# Smallest perfect Cube divisible by all elements of an array

Given an array arr[], the task is to find the smallest perfect cube which is divisible by all the elements of the given array.

Examples:

Input: arr[] = {20, 4, 128, 7}
Output: 21952000

Input: arr[] = {10, 125, 14, 42, 100}
Output: 9261000

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

Naive Approach: Check all perfect cubes one by one starting from 1 and select the one which is divisible by all the elements of the array.

Efficient Approach: Find the least common multiple of all the elements of the array and store it in a variable lcm. Find all prime factor of the found LCM.
Now for every prime factor fact which divides the lcm ‘x’ number of times where x % 3 != 0:

• If x % 3 = 2 then update lcm = lcm * fact.
• If x % 3 = 1 then update lcm = lcm * fact2.

Print the updated LCM in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define ll long long int ` ` `  `// Function to return the gcd of two numbers ` `ll gcd(ll a, ll b) ` `{ ` `    ``if` `(b == 0) ` `        ``return` `a; ` `    ``else` `        ``return` `gcd(b, a % b); ` `} ` ` `  `// Function to return the lcm of ` `// all the elements of the array ` `ll lcmOfArray(``int` `arr[], ``int` `n) ` `{ ` `    ``if` `(n < 1) ` `        ``return` `0; ` ` `  `    ``ll lcm = arr; ` ` `  `    ``// To calculate lcm of two numbers ` `    ``// multiply them and divide the result ` `    ``// by gcd of both the numbers ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``lcm = (lcm * arr[i]) / gcd(lcm, arr[i]); ` ` `  `    ``// Return the LCM of the array elements ` `    ``return` `lcm; ` `} ` ` `  `// Function to return the smallest perfect cube ` `// divisible by all the elements of arr[] ` `int` `minPerfectCube(``int` `arr[], ``int` `n) ` `{ ` `    ``ll minPerfectCube; ` ` `  `    ``// LCM of all the elements of arr[] ` `    ``ll lcm = lcmOfArray(arr, n); ` `    ``minPerfectCube = (``long` `long``)lcm; ` ` `  `    ``int` `cnt = 0; ` `    ``while` `(lcm > 1 && lcm % 2 == 0) { ` `        ``cnt++; ` `        ``lcm /= 2; ` `    ``} ` ` `  `    ``// If 2 divides lcm cnt number of times ` `    ``if` `(cnt % 3 == 2) ` `        ``minPerfectCube *= 2; ` `    ``else` `if` `(cnt % 3 == 1) ` `        ``minPerfectCube *= 4; ` ` `  `    ``int` `i = 3; ` ` `  `    ``// Check all the numbers that divide lcm ` `    ``while` `(lcm > 1) { ` `        ``cnt = 0; ` `        ``while` `(lcm % i == 0) { ` `            ``cnt++; ` `            ``lcm /= i; ` `        ``} ` ` `  `        ``if` `(cnt % 3 == 1) ` `            ``minPerfectCube *= i * i; ` `        ``else` `if` `(cnt % 3 == 2) ` `            ``minPerfectCube *= i; ` ` `  `        ``i += 2; ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `minPerfectCube; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 125, 14, 42, 100 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << minPerfectCube(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the gcd of two numbers ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == ``0``) ` `        ``return` `a; ` `    ``else` `        ``return` `gcd(b, a % b); ` `} ` ` `  `// Function to return the lcm of ` `// aint the elements of the array ` `static` `int` `lcmOfArray(``int` `arr[], ``int` `n) ` `{ ` `    ``if` `(n < ``1``) ` `        ``return` `0``; ` ` `  `    ``int` `lcm = arr[``0``]; ` ` `  `    ``// To calculate lcm of two numbers ` `    ``// multiply them and divide the result ` `    ``// by gcd of both the numbers ` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `        ``lcm = (lcm * arr[i]) / gcd(lcm, arr[i]); ` ` `  `    ``// Return the LCM of the array elements ` `    ``return` `lcm; ` `} ` ` `  `// Function to return the smaintest perfect cube ` `// divisible by aint the elements of arr[] ` `static` `int` `minPerfectCube(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `minPerfectCube; ` ` `  `    ``// LCM of all the elements of arr[] ` `    ``int` `lcm = lcmOfArray(arr, n); ` `    ``minPerfectCube = lcm; ` ` `  `    ``int` `cnt = ``0``; ` `    ``while` `(lcm > ``1` `&& lcm % ``2` `== ``0``) ` `    ``{ ` `        ``cnt++; ` `        ``lcm /= ``2``; ` `    ``} ` ` `  `    ``// If 2 divides lcm cnt number of times ` `    ``if` `(cnt % ``3` `== ``2``) ` `        ``minPerfectCube *= ``2``; ` `    ``else` `if` `(cnt % ``3` `== ``1``) ` `        ``minPerfectCube *= ``4``; ` ` `  `    ``int` `i = ``3``; ` ` `  `    ``// Check aint the numbers that divide lcm ` `    ``while` `(lcm > ``1``)  ` `    ``{ ` `        ``cnt = ``0``; ` `        ``while` `(lcm % i == ``0``) ` `        ``{ ` `            ``cnt++; ` `            ``lcm /= i; ` `        ``} ` ` `  `        ``if` `(cnt % ``3` `== ``1``) ` `            ``minPerfectCube *= i * i; ` `        ``else` `if` `(cnt % ``3` `== ``2``) ` `            ``minPerfectCube *= i; ` ` `  `        ``i += ``2``; ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `minPerfectCube; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = { ``10``, ``125``, ``14``, ``42``, ``100` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(minPerfectCube(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Surendra_Gangwar `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the gcd of two numbers  ` `def` `gcd(a, b) : ` `     `  `    ``if` `(b ``=``=` `0``) : ` `        ``return` `a ` `    ``else` `: ` `        ``return` `gcd(b, a ``%` `b) ` ` `  `# Function to return the lcm of  ` `# all the elements of the array  ` `def` `lcmOfArray(arr, n) : ` `     `  `    ``if` `(n < ``1``) : ` `        ``return` `0` ` `  `    ``lcm ``=` `arr[``0``] ` ` `  `    ``# To calculate lcm of two numbers  ` `    ``# multiply them and divide the result  ` `    ``# by gcd of both the numbers  ` `    ``for` `i ``in` `range``(n) :  ` `        ``lcm ``=` `(lcm ``*` `arr[i]) ``/``/` `gcd(lcm, arr[i]);  ` ` `  `    ``# Return the LCM of the array elements  ` `    ``return` `lcm ` ` `  `# Function to return the smallest perfect cube  ` `# divisible by all the elements of arr[]  ` `def` `minPerfectCube(arr, n) : ` `     `  `    ``# LCM of all the elements of arr[]  ` `    ``lcm ``=` `lcmOfArray(arr, n) ` `    ``minPerfectCube ``=` `lcm ` ` `  `    ``cnt ``=` `0` `    ``while` `(lcm > ``1` `and` `lcm ``%` `2` `=``=` `0``) : ` `        ``cnt ``+``=` `1` `        ``lcm ``/``/``=` `2` `     `  `    ``# If 2 divides lcm cnt number of times  ` `    ``if` `(cnt ``%` `3` `=``=` `2``) : ` `        ``minPerfectCube ``*``=` `2` `         `  `    ``elif` `(cnt ``%` `3` `=``=` `1``) : ` `        ``minPerfectCube ``*``=` `4` ` `  `    ``i ``=` `3` `     `  `    ``# Check all the numbers that divide lcm  ` `    ``while` `(lcm > ``1``) : ` `        ``cnt ``=` `0` `         `  `        ``while` `(lcm ``%` `i ``=``=` `0``) : ` `            ``cnt ``+``=` `1` `            ``lcm ``/``/``=` `i  ` `         `  `        ``if` `(cnt ``%` `3` `=``=` `1``) : ` `            ``minPerfectCube ``*``=` `i ``*` `i ` `             `  `        ``elif` `(cnt ``%` `3` `=``=` `2``) : ` `            ``minPerfectCube ``*``=` `i ` ` `  `        ``i ``+``=` `2` ` `  `    ``# Return the answer  ` `    ``return` `minPerfectCube  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``10``, ``125``, ``14``, ``42``, ``100` `]  ` `     `  `    ``n ``=` `len``(arr) ` `    ``print``(minPerfectCube(arr, n))  ` `     `  `# This code is contributed by Ryuga `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the gcd of two numbers ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == 0) ` `        ``return` `a; ` `    ``else` `        ``return` `gcd(b, a % b); ` `} ` ` `  `// Function to return the lcm of ` `// aint the elements of the array ` `static` `int` `lcmOfArray(``int` `[]arr, ``int` `n) ` `{ ` `    ``if` `(n < 1) ` `        ``return` `0; ` ` `  `    ``int` `lcm = arr; ` ` `  `    ``// To calculate lcm of two numbers ` `    ``// multiply them and divide the result ` `    ``// by gcd of both the numbers ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``lcm = (lcm * arr[i]) / gcd(lcm, arr[i]); ` ` `  `    ``// Return the LCM of the array elements ` `    ``return` `lcm; ` `} ` ` `  `// Function to return the smaintest perfect cube ` `// divisible by aint the elements of arr[] ` `static` `int` `minPerfectCube(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `minPerfectCube; ` ` `  `    ``// LCM of all the elements of arr[] ` `    ``int` `lcm = lcmOfArray(arr, n); ` `    ``minPerfectCube = lcm; ` ` `  `    ``int` `cnt = 0; ` `    ``while` `(lcm > 1 && lcm % 2 == 0) ` `    ``{ ` `        ``cnt++; ` `        ``lcm /= 2; ` `    ``} ` ` `  `    ``// If 2 divides lcm cnt number of times ` `    ``if` `(cnt % 3 == 2) ` `        ``minPerfectCube *= 2; ` `    ``else` `if` `(cnt % 3 == 1) ` `        ``minPerfectCube *= 4; ` ` `  `    ``int` `i = 3; ` ` `  `    ``// Check aint the numbers that divide lcm ` `    ``while` `(lcm > 1)  ` `    ``{ ` `        ``cnt = 0; ` `        ``while` `(lcm % i == 0) ` `        ``{ ` `            ``cnt++; ` `            ``lcm /= i; ` `        ``} ` ` `  `        ``if` `(cnt % 3 == 1) ` `            ``minPerfectCube *= i * i; ` `        ``else` `if` `(cnt % 3 == 2) ` `            ``minPerfectCube *= i; ` ` `  `        ``i += 2; ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `minPerfectCube; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 10, 125, 14, 42, 100 }; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(minPerfectCube(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

