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Find A and B from list of divisors

  • Difficulty Level : Hard
  • Last Updated : 29 Apr, 2021

Given an array arr[] which consists of all the divisors of two integers A and B (along with A, B, and 1). The task is to find A and B from the given array. 
Note: If a number is a divisor of both A and B then it will be present twice in the array.

Examples: 

Input: arr[] = {1, 2, 4, 8, 16, 1, 2, 3, 6} 
Output: A = 16, B = 6 
1, 2, 4, 8 and 16 are the divisors of 16 
1, 2, 3 and 6 are the divisors of 6

Input: arr[] = {1, 2, 4, 8, 16, 1, 2, 4} 
Output: A = 16, B = 4 

Approach: From the given array, as all the elements are divisors of either A or B then it is compulsory that the largest element of the array will be either A or B. For simplicity take it as A. Now, there are two cases which is applicable for an element to be B:  

  1. B can be the largest element which is not a divisor of A.
  2. B can be the largest element smaller than A whose frequency is 2.

In order to find the value of A and B, sort the array. Print largest element as A and then the largest element which is either not a divisor of A or having frequency 2 will be B.
Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print A and B all of whose
// divisors are present in the given array
void printNumbers(int arr[], int n)
{
 
    // Sort the array
    sort(arr, arr + n);
 
    // A is the largest element from the array
    int A = arr[n - 1], B = -1;
 
    // Iterate from the second largest element
    for (int i = n - 2; i >= 0; i--) {
 
        // If current element is not a divisor
        // of A then it must be B
        if (A % arr[i] != 0) {
            B = arr[i];
            break;
        }
 
        // If current element occurs more than once
        if (i - 1 >= 0 && arr[i] == arr[i - 1]) {
            B = arr[i];
            break;
        }
    }
 
    // Print A and B
    cout << "A = " << A << ", B = " << B;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 4, 8, 16, 1, 2, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printNumbers(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
class GfG {
 
    // Function to print A and B all of whose
    // divisors are present in the given array
    static void printNumbers(int arr[], int n)
    {
 
        // Sort the array
        Arrays.sort(arr);
 
        // A is the largest element from the array
        int A = arr[n - 1], B = -1;
 
        // Iterate from the second largest element
        for (int i = n - 2; i >= 0; i--) {
 
            // If current element is not a divisor
            // of A then it must be B
            if (A % arr[i] != 0) {
                B = arr[i];
                break;
            }
 
            // If current element occurs more than once
            if (i - 1 >= 0 && arr[i] == arr[i - 1]) {
                B = arr[i];
                break;
            }
        }
 
        // Print A and B
        System.out.print("A = " + A + ", B = " + B);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 4, 8, 16, 1, 2, 4 };
        int n = arr.length;
        printNumbers(arr, n);
    }
}

Python3




# Python3 implementation of the approach
 
# Function to print A and B all of whose
# divisors are present in the given array
def printNumbers(arr, n):
 
    # Sort the array
    arr.sort()
 
    # A is the largest element from the array
    A, B = arr[n - 1], -1
 
    # Iterate from the second largest element
    for i in range(n - 2, -1, -1):
 
        # If current element is not a divisor
        # of A then it must be B
        if A % arr[i] != 0:
            B = arr[i]
            break
 
        # If current element occurs more than once
        if i - 1 >= 0 and arr[i] == arr[i - 1]:
            B = arr[i]
            break
 
    # Print A and B
    print("A =", A, ", B =", B)
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 4, 8, 16, 1, 2, 4]
    n = len(arr)
    printNumbers(arr, n)
 
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System.Collections;
using System;
 
class GfG
{
 
    // Function to print A and B all of whose
    // divisors are present in the given array
    static void printNumbers(int []arr, int n)
    {
 
        // Sort the array
        Array.Sort(arr);
 
        // A is the largest element from the array
        int A = arr[n - 1], B = -1;
 
        // Iterate from the second largest element
        for (int i = n - 2; i >= 0; i--)
        {
 
            // If current element is not a divisor
            // of A then it must be B
            if (A % arr[i] != 0)
            {
                B = arr[i];
                break;
            }
 
            // If current element occurs more than once
            if (i - 1 >= 0 && arr[i] == arr[i - 1])
            {
                B = arr[i];
                break;
            }
        }
 
        // Print A and B
        Console.WriteLine("A = " + A + ", B = " + B);
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = { 1, 2, 4, 8, 16, 1, 2, 4 };
        int n = arr.Length;
        printNumbers(arr, n);
    }
}
 
// This code is contributed by mits

PHP




<?php
// PHP implementation of the approach
 
// Function to print A and B all of whose
// divisors are present in the given array
function printNumbers($arr, $n)
{
 
    // Sort the array
    sort($arr);
 
    // A is the largest element from the array
    $A = $arr[$n - 1]; $B = -1;
 
    // Iterate from the second largest element
    for ($i = $n - 2; $i >= 0; $i--)
    {
 
        // If current element is not a divisor
        // of A then it must be B
        if ($A % $arr[$i] != 0)
        {
            $B = $arr[$i];
            break;
        }
 
        // If current element occurs more than once
        if ($i - 1 >= 0 && $arr[$i] == $arr[$i - 1])
        {
            $B = $arr[$i];
            break;
        }
    }
 
    // Print A and B
    echo("A = " . $A . ", B = " . $B);
}
 
// Driver code
$arr = array( 1, 2, 4, 8, 16, 1, 2, 4 );
$n = sizeof($arr);
printNumbers($arr, $n);
 
// This code is contributed by Code_Mech.

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to print A and B all of whose
// divisors are present in the given array
function printNumbers(arr, n)
{
 
    // Sort the array
    arr.sort((a,b)=>{return a-b;});
 
    // A is the largest element from the array
    var A = arr[n - 1], B = -1;
 
    // Iterate from the second largest element
    for (var i = n - 2; i >= 0; i--) {
 
        // If current element is not a divisor
        // of A then it must be B
        if ((A % arr[i]) != 0) {
            B = arr[i];
            break;
        }
 
        // If current element occurs more than once
        if (i - 1 >= 0 && arr[i] == arr[i - 1]) {
            B = arr[i];
            break;
        }
    }
 
    // Print A and B
    document.write( "A = " + A + ", B = " + B);
}
 
// Driver code
var arr = [ 1, 2, 4, 8, 16, 1, 2, 4 ];
var n = arr.length;
printNumbers(arr, n);
 
</script>
Output: 
A = 16, B = 4

 


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