# Find two numbers whose divisors are given in a random order

Given an array of N numbers which has all divisors of two numbers in any order. The task is to find the two numbers whose divisors are given in the array.

Examples:

Input: a[] = {10, 2, 8, 1, 2, 4, 1, 20, 4, 5}
Output: 20 8
The divisors of 20 and 8 are given in the array.

Input: a[] = {2, 2, 1, 1}
Output: 1 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The following steps are followed to solve the above problem:

• Find the maximum element in the array, which is one of the two numbers.
• Keep a frequency array which counts the array element’s occurrence.
• Decrease the count of all divisors of the maximum element in the frequency array.
• The next max element will be the second number.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the two numbers ` `// when divisors are given in a random order ` `#include ` `using` `namespace` `std; ` ` `  `// Function to get the two numbers ` `void` `getNumbers(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// Mark the frequency of all elements ` `    ``unordered_map<``int``, ``int``> freq; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``freq[a[i]]++; ` ` `  `    ``// Get the first maximum element ` `    ``int` `maxi1 = *max_element(a, a+n); ` ` `  `    ``// Decrease the frequency of all divisors ` `    ``// of the maximum number that are present ` `    ``// int a[] ` `    ``for` `(``int` `i = 1; i * i <= maxi1; i++) { ` ` `  `        ``if` `(maxi1 % i == 0 &&  ` `            ``freq.find(i) != freq.end() &&  ` `            ``freq[i] != 0) { ` `            ``freq[i]--; ` ` `  `            ``if` `(i != (maxi1 / i) &&  ` `               ``freq.find(maxi1 / i) != freq.end() &&  ` `               ``freq[maxi1 / i] != 0) ` `                ``freq[maxi1 / i]--; ` `        ``} ` `    ``} ` ` `  `    ``// The second number is the largest number ` `    ``// present in remaining numbers. ` `    ``int` `maxi2 = -1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(freq[a[i]] != 0) ` `            ``maxi2 = max(maxi2, a[i]); ` `    ``} ` ` `  `    ``cout << maxi1 << ``" "` `<< maxi2; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 10, 2, 8, 1, 2, 4, 1, 20, 4, 5 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` `    ``getNumbers(a, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find the two numbers ` `// when divisors are given in a random order ` ` `  `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to get the two numbers ` `static` `void` `getNumbers(``int` `a[], ``int` `n) ` `{ ` `  `  `    ``// Mark the frequency of all elements ` `    ``HashMap freq = ``new` `HashMap(); ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``if``(freq.containsKey(a[i])){ ` `            ``freq.put(a[i], freq.get(a[i])+``1``); ` `        ``} ` `        ``else``{ ` `            ``freq.put(a[i], ``1``); ` `        ``} ` `  `  `    ``// Get the first maximum element ` `    ``int` `maxi1 = Arrays.stream(a).max().getAsInt(); ` `  `  `    ``// Decrease the frequency of all divisors ` `    ``// of the maximum number that are present ` `    ``// int a[] ` `    ``for` `(``int` `i = ``1``; i * i <= maxi1; i++) { ` `  `  `        ``if` `(maxi1 % i == ``0` `&&  ` `            ``freq.containsKey(i)&&  ` `            ``freq.get(i)!= ``0``) { ` `            ``freq.put(i, freq.get(i)- ``1``); ` `  `  `            ``if` `(i != (maxi1 / i) &&  ` `               ``freq.containsKey(maxi1 / i) &&  ` `               ``freq.get(maxi1 / i)!= ``0``) ` `                ``freq.put(maxi1 / i, freq.get(maxi1 / i) - ``1``); ` `        ``} ` `    ``} ` `  `  `    ``// The second number is the largest number ` `    ``// present in remaining numbers. ` `    ``int` `maxi2 = -``1``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``if` `(freq.get(a[i]) != ``0``) ` `            ``maxi2 = Math.max(maxi2, a[i]); ` `    ``} ` `  `  `    ``System.out.print(maxi1+ ``" "` `+  maxi2); ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `a[] = { ``10``, ``2``, ``8``, ``1``, ``2``, ``4``, ``1``, ``20``, ``4``, ``5` `}; ` `    ``int` `n = a.length; ` `    ``getNumbers(a, n); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python3

