Find elements of original array from doubled array

• Given an array arr[] of 2*N integers such that it consists of all elements along with the twice values of another array, say A[], the task is to find the array A[].

Examples:

Input: arr[] = {4, 1, 18, 2, 9, 8}
Output: 1 4 9
Explanation:
After taking double values of 1, 4, and 9 then adding them to the original array, All elements of the given array arr[] are obtained 

Input: arr[] = {4, 1, 2, 2, 8, 2, 4, 4}
Output: 1 2 2 4

Approach: The given problem can be solved by counting the frequency of array elements in the HashMap array elements and observation can be made that, the smallest element in the array will always be a part of the original array, therefore it can be included in the result list res[]. The element with a double value of the smallest element will be the duplicate element that is not part of the original array so its frequency can be reduced from the map. Below steps can be followed to solve the problem: 

• Sort the given array arr[] in ascending order
• Iterate through the array elements and store the numbers and their frequencies in a hashmap
• Create a result list res[] to store the elements present in the original list
• Add the first element in the result list and reduce the frequency of the element which has a double value of the first element.
• Traverse the array and check for the frequency of every element in the map:
  • If the frequency is greater than 0, then add the element in the result list and decrement the frequency.
  • Otherwise, skip the element and move ahead because it is a double value and not a part of the original array.
• After completing the above steps, print the elements in the list res[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the original array
// from the doubled array
vector<int> findOriginal(vector<int>& arr)
{

    // Stores the numbers and
    // their frequency
    map<int, int> numFreq;

    // Add number with their frequencies
    // in the hashmap
    for (int i = 0; i < arr.size(); i++) {
        numFreq[arr[i]]++;
    }

    // Sort the array
    sort(arr.begin(), arr.end());

    // Initialize an arraylist
    vector<int> res;

    for (int i = 0; i < arr.size(); i++) {

        // Get the frequency of the number
        int freq = numFreq[arr[i]];
        if (freq > 0) {

            // Element is of original array
            res.push_back(arr[i]);

            // Decrement the frequency of
            // the number
            numFreq[arr[i]]--;

            int twice = 2 * arr[i];

            // Decrement the frequency of
            // the number having double value
            numFreq[twice]--;
        }
    }

    // Return the resultant string
    return res;
}

// Driver Code
int main()
{

    vector<int> arr = { 4, 1, 2, 2, 2, 4, 8, 4 };
    vector<int> res = findOriginal(arr);

    // Print the result list
    for (int i = 0; i < res.size(); i++) {
        cout << res[i] << " ";
    }

    return 0;
}

    // This code is contributed by rakeshsahni

Java

// Java program for the above approach

import java.io.*;
import java.util.*;
class GFG {

    // Function to find the original array
    // from the doubled array
    public static List<Integer>
    findOriginal(int[] arr)
    {

        // Stores the numbers and
        // their frequency
        Map<Integer, Integer> numFreq
            = new HashMap<>();

        // Add number with their frequencies
        // in the hashmap
        for (int i = 0; i < arr.length; i++) {
            numFreq.put(
                arr[i],
                numFreq.getOrDefault(arr[i], 0)
                    + 1);
        }

        // Sort the array
        Arrays.sort(arr);

        // Initialize an arraylist
        List<Integer> res = new ArrayList<>();

        for (int i = 0; i < arr.length; i++) {

            // Get the frequency of the number
            int freq = numFreq.get(arr[i]);
            if (freq > 0) {

                // Element is of original array
                res.add(arr[i]);

                // Decrement the frequency of
                // the number
                numFreq.put(arr[i], freq - 1);

                int twice = 2 * arr[i];

                // Decrement the frequency of
                // the number having double value
                numFreq.put(
                    twice,
                    numFreq.get(twice) - 1);
            }
        }

        // Return the resultant string
        return res;
    }

    // Driver Code
    public static void main(String[] args)
    {

        List<Integer> res = findOriginal(
            new int[] { 4, 1, 2, 2, 2, 4, 8, 4 });

        // Print the result list
        for (int i = 0; i < res.size(); i++) {
            System.out.print(
                res.get(i) + " ");
        }
    }
}

