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.
- Take two vectors even and odd to store the even and odd elements of the array separately.
- 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.
- 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++
#include <bits/stdc++.h>
using namespace std;
void printPermutation( int arr[], int n)
{
vector< int > odd, 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();
if (size_odd > size_even + 1)
cout << -1 << endl;
else {
int i = 0;
int j = 0;
while (i < size_odd && j < size_even) {
cout << odd[i] << " " ;
++i;
cout << even[j] << " " ;
++j;
}
while (i < size_odd) {
cout << odd[i] << " " ;
++i;
}
while (j < size_even) {
cout << even[j] << " " ;
}
}
}
int main()
{
int arr[] = { 6, 7, 9, 8, 10, 11 };
int N = sizeof (arr) / sizeof (arr[0]);
printPermutation(arr, N);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG{
static void printPermutation( int arr[], int n)
{
ArrayList<Integer> odd = new ArrayList<Integer>();
ArrayList<Integer> even = new ArrayList<Integer>();
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();
if (size_odd > size_even + 1 )
System.out.println( "-1" );
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;
}
while (i < size_odd)
{
System.out.print(odd.get(i) + " " );
++i;
}
while (j < size_even)
{
System.out.print(even.get(j) + " " );
}
}
}
public static void main (String[] args)
{
int arr[] = { 6 , 7 , 9 , 8 , 10 , 11 };
int N = arr.length;
printPermutation(arr, N);
}
}
|
Python3
def printPermutation(arr, n):
odd, 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)
if (size_odd > size_even + 1 ):
print ( - 1 )
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
while (i < size_odd):
print (odd[i], end = " " )
i + = 1
while (j < size_even):
print (even[j], end = " " )
j + = 1
arr = [ 6 , 7 , 9 , 8 , 10 , 11 ]
N = len (arr)
printPermutation(arr, N)
|
C#
using System;
using System.Collections.Generic;
class GFG{
static void printPermutation( int []arr, int n)
{
List< int > odd = new List< int >();
List< int > even = new List< int >();
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;
if (size_odd > size_even + 1)
Console.WriteLine( "-1" );
else
{
int i = 0;
int j = 0;
while (i < size_odd && j < size_even)
{
Console.Write(odd[i] + " " );
++i;
Console.Write(even[j] + " " );
++j;
}
while (i < size_odd)
{
Console.Write(odd[i] + " " );
++i;
}
while (j < size_even)
{
Console.Write(even[j] + " " );
}
}
}
public static void Main(String[] args)
{
int []arr = { 6, 7, 9, 8, 10, 11 };
int N = arr.Length;
printPermutation(arr, N);
}
}
|
Javascript
<script>
function printPermutation(arr, n)
{
let odd = [];
let 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;
if (size_odd > size_even + 1)
document.write( "-1" );
else
{
let i = 0;
let j = 0;
while (i < size_odd && j < size_even)
{
document.write(odd[i] + " " );
++i;
document.write(even[j] + " " );
++j;
}
while (i < size_odd)
{
document.write(odd[i] + " " );
++i;
}
while (j < size_even)
{
document.write(even[j] + " " );
}
}
}
let arr = [ 6, 7, 9, 8, 10, 11 ];
let N = arr.length;
printPermutation(arr, N);
</script>
|
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.
Last Updated :
07 Mar, 2022
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