Construct an array from its pair-product

Given a pair-product array pair[], the task is to find the original array. A pair-product array for an array arr[] is the array that contains product of all the pairs in ordered form i.e. {(arr[0] * arr[1]), (arr[0] * arr[2]), …, (arr[1] * arr[2]), (arr[1] * arr[3]), …, (arr[n – 2] * arr[n – 1])}.
Examples:

Input: pair[] = {2, 3, 6} 
Output: 1 2 3
Input: pair[] = {48, 18, 24, 24, 32, 12} 
Output: 6 8 3 4 

Approach: First find the size of the required array from the given pair-product array. Assuming the size of the original array to be N and size of the pair-product array be X. Therefore, by solving (N * (N – 1)) / 2 = X, the value of N can be calculated as N = (1 + (int)sqrt(1 + 8 * X)) / 2. 
Now lets see the solution with an example, lets say the pair-product array of [A, B, C, D] be arr[AB, AC, AD, BC, BD, CD] then by taking sqrt((arr[0] * arr[1]) / arr[n – 1]) -> sqrt((AB * AC) / BC) will give the value A. 
Once the value of the first element has been recovered then all the remaining elements of the pair-product array can be divided by it to get the remaining elements of the original array.
Below is the implementation of the above approach: 

C++

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

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

// Function to generate the original
// array from the pair-product array
void constructArr(int pair[], int n)
{
    int size = (1 + (int)sqrt(1 + 8 * n)) / 2;
    int arr[size];

    // First element of the resulting array
    arr[0] = sqrt((pair[0] * pair[1]) / pair[size - 1]);

    // Find all the other elements
    for (int i = 1; i < size; i++)
        arr[i] = pair[i - 1] / arr[0];

    // Print the elements of the generated array
    printArr(arr, size);
}

// Driver code
int main()
{
    int pair[] = { 48, 18, 24, 24, 32, 12 };
    int n = sizeof(pair) / sizeof(int);

    constructArr(pair, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

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

// Function to generate the original
// array from the pair-product array
static void constructArr(int pair[], int n)
{
    int size = (1 + (int)Math.sqrt(1 + 8 * n)) / 2;
    int []arr = new int[size];

    // First element of the resulting array
    arr[0] = (int) Math.sqrt((pair[0] * pair[1]) /
                                        pair[size - 1]);

    // Find all the other elements
    for (int i = 1; i < size; i++)
        arr[i] = pair[i - 1] / arr[0];

    // Print the elements of the generated array
    printArr(arr, size);
}

// Driver code
public static void main(String[] args)
{
    int pair[] = { 48, 18, 24, 24, 32, 12 };
    int n = pair.length;

    constructArr(pair, n);
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
from math import sqrt

# Utility function to print the array
def printArr(arr, n) :

    for i in range(n) :
        print(arr[i], end = " ");

# Function to generate the original
# array from the pair-product array
def constructArr(pair, n) :

    size = int((1 + sqrt(1 + 8 * n)) // 2);
    arr = [0] * (size);

    # First element of the resulting array
    arr[0] = int(sqrt((pair[0] * pair[1]) /
                       pair[size - 1]));

    # Find all the other elements
    for i in range(1, size) :
        arr[i] = pair[i - 1] // arr[0];

    # Print the elements of the generated array
    printArr(arr, size);

# Driver code
if __name__ == "__main__" :

    pair = [ 48, 18, 24, 24, 32, 12 ];
    n = len(pair);

    constructArr(pair, n);

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

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

// Function to generate the original
// array from the pair-product array
static void constructArr(int []pair, int n)
{
    int size = (1 + (int)Math.Sqrt(1 + 8 * n)) / 2;
    int []arr = new int[size];

    // First element of the resulting array
    arr[0] = (int) Math.Sqrt((pair[0] * pair[1]) /
                                        pair[size - 1]);

    // Find all the other elements
    for (int i = 1; i < size; i++)
        arr[i] = pair[i - 1] / arr[0];

    // Print the elements of the generated array
    printArr(arr, size);
}

// Driver code
public static void Main(String[] args)
{
    int []pair = { 48, 18, 24, 24, 32, 12 };
    int n = pair.Length;

    constructArr(pair, n);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript implementation of the approach


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

// Function to generate the original
// array from the pair-product array
function constructArr(pair, n) {
    let size = Math.floor((1 + Math.sqrt(1 + 8 * n)) / 2);
    let arr = new Array(size);

    // First element of the resulting array

    arr[0] =
    Math.floor(Math.sqrt((pair[0] * pair[1]) / pair[size - 1]));

    // Find all the other elements
    for (let i = 1; i < size; i++)
        arr[i] = Math.floor(pair[i - 1] / arr[0]);

    // Print the elements of the generated array
    printArr(arr, size);
}

// Driver code

let pair = [48, 18, 24, 24, 32, 12];
let n = pair.length;
constructArr(pair, n);

</script>

Output:

6 8 3 4

Time Complexity: O(N).
Auxiliary Space: O(N).