Find original numbers from gcd() every pair

Given an array arr[] containing GCD of every possible pair of elements of another array. The task is to find the original numbers which are used to calculate the GCD array. For example, if original numbers are {4, 6, 8} then the given array will be {4, 2, 4, 2, 6, 2, 4, 2, 8}. 

Examples:

Input: arr[] = {5, 1, 1, 12} 
Output: 12 5 
gcd(12, 12) = 12 
gcd(12, 5) = 1 
gcd(5, 12) = 1 
gcd(5, 5) = 5

Input: arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2} 
Output: 12 10 7 5 1 

Approach:

1. Sort the array in decreasing order.
2. Highest element will always be one of the original numbers. Keep that number and remove it from the array.
3. Compute GCD of the element taken in the previous step with the current element starting from the greatest and discard the GCD value from the given array.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to print
// the contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Function to find the required numbers
void findNumbers(int arr[], int n)
{

    // Sort array in decreasing order
    sort(arr, arr + n, greater<int>());

    int freq[arr[0] + 1] = { 0 };

    // Count frequency of each element
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;

    // Size of the resultant array
    int size = sqrt(n);
    int brr[size] = { 0 }, x, l = 0;

    for (int i = 0; i < n; i++) {
        if (freq[arr[i]] > 0) {

            // Store the highest element in
            // the resultant array
            brr[l] = arr[i];

            // Decrement the frequency of that element
            freq[brr[l]]--;
            l++;
            for (int j = 0; j < l; j++) {
                if (i != j) {

                    // Compute GCD
                    x = __gcd(arr[i], brr[j]);

                    // Decrement GCD value by 2
                    freq[x] -= 2;
                }
            }
        }
    }

    printArr(brr, size);
}

// Driver code
int main()
{
    int arr[] = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                  1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findNumbers(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;

class GFG
{

    // Utility function to print
    // the contents of an array
    static void printArr(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
        {
            System.out.print(arr[i] + " ");
        }
    }

    // Function to find the required numbers
    static void findNumbers(int arr[], int n)
    {

        // Sort array in decreasing order
        Arrays.sort(arr);
        reverse(arr);
        int freq[] = new int[arr[0] + 1];

        // Count frequency of each element
        for (int i = 0; i < n; i++)
        {
            freq[arr[i]]++;
        }

        // Size of the resultant array
        int size = (int) Math.sqrt(n);
        int brr[] = new int[size], x, l = 0;

        for (int i = 0; i < n; i++)
        {
            if (freq[arr[i]] > 0 && l < size)
            {

                // Store the highest element in
                // the resultant array
                brr[l] = arr[i];

                // Decrement the frequency of that element
                freq[brr[l]]--;
                l++;
                for (int j = 0; j < l; j++)
                {
                    if (i != j)
                    {

                        // Compute GCD
                        x = __gcd(arr[i], brr[j]);

                        // Decrement GCD value by 2
                        freq[x] -= 2;
                    }
                }
            }
        }

        printArr(brr, size);
    }

    // reverse array
    public static void reverse(int[] input)
    {
        int last = input.length - 1;
        int middle = input.length / 2;
        for (int i = 0; i <= middle; i++)
        {
            int temp = input[i];
            input[i] = input[last - i];
            input[last - i] = temp;
        }
    }

    static int __gcd(int a, int b)
    {
        if (b == 0)
        {
            return a;
        }
        return __gcd(b, a % b);

    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
            1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2};
        int n = arr.length;
        findNumbers(arr, n);
    }
}

// This code has been contributed by 29AjayKumar

Python3

# Python 3 implementation of the approach
from math import sqrt, gcd

# Utility function to print
# the contents of an array
def printArr(arr, n):
    for i in range(n):
        print(arr[i], end = " ")

# Function to find the required numbers
def findNumbers(arr, n):

    # Sort array in decreasing order
    arr.sort(reverse = True)

    freq = [0 for i in range(arr[0] + 1)]

    # Count frequency of each element
    for i in range(n):
        freq[arr[i]] += 1

    # Size of the resultant array
    size = int(sqrt(n))
    brr = [0 for i in range(len(arr))]
    l = 0

    for i in range(n):
        if (freq[arr[i]] > 0):

            # Store the highest element in
            # the resultant array
            brr[l] = arr[i]

            # Decrement the frequency of that element
            freq[brr[l]] -= 1
            l += 1
            for j in range(l):
                if (i != j):

                    # Compute GCD
                    x = gcd(arr[i], brr[j])

                    # Decrement GCD value by 2
                    freq[x] -= 2

    printArr(brr, size)

# Driver code
if __name__ == '__main__':
    arr = [1, 1, 1, 1, 1, 1, 1, 1,
           1, 1, 1, 1, 1, 1, 1, 1,
           1, 5, 5, 5, 7, 10, 12, 2, 2]
    n = len(arr)
    findNumbers(arr, n)

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation for above approach
using System;

class GFG
{

    // Utility function to print
    // the contents of an array
    static void printArr(int []arr, int n)
    {
        for (int i = 0; i < n; i++)
        {
            Console.Write(arr[i] + " ");
        }
    }

    // Function to find the required numbers
    static void findNumbers(int []arr, int n)
    {

