Given an array arr of N (even) integer elements. The task is to check if it is possible to reorder the elements of the array such that:
arr[2*i + 1] = 2 * A[2 * i] for i = 0 ... N-1.
Print True if it is possible, otherwise print False.
Examples:
Input: arr[] = {4, -2, 2, -4}
Output: True
{-2, -4, 2, 4} is a valid arrangement, -2 * 2 = -4 and 2 * 2 = 4Input: arr[] = {1, 2, 4, 16, 8, 4}
Output: False
Approach: The idea is that, if k is current minimum element in the array then it must pair with 2 * k as there does not exist any other element k / 2 to pair it with.
We check elements in ascending order. When we check an element k and it isn’t used, it must pair with 2 * k. We will attempt to arrange k followed by 2 * k however if we can’t, then the answer is False. In the end, if all the operations are successful, then print True.
We will store a count of each element to keep track of what we have not yet considered.
Below is the implementation of above approach:
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std;
// Function to return true if the elements // can be arranged in the desired order string canReorder( int A[], int n)
{ map< int , int > m;
for ( int i=0;i<n;i++)
m[A[i]]++;
sort(A,A+n);
int count = 0;
for ( int i=0;i<n;i++)
{
if (m[A[i]] == 0)
continue ;
// If 2 * x is not found to pair
if (m[2 * A[i]]){
count+=2;
// Remove an occurrence of x
// and an occurrence of 2 * x
m[A[i]] -= 1;
m[2 * A[i]] -= 1;
}
}
if (count ==n)
return "true" ;
else
return "false" ;
} // Driver Code int main()
{ int A[] = {4, -2, 2, -4};
int n= sizeof (A)/ sizeof ( int );
// Function call to print required answer cout<<(canReorder(A,n)); return 0;
} //contributed by Arnab Kundu |
// Java implementation of the approach import java.util.HashMap;
import java.util.Map;
import java.util.Arrays;
class GfG
{ // Function to return true if the elements
// can be arranged in the desired order
static String canReorder( int A[], int n)
{
HashMap<Integer,Integer> m = new HashMap<>();
for ( int i = 0 ; i < n; i++)
{
if (m.containsKey(A[i]))
m.put(A[i], m.get(A[i]) + 1 );
else
m.put(A[i], 1 );
}
Arrays.sort(A);
int count = 0 ;
for ( int i = 0 ; i < n; i++)
{
if (m.get(A[i]) == 0 )
continue ;
// If 2 * x is not found to pair
if (m.containsKey( 2 * A[i]))
{
count += 2 ;
// Remove an occurrence of x
// and an occurrence of 2 * x
m.put(A[i], m.get(A[i]) - 1 );
m.put( 2 * A[i], m.get( 2 * A[i]) - 1 );
}
}
if (count == n)
return "true" ;
else
return "false" ;
}
// Driver code
public static void main(String []args)
{
int A[] = { 4 , - 2 , 2 , - 4 };
int n = A.length;
// Function call to print required answer
System.out.println(canReorder(A,n));
}
} // This code is contributed by Rituraj Jain |
# Python implementation of the approach import collections
# Function to return true if the elements # can be arranged in the desired order def canReorder(A):
count = collections.Counter(A)
for x in sorted (A, key = abs ):
if count[x] = = 0 :
continue
# If 2 * x is not found to pair
if count[ 2 * x] = = 0 :
return False
# Remove an occurrence of x
# and an occurrence of 2 * x
count[x] - = 1
count[ 2 * x] - = 1
return True
# Driver Code A = [ 4 , - 2 , 2 , - 4 ]
# Function call to print required answer print (canReorder(A))
|
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG
{ // Function to return true if the elements // can be arranged in the desired order static String canReorder( int []A, int n)
{ Dictionary< int ,
int > m = new Dictionary< int ,
int >();
for ( int i = 0; i < n; i++)
{
if (m.ContainsKey(A[i]))
m[A[i]]= m[A[i]] + 1;
else
m.Add(A[i], 1);
}
Array.Sort(A);
int count = 0;
for ( int i = 0; i < n; i++)
{
if (m[A[i]] == 0)
continue ;
// If 2 * x is not found to pair
if (m.ContainsKey(2 * A[i]))
{
count += 2;
// Remove an occurrence of x
// and an occurrence of 2 * x
m[A[i]]= m[A[i]] - 1;
if (m.ContainsKey(2 * A[i]))
m[2 * A[i]]= m[2 * A[i]] - 1;
else
m.Add(2 * A[i], m[2 * A[i]] - 1);
}
}
if (count == n)
return "True" ;
else
return "False" ;
} // Driver code public static void Main(String []args)
{ int []A = {4, -2, 2, -4};
int n = A.Length;
// Function call to print required answer
Console.WriteLine(canReorder(A,n));
} } // This code is contributed by Rajput-Ji |
<script> // JavaScript implementation of the approach // Function to return true if the elements // can be arranged in the desired order function canReorder(A, n)
{ let m = new Map();
for (let i=0;i<n;i++){
if (m.has(A[i])){
m.set(A[i], m.get(A[i]) + 1)
} else {
m.set(A[i], 1)
}
}
A.sort((a, b) => a - b);
let count = 0;
for (let i=0;i<n;i++)
{
if (m.get(A[i]) == 0)
continue ;
// If 2 * x is not found to pair
if (m.get(2 * A[i])){
count+=2;
// Remove an occurrence of x
// and an occurrence of 2 * x
m.set(A[i], m.get(A[i]) - 1);
m.set(2 * A[i], m.get(2 * A[i])- 1);
}
}
if (count ==n)
return "True" ;
else
return "False" ;
} // Driver Code let A = [4, -2, 2, -4]; let n= A.length; // Function call to print required answer document.write((canReorder(A,n))); // This code is contributed by _saurabh_jaiswal </script> |
true
Complexity Analysis:
- Time Complexity: O(n * log n)
- Auxiliary Space: O(n)