## PHP

 ` 1 && ``\$lcm` `% 2 == 0) ` `    ``{ ` `        ``\$cnt``++; ` `        ``\$lcm` `/= 2; ` `    ``} ` ` `  `    ``// If 2 divides lcm cnt number of times ` `    ``if` `(``\$cnt` `% 3 == 2) ` `        ``\$minPerfectCube` `*= 2; ` `    ``else` `if` `(``\$cnt` `% 3 == 1) ` `        ``\$minPerfectCube` `*= 4; ` ` `  `    ``\$i` `= 3; ` ` `  `    ``// Check all the numbers that divide lcm ` `    ``while` `(``\$lcm` `> 1)  ` `    ``{ ` `        ``\$cnt` `= 0; ` `        ``while` `(``\$lcm` `% ``\$i` `== 0) ` `        ``{ ` `            ``\$cnt``++; ` `            ``\$lcm` `/= ``\$i``; ` `        ``} ` ` `  `        ``if` `(``\$cnt` `% 3 == 1) ` `            ``\$minPerfectCube` `*= ``\$i` `* ``\$i``; ` `        ``else` `if` `(``\$cnt` `% 3 == 2) ` `            ``\$minPerfectCube` `*= ``\$i``; ` ` `  `        ``\$i` `+= 2; ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `\$minPerfectCube``; ` `} ` ` `  `// Driver code ` `\$arr` `= ``array``(10, 125, 14, 42, 100 ); ` `\$n` `= sizeof(``\$arr``); ` `echo``(minPerfectCube(``\$arr``, ``\$n``)); ` ` `  `// This code is contributed by Shivi_Aggarwal ` `?> `

Output:

```9261000
```

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