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Check if elements of an array can be arranged satisfying the given condition

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  • Difficulty Level : Easy
  • Last Updated : 12 Sep, 2022
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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 = 4

Input: 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++




// 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




// 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




# 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#




// 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

Javascript




<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>

Output

true

Complexity Analysis:

  • Time Complexity: O(n * log n)
  • Auxiliary Space: O(n)

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