 `# Python3 program to find the two numbers  ` `# when divisors are given in a random order  ` ` `  `# from math lib import sqrt method ` `from` `math ``import` `sqrt ` ` `  `# Function to get the two numbers  ` `def` `getNumbers(a, n) :  ` ` `  `    ``# Mark the frequency of all elements  ` `    ``freq ``=` `{};  ` `    ``for` `i ``in` `range``(n) : ` `        ``if` `a[i] ``not` `in` `freq.keys() : ` `            ``freq[a[i]] ``=` `0` `             `  `        ``freq[a[i]] ``+``=` `1` ` `  `    ``# Get the first maximum element  ` `    ``maxi1 ``=` `max``(a) ` ` `  `    ``# Decrease the frequency of all divisors  ` `    ``# of the maximum number that are present  ` `    ``# int a[]  ` `    ``for` `i ``in` `range``(``1``, ``int``(sqrt(maxi1)) ``+` `1``) : ` ` `  `        ``if` `(maxi1 ``%` `i ``=``=` `0` `and` `            ``freq[i] ``in` `freq.keys() ``and` `            ``freq[i] !``=` `0``) :  ` `            ``freq[i] ``-``=` `1` ` `  `            ``if` `(i !``=` `(maxi1 ``/``/` `i) ``and` `                ``freq[maxi1 ``/``/` `i] ``in` `freq.keys() ``and` `                ``freq[maxi1 ``/``/` `i] !``=` `0``) :  ` `                ``freq[maxi1 ``/``/` `i] ``-``=` `1` `     `  `    ``# The second number is the largest number  ` `    ``# present in remaining numbers.  ` `    ``maxi2 ``=` `-``1` `    ``for` `i ``in` `range``(n) : ` `        ``if` `(freq[a[i]] !``=` `0``) : ` `            ``maxi2 ``=` `max``(maxi2, a[i])  ` ` `  `    ``print``(maxi1, maxi2)  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``a ``=` `[ ``10``, ``2``, ``8``, ``1``, ``2``, ``4``, ``1``, ``20``, ``4``, ``5` `]  ` `    ``n ``=` `len``(a) ` `    ``getNumbers(a, n) ` `     `  `# This code is contributed by Ryuga `

## C#

 `     `  `// C# program to find the two numbers ` `// when divisors are given in a random order ` `  `  `using` `System; ` `using` `System.Collections.Generic; ` `using` `System.Linq; ` `public` `class` `GFG{ ` `   `  `// Function to get the two numbers ` `static` `void` `getNumbers(``int` `[]a, ``int` `n) ` `{ ` `   `  `    ``// Mark the frequency of all elements ` `    ``Dictionary<``int``,``int``> freq = ``new` `Dictionary<``int``,``int``>(); ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if``(freq.ContainsKey(a[i])){ ` `            ``freq[a[i]] = freq[a[i]]+1; ` `        ``} ` `        ``else``{ ` `            ``freq.Add(a[i], 1); ` `        ``} ` `   `  `    ``// Get the first maximum element ` `    ``int` `maxi1 = a.Max(); ` `   `  `    ``// Decrease the frequency of all divisors ` `    ``// of the maximum number that are present ` `    ``// int []a ` `    ``for` `(``int` `i = 1; i * i <= maxi1; i++) { ` `   `  `        ``if` `(maxi1 % i == 0 &&  ` `            ``freq.ContainsKey(i)&&  ` `            ``freq[i]!= 0) { ` `            ``freq[i] = freq[i]- 1; ` `   `  `            ``if` `(i != (maxi1 / i) &&  ` `               ``freq.ContainsKey(maxi1 / i) &&  ` `               ``freq[maxi1 / i]!= 0) ` `                ``freq[maxi1 / i] = freq[maxi1 / i] - 1; ` `        ``} ` `    ``} ` `   `  `    ``// The second number is the largest number ` `    ``// present in remaining numbers. ` `    ``int` `maxi2 = -1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(freq[a[i]] != 0) ` `            ``maxi2 = Math.Max(maxi2, a[i]); ` `    ``} ` `   `  `    ``Console.Write(maxi1+ ``" "` `+  maxi2); ` `} ` `   `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]a = { 10, 2, 8, 1, 2, 4, 1, 20, 4, 5 }; ` `    ``int` `n = a.Length; ` `    ``getNumbers(a, n); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```20 8
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