Here's a Java code that solves the problem:
```
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = input.nextInt();
}
int oddCount = 0;
for (int i = 0; i < n; i++) {
if (arr[i] % 2 == 1) {
oddCount++;
}
}
int evenCount = n - oddCount;
if (oddCount == 0 || evenCount == 0) {
System.out.println(0);
} else {
System.out.println(Math.min(oddCount, evenCount));
}
}
}
```
The program reads in the size of the sequence and the array elements from the user, and then counts the number of odd and even elements in the array. If the array contains only odd or even elements, then no moves are needed, and the program prints 0. Otherwise, the program prints the minimum number of moves needed to make all the array elements even. This is equal to the minimum of the number of odd elements and the number of even elements in the array.