size_odd && j < size_even) 
        { 
            System.out.print(odd.get(i) + " "); 
            ++i; 
            System.out.print(even.get(j) + " "); 
            ++j; 
        } 

        // Print remaining odds are even. 
        // and even elements 
        while (i < size_odd) 
        { 
            System.out.print(odd.get(i) + " "); 
            ++i; 
        } 
        while (j < size_even) 
        { 
            System.out.print(even.get(j) + " "); 
        } 
    } 
}

// Driver Code
public static void main (String[] args)
{
    int arr[] = { 6, 7, 9, 8, 10, 11 }; 
    int N = arr.length; 

    printPermutation(arr, N); 
}
}

// This code is contributed by offbeat

# Python3 program to Permutation of Array 
# such that product of all 
# adjacent elements is even 

# Function to print 
# the required permutation 
def printPermutation(arr, n): 

    odd, even = [], [] 

    # push odd elements in 'odd' 
    # and even elements in 'even' 
    for i in range(n): 
        if(arr[i] % 2 == 0): 
            even.append(arr[i]) 
        else: 
            odd.append(arr[i]) 

    size_odd = len(odd) 
    size_even = len(even) 

    # Check if it possible to 
    # arrange the elements 
    if(size_odd > size_even + 1): 
        print(-1) 

    # else print the permutation 
    else: 
        i, j = 0, 0

        while(i < size_odd and j < size_even): 
            
            print(odd[i], end = " ") 
            i += 1
            print(even[j], end = " ") 
            j += 1

    # Print remaining odds are even. 
    # and even elements 
    while(i < size_odd): 
        print(odd[i], end = " ") 
        i += 1

    while(j < size_even): 
        print(even[j], end = " ") 
        j += 1

# Driver Code 
arr = [ 6, 7, 9, 8, 10, 11 ] 

N = len(arr) 

# Function call 
printPermutation(arr, N) 

# This code is contributed by Shivam Singh 

// C# program to permutation of array 
// such that product of all adjacent 
// elements is even 
using System;
using System.Collections.Generic;

class GFG{
    
// Function to print 
// the required permutation 
static void printPermutation(int []arr, int n) 
{ 
    List<int> odd = new List<int>();
    List<int> even = new List<int>();
    
    // push odd elements in 'odd' 
    // and even elements in 'even' 
    for(int i = 0; i < n; i++)
    { 
        if (arr[i] % 2 == 0) 
            even.Add(arr[i]); 
        else
            odd.Add(arr[i]); 
    } 

    int size_odd = odd.Count; 
    int size_even = even.Count; 

    // Check if it possible to 
    // arrange the elements 
    if (size_odd > size_even + 1) 
        Console.WriteLine("-1"); 

    // Else print the permutation 
    else
    { 
        int i = 0; 
        int j = 0; 

        while (i < size_odd && j < size_even) 
        { 
            Console.Write(odd[i] + " "); 
            ++i; 
            Console.Write(even[j] + " "); 
            ++j; 
        } 

        // Print remaining odds are even. 
        // and even elements 
        while (i < size_odd) 
        { 
            Console.Write(odd[i] + " "); 
            ++i; 
        } 
        while (j < size_even) 
        { 
            Console.Write(even[j] + " "); 
        } 
    } 
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 6, 7, 9, 8, 10, 11 }; 
    int N = arr.Length; 

    printPermutation(arr, N); 
}
}

// This code is contributed by PrinciRaj1992

<script>

// JavaScript program to permutation of array 
// such that product of all adjacent 
// elements is even 

// Function to print
// the required permutation 
function printPermutation(arr, n) 
{ 
    let odd = [];
    let even = [];
      
    // push odd elements in 'odd' 
    // and even elements in 'even' 
    for(let i = 0; i < n; i++)
    { 
        if (arr[i] % 2 == 0) 
            even.push(arr[i]); 
        else
            odd.push(arr[i]); 
    } 
  
    let size_odd = odd.length; 
    let size_even = even.length; 
  
    // Check if it possible to 
    // arrange the elements 
    if (size_odd > size_even + 1) 
        document.write("-1"); 
  
    // Else print the permutation 
    else
    { 
        let i = 0; 
        let j = 0; 
  
        while (i < size_odd && j < size_even) 
        { 
            document.write(odd[i] + " "); 
            ++i; 
            document.write(even[j] + " "); 
            ++j; 
        } 
  
        // Print remaining odds are even. 
        // and even elements 
        while (i < size_odd) 
        { 
            document.write(odd[i] + " "); 
            ++i; 
        } 
        while (j < size_even) 
        { 
            document.write(even[j] + " "); 
        } 
    } 
}


// Driver Code

    let arr = [ 6, 7, 9, 8, 10, 11 ]; 
    let N = arr.length; 
  
    printPermutation(arr, N);
     
</script>

7 6 9 8 11 10 

Time Complexity: O(N) where N the number of elements. O(N) time is required to traverse the given array and form the odd & even vectors and O(N) is required to print the permutation.
Auxiliary Space: O(N) because the given array elements are distributed among the two vectors.