A good answer might be:

The beginning (index 0) element in an array will never be larger than the maximum element in that array. (It might be the maximum element of the array; but this is fine.) 

max = array[0];

This way of initializing max assumes that there is at least one element in the array. If there is not, the java system will detect that the index "0" is out of bounds and throw an exception.

Initializing Max

Here is the program so far. It has initialized max to the first element in the array. The for loop is set up to look at every element in the array, starting with the beginning element, to see if that element "beats" the current maximum. 

class MaxAlgorithm
{

  public static void main ( String[] args ) 
  {

    int[] array =  { -20, 19, 1, 5, -1, 27, 19, 5 } ;
    int   max;

    // initialize the current maximum
 
    max = array[0];

    // scan the array
    for ( int index=0; index < array.length; index++ )
    {
     
      if ( ____________________ )  // examine the current element

        max = array[ index ];        // if it is the largest so far, change max

    }
      
    System.out.println("The maximum of this array is: " + max );

  }
}

It would be OK (and maybe preferable) to initialize index at 1. But the program as written works fine (as soon as you fill in the blank.)

QUESTION 9:

Complete the program.