Find elements of original array from doubled array
Last Updated :
22 Dec, 2021
- 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 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) + " ");
}
}
}
# 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# 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.
<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>
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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