Smallest perfect Cube divisible by all elements of an array
Last Updated :
24 Apr, 2023
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
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++
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
ll lcmOfArray( int arr[], int n)
{
if (n < 1)
return 0;
ll lcm = arr[0];
for ( int i = 1; i < n; i++)
lcm = (lcm * arr[i]) / gcd(lcm, arr[i]);
return lcm;
}
int minPerfectCube( int arr[], int n)
{
ll minPerfectCube;
ll lcm = lcmOfArray(arr, n);
minPerfectCube = ( long long )lcm;
int cnt = 0;
while (lcm > 1 && lcm % 2 == 0) {
cnt++;
lcm /= 2;
}
if (cnt % 3 == 2)
minPerfectCube *= 2;
else if (cnt % 3 == 1)
minPerfectCube *= 4;
int i = 3;
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 minPerfectCube;
}
int main()
{
int arr[] = { 10, 125, 14, 42, 100 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << minPerfectCube(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int gcd( int a, int b)
{
if (b == 0 )
return a;
else
return gcd(b, a % b);
}
static int lcmOfArray( int arr[], int n)
{
if (n < 1 )
return 0 ;
int lcm = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
lcm = (lcm * arr[i]) / gcd(lcm, arr[i]);
return lcm;
}
static int minPerfectCube( int arr[], int n)
{
int minPerfectCube;
int lcm = lcmOfArray(arr, n);
minPerfectCube = lcm;
int cnt = 0 ;
while (lcm > 1 && lcm % 2 == 0 )
{
cnt++;
lcm /= 2 ;
}
if (cnt % 3 == 2 )
minPerfectCube *= 2 ;
else if (cnt % 3 == 1 )
minPerfectCube *= 4 ;
int i = 3 ;
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 minPerfectCube;
}
public static void main(String args[])
{
int arr[] = { 10 , 125 , 14 , 42 , 100 };
int n = arr.length;
System.out.println(minPerfectCube(arr, n));
}
}
|
Python3
def gcd(a, b) :
if (b = = 0 ) :
return a
else :
return gcd(b, a % b)
def lcmOfArray(arr, n) :
if (n < 1 ) :
return 0
lcm = arr[ 0 ]
for i in range (n) :
lcm = (lcm * arr[i]) / / gcd(lcm, arr[i]);
return lcm
def minPerfectCube(arr, n) :
lcm = lcmOfArray(arr, n)
minPerfectCube = lcm
cnt = 0
while (lcm > 1 and lcm % 2 = = 0 ) :
cnt + = 1
lcm / / = 2
if (cnt % 3 = = 2 ) :
minPerfectCube * = 2
elif (cnt % 3 = = 1 ) :
minPerfectCube * = 4
i = 3
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 minPerfectCube
if __name__ = = "__main__" :
arr = [ 10 , 125 , 14 , 42 , 100 ]
n = len (arr)
print (minPerfectCube(arr, n))
|
C#
using System;
class GFG
{
static int gcd( int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
static int lcmOfArray( int []arr, int n)
{
if (n < 1)
return 0;
int lcm = arr[0];
for ( int i = 1; i < n; i++)
lcm = (lcm * arr[i]) / gcd(lcm, arr[i]);
return lcm;
}
static int minPerfectCube( int []arr, int n)
{
int minPerfectCube;
int lcm = lcmOfArray(arr, n);
minPerfectCube = lcm;
int cnt = 0;
while (lcm > 1 && lcm % 2 == 0)
{
cnt++;
lcm /= 2;
}
if (cnt % 3 == 2)
minPerfectCube *= 2;
else if (cnt % 3 == 1)
minPerfectCube *= 4;
int i = 3;
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 minPerfectCube;
}
public static void Main()
{
int []arr = { 10, 125, 14, 42, 100 };
int n = arr.Length;
Console.WriteLine(minPerfectCube(arr, n));
}
}
|
PHP
<?php
function gcd( $a , $b )
{
if ( $b == 0)
return $a ;
else
return gcd( $b , $a % $b );
}
function lcmOfArray(& $arr , $n )
{
if ( $n < 1)
return 0;
$lcm = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
$lcm = ( $lcm * $arr [ $i ]) /
gcd( $lcm , $arr [ $i ]);
return $lcm ;
}
function minPerfectCube(& $arr , $n )
{
$lcm = lcmOfArray( $arr , $n );
$minPerfectCube = $lcm ;
$cnt = 0;
while ( $lcm > 1 && $lcm % 2 == 0)
{
$cnt ++;
$lcm /= 2;
}
if ( $cnt % 3 == 2)
$minPerfectCube *= 2;
else if ( $cnt % 3 == 1)
$minPerfectCube *= 4;
$i = 3;
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 $minPerfectCube ;
}
$arr = array (10, 125, 14, 42, 100 );
$n = sizeof( $arr );
echo (minPerfectCube( $arr , $n ));
?>
|
Javascript
<script>
function gcd(a, b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
function lcmOfArray(arr, n)
{
if (n < 1)
return 0;
let lcm = arr[0];
for (let i = 1; i < n; i++)
lcm = parseInt((lcm * arr[i]) / gcd(lcm, arr[i]));
return lcm;
}
function minPerfectCube(arr, n)
{
let minPerfectCube;
let lcm = lcmOfArray(arr, n);
minPerfectCube = lcm;
let cnt = 0;
while (lcm > 1 && lcm % 2 == 0) {
cnt++;
lcm = parseInt(lcm/2);
}
if (cnt % 3 == 2)
minPerfectCube *= 2;
else if (cnt % 3 == 1)
minPerfectCube *= 4;
let i = 3;
while (lcm > 1) {
cnt = 0;
while (lcm % i == 0) {
cnt++;
lcm = parseInt(lcm/i);
}
if (cnt % 3 == 1)
minPerfectCube *= i * i;
else if (cnt % 3 == 2)
minPerfectCube *= i;
i += 2;
}
return minPerfectCube;
}
let arr = [ 10, 125, 14, 42, 100 ];
let n = arr.length;
document.write(minPerfectCube(arr, n));
</script>
|
Complexity Analysis:
- Time Complexity: O(n * log(arr[i])
- Auxiliary Space: O(1), since no extra space has been taken.
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