        // Sort array in decreasing order
        Array.Sort(arr);
        reverse(arr);
        int []freq = new int[arr[0] + 1];

        // Count frequency of each element
        for (int i = 0; i < n; i++)
        {
            freq[arr[i]]++;
        }

        // Size of the resultant array
        int size = (int) Math.Sqrt(n);
        int []brr = new int[size];int x, l = 0;

        for (int i = 0; i < n; i++)
        {
            if (freq[arr[i]] > 0 && l < size)
            {

                // Store the highest element in
                // the resultant array
                brr[l] = arr[i];

                // Decrement the frequency of that element
                freq[brr[l]]--;
                l++;
                for (int j = 0; j < l; j++)
                {
                    if (i != j)
                    {

                        // Compute GCD
                        x = __gcd(arr[i], brr[j]);

                        // Decrement GCD value by 2
                        freq[x] -= 2;
                    }
                }
            }
        }

        printArr(brr, size);
    }

    // reverse array
    public static void reverse(int []input)
    {
        int last = input.Length - 1;
        int middle = input.Length / 2;
        for (int i = 0; i <= middle; i++)
        {
            int temp = input[i];
            input[i] = input[last - i];
            input[last - i] = temp;
        }
    }

    static int __gcd(int a, int b)
    {
        if (b == 0)
        {
            return a;
        }
        return __gcd(b, a % b);

    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
            1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2};
        int n = arr.Length;
        findNumbers(arr, n);
    }
}

/* This code contributed by PrinciRaj1992 */

PHP

<?php
// PHP implementation of the approach
function gcd($a, $b)
{
    return ($a % $b) ? gcd($b, $a % $b) : $b;
}


// Utility function to print
// the contents of an array
function printArr($arr, $n)
{
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i], " ";
}

// Function to find the required numbers
function findNumbers($arr, $n)
{

    // Sort array in decreasing order
    rsort($arr);

    $freq = array_fill(0, $arr[0] + 1, 0);

    // Count frequency of each element
    for ($i = 0; $i < $n; $i++)
        $freq[$arr[$i]]++;

    // Size of the resultant array
    $size = floor(sqrt($n));
    $brr = array_fill(0, $size, 0);
    $l = 0;

    for ($i = 0; $i < $n; $i++)
    {
        if ($freq[$arr[$i]] > 0)
        {

            // Store the highest element in
            // the resultant array
            $brr[$l] = $arr[$i];

            // Decrement the frequency of that element
            $freq[$brr[$l]]--;
            $l++;
            for ($j = 0; $j < $l; $j++)
            {
                if ($i != $j)
                {

                    // Compute GCD
                    $x = gcd($arr[$i], $brr[$j]);

                    // Decrement GCD value by 2
                    $freq[$x] -= 2;
                }
            }
        }
    }

    printArr($brr, $size);
}

// Driver code
$arr = array(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
             1, 1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2 );
$n = count($arr) ;
findNumbers($arr, $n);

// This code is contributed by Ryuga
?>

Javascript

<script>

// Javascript implementation for above approach

// Utility function to print
// the contents of an array
function printArr(arr, n)
{
    for(let i = 0; i < n; i++)
    {
        document.write(arr[i] + " ");
    }
}

// Function to find the required numbers
function findNumbers(arr, n)
{

    // Sort array in decreasing order
    arr.sort(function(a, b){return a - b});
    reverse(arr);
    let freq = new Array(arr[0] + 1);
    freq.fill(0);

    // Count frequency of each element
    for(let i = 0; i < n; i++)
    {
        freq[arr[i]]++;
    }

    // Size of the resultant array
    let size = parseInt(Math.sqrt(n), 10);
    let brr = new Array(size);
    brr.fill(0);
    let x, l = 0;

    for(let i = 0; i < n; i++)
    {
        if (freq[arr[i]] > 0 && l < size)
        {

            // Store the highest element in
            // the resultant array
            brr[l] = arr[i];

            // Decrement the frequency of that element
            freq[brr[l]]--;
            l++;

            for(let j = 0; j < l; j++)
            {
                if (i != j)
                {

                    // Compute GCD
                    x = __gcd(arr[i], brr[j]);

                    // Decrement GCD value by 2
                    freq[x] -= 2;
                }
            }
        }
    }

    printArr(brr, size);
}

// Reverse array
function reverse(input)
{
    let last = input.length - 1;
    let middle = parseInt(input.length / 2, 10);
    for(let i = 0; i <= middle; i++)
    {
        let temp = input[i];
        input[i] = input[last - i];
        input[last - i] = temp;
    }
}

function __gcd(a, b)
{
    if (b == 0)
    {
        return a;
    }
    return __gcd(b, a % b);
}

// Driver code
let arr = [ 1, 1, 1, 1, 1, 1, 1, 1,
            1, 1, 1, 1, 1, 1, 1, 1,
            1, 5, 5, 5, 7, 10, 12, 2, 2];
let n = arr.length;
findNumbers(arr, n);

// This code is contributed by divyeshrabadiya07

</script>

Output:

12 10 7 5 1

Time Complexity: O(n²)
Auxiliary Space: O(?n+k) where n is the size of array and k is the maximum element of the array.