Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

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

  • Difficulty Level : Easy
  • Last Updated : 09 Jun, 2021

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: 
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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
 



My Personal Notes
arrow_drop_up
Previous
Count valid pairs in the array satisfying given conditions
Next

Largest Divisor of a Number not divisible by a perfect square
Recommended Articles
Page :
Check if elements of an array can be arranged in a Circle with consecutive difference as 1
11, Oct 18
Check if elements of array can be arranged in AP, GP or HP
24, Mar 20
Check if an array can be Arranged in Left or Right Positioned Array
17, Oct 16
Find the maximum sum (a+b) for a given input integer N satisfying the given condition
14, Apr 20
Maximum length of same indexed subarrays from two given arrays satisfying the given condition
09, Dec 20
Largest sub-set possible for an array satisfying the given condition
09, Jan 19
Length of longest strictly increasing subset with each pair of adjacent elements satisfying the condition 2 * A[i] ≥ A[i + 1]
22, Mar 21
Count of sub-arrays whose elements can be re-arranged to form palindromes
20, Mar 19
Count of pairs satisfying the given condition
22, Jan 20
Smallest value of X satisfying the condition X % A[i] = B[i] for two given arrays
15, Mar 21
Number of K's such that the given array can be divided into two sets satisfying the given conditions
29, Oct 19
Lexicographically smallest permutation of length 2N that can be obtained from an N-length array satisfying given conditions
17, Jun 21
Python Program for Maximum height when coins are arranged in a triangle
12, Nov 16
Count non-adjacent subsets from numbers arranged in Circular fashion
29, Oct 19
Maximize count of subsets into which the given array can be split such that it satisfies the given condition
03, Feb 21
Count possible permutations of given array satisfying the given conditions
06, Oct 20
Maximum possible sum of squares of stack elements satisfying the given properties
09, Oct 20
Count number of coordinates from an array satisfying the given conditions
30, Apr 21
Find range of values for S in given Array with values satisfying [ arr[i] = floor((i*S)/K) ]
01, Sep 21
Count valid pairs in the array satisfying given conditions
11, Dec 18
Check if the array can be sorted only if the elements on given positions can be swapped
07, May 20
Choose X elements from A[] and Y elements from B[] which satisfy the given condition
04, Oct 19
Generate an array of K elements such that sum of elements is N and the condition a[i] < a[i+1] <= 2*a[i] is met | Set 2
09, May 19
Check if given Binary string follows then given condition or not
15, Jul 20
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
Sanjit_Prasad
@Sanjit_Prasad
Vote for difficulty
Current difficulty :
Easy





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!