changed here as in case of even elements there will be 2 medians
int median2 = arr [ ( ( n * m ) / 2 ) - 1 ] ; int minOperations2 = 0 ; for ( int i = 0 ; i < n * m ; ++ i ) minOperations2 += abs ( arr [ i ] - median2 ) / k ; minOperations = min ( minOperations , minOperations2 ) ; }