Minimum operations of given type to make all elements of a matrix equal | Java implementation of the approach ; Function to return the minimum number of operations required ; Create another array to store the elements of matrix ; If not possible ; Sort the array to get median ; To count the minimum operations ; If there are even elements , then there are two medians . We consider the best of two as answer . ; Return minimum operations required ; Driver code
import java . util . * ; class GFG { static int minOperations ( int n , int m , int k , int matrix [ ] [ ] ) { int [ ] arr = new int [ n * m ] ; int mod = matrix [ 0 ] [ 0 ] % k ; for ( int i = 0 ; i < n ; ++ i ) { for ( int j = 0 ; j < m ; ++ j ) { arr [ i * m + j ] = matrix [ i ] [ j ] ; if ( matrix [ i ] [ j ] % k != mod ) { return - 1 ; } } } Arrays . sort ( arr ) ; int median = arr [ ( n * m ) / 2 ] ; int minOperations = 0 ; for ( int i = 0 ; i < n * m ; ++ i ) minOperations += Math . abs ( arr [ i ] - median ) / k ; if ( ( n * m ) % 2 == 0 ) { int median2 = arr [ ( ( n * m ) / 2 ) - 1 ] ; int minOperations2 = 0 ; for ( int i = 0 ; i < n * m ; ++ i ) minOperations2 += Math . abs ( arr [ i ] - median2 ) / k ; minOperations = Math . min ( minOperations , minOperations2 ) ; } return minOperations ; } public static void main ( String [ ] args ) { int matrix [ ] [ ] = { { 2 , 4 , 6 } , { 8 , 10 , 12 } , { 14 , 16 , 18 } , { 20 , 22 , 24 } } ; int n = matrix . length ; int m = matrix [ 0 ] . length ; int k = 2 ; System . out . println ( minOperations ( n , m , k , matrix ) ) ; } }