Construct an array from GCDs of consecutive elements in given array

Given an array a[] of n elements. The task is to find the array (say b[]) of n + 1 such that Greatest Common Divisor of b[i] and b[i + 1] is equals to a[i]. If multiple solutions exist, print the one whose array sum is minimum.

Examples: 

Input : a[] = { 1, 2, 3 }
Output : 1 2 6 3
GCD(1, 2) = 1
GCD(2, 6) = 2
GCD(6, 3) = 3
Also, 1 + 2 + 6 + 3 = 12 which is smallest among all
possible value of array that can be constructed.

Input : a[] = { 5, 10, 5 }
Output : 5 10 10 5

Suppose there is only one number in the given array a[]. Let it be K, and then both numbers in the constructed array (say b[]) will be K and K.
So, the value of the b[0] will be a[0] only. Now consider that, we are done up to index i i.e we have already processed upto index i and calculated b[i + 1].
Now the gcd(b[i + 1], b[i + 2]) = a[i + 1] and gcd(b[i + 2], b[i + 3]) = a[i + 2]. So, b[i + 2] >= lcm(a[i + 1], a[i + 2]). Or, b[i + 2] will be multiple of lcm(a[i + 1], a[i + 2]). As we want the minimum sum, so we want the minimum value of b[i + 2]. So, b[i + 2] = lcm(a[i + 2], a[i + 3]).

Below is the implementation of this approach: 

C++

// CPP Program to construct an array whose GCD of
// every consecutive element is the given array
#include <bits/stdc++.h>
using namespace std;

// Return the LCM of two numbers.
int lcm(int a, int b)
{
    return (a * b) / __gcd(a, b);
}

// Print the required constructed array
void printArray(int a[], int n)
{
    // printing the first element.
    cout << a[0] << " ";

    // finding and printing the LCM of consecutive
    // element of given array.
    for (int i = 0; i < n - 1; i++)
        cout << lcm(a[i], a[i + 1]) << " ";

    // printing the last element of the given array.
    cout << a[n - 1] << endl;
}

// Driven Program
int main()
{
    int a[] = { 1, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    printArray(a, n);
    return 0;
}

Java

// Java Program to construct an array whose
// GCD of every consecutive element is the
// given array

import java.io.*;

class GFG {

    // Recursive function to return gcd of
    // a and b
    static int __gcd(int a, int b)
    {

        // Everything divides 0
        if (a == 0 || b == 0)
        return 0;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return __gcd(a - b, b);

        return __gcd(a, b - a);
    }

    // Return the LCM of two numbers.
    static int lcm(int a, int b)
    {
        return (a * b) / __gcd(a, b);
    }

    // Print the required constructed array
    static void printArray(int a[], int n)
    {

        // printing the first element.
        System.out.print( a[0] + " ");

        // finding and printing the LCM of
        // consecutive element of given array.
        for (int i = 0; i < n - 1; i++)
            System.out.print(lcm(a[i],
                            a[i + 1]) + " ");

        // printing the last element of the
        // given array.
        System.out.print(a[n - 1]);
    }

    // Driven Program
    public static void main (String[] args)
    {
        int a[] = { 1, 2, 3 };
        int n = a.length;
        printArray(a, n);
    }
}

// This code is contributed by anuj_67.

Python3

# Python Program to construct an array whose
# GCD of every consecutive element is the
# given array

# Recursive function to return gcd of
# a and b
def __gcd( a, b):

    # Everything divides 0
    if (a == 0 or b == 0):
        return 0

    # base case
    if (a == b):
        return a

    # a is greater
    if (a > b):
        return __gcd(a - b, b)
    return __gcd(a, b - a)

# Return the LCM of two numbers.
def lcm(a, b):
    return (a * b) / __gcd(a, b)

# Print the required constructed array
def printArray(a, n):

    # printing the first element.
    print ( str(a[0]) + " ")

    # finding and printing the LCM of
    # consecutive element of given array.
    for i in range(0,n-1):
        print (str(lcm(a[i],a[i + 1])) + " ")

    # printing the last element of the
    # given array.
    print (a[n - 1])

# Driver code
a = [1, 2, 3 ]
n = len(a)
printArray(a, n)

# This code is contributed by Prateek Bajaj

C#

// C# Program to construct an array whose
// GCD of every consecutive element is the
// given array
using System;
class GFG {

    // Recursive function to return
    // gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0 || b == 0)
        return 0;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return __gcd(a - b, b);

        return __gcd(a, b - a);
    }

    // Return the LCM of two numbers.
    static int lcm(int a, int b)
    {
        return (a * b) / __gcd(a, b);
    }

    // Print the required constructed array
    static void printArray(int []a, int n)
    {

        // printing the first element.
        Console.Write( a[0] + " ");

        // finding and printing the LCM of
        // consecutive element of given array.
        for (int i = 0; i < n - 1; i++)
            Console.Write(lcm(a[i],
                  a[i + 1]) + " ");

        // printing the last element
        // of the given array.
        Console.Write(a[n - 1]);
    }

    // Driver Code
    public static void Main ()
    {
        int []a = {1, 2, 3};
        int n = a.Length;
        printArray(a, n);
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP Program to construct
// an array whose GCD of
// every consecutive element
// is the given array

// Recursive function to
// return gcd of a and b
function __gcd($a, $b)
{

    // Everything divides 0
    if($a == 0 or $b == 0)
        return 0 ;

    // base case
    if($a == $b)
        return $a ;

    // a is greater
    if($a > $b)
        return __gcd( $a - $b , $b ) ;

    return __gcd( $a , $b - $a ) ;
}

// Return the LCM of two numbers.
function lcm($a, $b)
{
    return ($a * $b) / __gcd($a, $b);
}

// Print the required constructed array
function printArray( $a, $n)
{

    // printing the first element.
    echo $a[0] , " ";

    // finding and printing
    // the LCM of consecutive
    // element of given array.
    for ( $i = 0; $i < $n - 1; $i++)
        echo lcm($a[$i], $a[$i + 1]) , " ";

    // printing the last element
    // of the given array.
    echo $a[$n - 1] ,"\n";
}

    // Driver Code
    $a = array(1, 2, 3);
    $n = count($a);
    printArray($a, $n);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// Javascript Program to construct an array whose GCD of
// every consecutive element is the given array

function __gcd(a, b)
    {
        // Everything divides 0
        if (a == 0 || b == 0)
        return 0;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return __gcd(a - b, b);

        return __gcd(a, b - a);
    }

// Return the LCM of two numbers.
function lcm(a, b)
{
    return (a * b) / __gcd(a, b);
}

// Print the required constructed array
function printArray(a, n)
{

    // printing the first element.
    document.write( a[0] + " ");

    // finding and printing the LCM of consecutive
    // element of given array.
    for (var i = 0; i < n - 1; i++)
        document.write( lcm(a[i], a[i + 1]) + " ");

    // printing the last element of the given array.
    document.write( a[n - 1] + "<br>");
}

// Driven Program
var a = [ 1, 2, 3 ];
var n = a.length;
printArray(a, n);

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

Output

1 2 6 3

Complexity Analysis:

• Time complexity: O(n * log(max(a, b)), where n represents the size of the given array.
• Auxiliary Space: O(1)