Python3

# Python program for the above approach

# Function to find the original array
# from the doubled array
def findOriginal(arr):

    # Stores the numbers and
    # their frequency
    numFreq = {}

    # Add number with their frequencies
    # in the hashmap
    for i in range(0, len(arr)):
        if (arr[i] in numFreq):
            numFreq[arr[i]] += 1
        else:
            numFreq[arr[i]] = 1

    # Sort the array
    arr.sort()

    # Initialize an arraylist
    res = []

    for i in range(0, len(arr)):

        # Get the frequency of the number
        freq = numFreq[arr[i]]
        if (freq > 0):

            # Element is of original array
            res.append(arr[i])

            # Decrement the frequency of
            # the number
            numFreq[arr[i]] -= 1

            twice = 2 * arr[i]

            # Decrement the frequency of
            # the number having double value
            numFreq[twice] -= 1

    # Return the resultant string
    return res

# Driver Code
arr = [4, 1, 2, 2, 2, 4, 8, 4]
res = findOriginal(arr)

# Print the result list
for i in range(0, len(res)):
    print(res[i], end=" ")


# This code is contributed by _Saurabh_Jaiswal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG {

    // Function to find the original array
    // from the doubled array
    public static List<int> findOriginal(int[] arr)
    {

        // Stores the numbers and
        // their frequency
        Dictionary<int, int> numFreq = new Dictionary<int, int>();

        // Add number with their frequencies
        // in the hashmap
        for (int i = 0; i < arr.Length; i++) {
             if(numFreq.ContainsKey(arr[i])){
                numFreq[arr[i]] = numFreq[arr[i]] + 1;
            }else{
                numFreq.Add(arr[i], 1);
            }
        }

        // Sort the array
        Array.Sort(arr);

        // Initialize an arraylist
        List<int> res = new List<int>();

        for (int i = 0; i < arr.Length; i++) {

            // Get the frequency of the number
            int freq = numFreq[arr[i]];
            if (freq > 0) {

                // Element is of original array
                res.Add(arr[i]);

                // Decrement the frequency of
                // the number
                numFreq[arr[i]] = freq - 1;

                int twice = 2 * arr[i];

                // Decrement the frequency of
                // the number having double value
                numFreq[twice] = numFreq[twice] - 1;
            }
        }

        // Return the resultant string
        return res;
    }

    // Driver Code
    public static void Main()
    {

        List<int> res = findOriginal(new int[] { 4, 1, 2, 2, 2, 4, 8, 4 });

        // Print the result list
        for (int i = 0; i < res.Count; i++) {
            Console.Write(res[i] + " ");
        }
    }
}

// This code is contributed by gfgking.

Javascript

<script>
// Javascript program for the above approach

// Function to find the original array
// from the doubled array
function findOriginal(arr)
{

  // Stores the numbers and
  // their frequency
  let numFreq = new Map();

  // Add number with their frequencies
  // in the hashmap
  for (let i = 0; i < arr.length; i++) {
    if (numFreq.has(arr[i])) {
      numFreq.set(arr[i], numFreq.get(arr[i]) + 1);
    } else {
      numFreq.set(arr[i], 1);
    }
  }

  // Sort the array
  arr.sort((a, b) => a - b);

  // Initialize an arraylist
  let res = [];

  for (let i = 0; i < arr.length; i++) {
    // Get the frequency of the number
    let freq = numFreq.get(arr[i]);
    if (freq > 0) {
      // Element is of original array
      res.push(arr[i]);

      // Decrement the frequency of
      // the number
      numFreq.set(arr[i], numFreq.get(arr[i]) - 1);

      let twice = 2 * arr[i];

      // Decrement the frequency of
      // the number having double value
      numFreq.set(twice, numFreq.get(twice) - 1);
    }
  }

  // Return the resultant string
  return res;
}

// Driver Code

let arr = [4, 1, 2, 2, 2, 4, 8, 4];
let res = findOriginal(arr);

// Print the result list
for (let i = 0; i < res.length; i++) {
  document.write(res[i] + " ");
}

// This code is contributed by gfgking.
</script>

Output:

1 2 2 4

Time Complexity: O(N*log N)
Auxiliary Space: O(N)