text: Smallest perfect Cube divisible by all elements of an array | Java implementation of the approach ; Function to return the gcd of two numbers ; Function to return the lcm of aint the elements of the array ; To calculate lcm of two numbers multiply them and divide the result by gcd of both the numbers ; Return the LCM of the array elements ; Function to return the smaintest perfect cube divisible by aint the elements of arr [ ] ; LCM of all the elements of arr [ ] ; If 2 divides lcm cnt number of times ; Check aint the numbers that divide lcm ; Return the answer ; Driver code
code: 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 ) ) ; } }