Check if elements of an array can be arranged satisfying the given condition
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++
// 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> |
true
Complexity Analysis:
- Time Complexity: O(n * log n)
- Auxiliary Space: O(n)
Please Login to comment...