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Array maximum problem (Java)

Started by M-E Mar 30, 2007 at 8:23 AM 9 replies 4.1k views

Original Post

Mar 30, 2007 at 8:23 AM

I have a problem with one of the exercises of my Java tutorial, i hope someone can give me a good hint and help me with this one. I would appreciate it. The exercise is this:

Quote:
Exercise 2 --- Two Largest Elements Complete the following program so that it computes and writes out the two largest elements in the array. import java.io.* ; class TwoLargest { public static void main ( String[] args ) throws IOException { int[] data = {3, 1, 5, 7, 4, 12, -3, 8, -2}; // declare and initialize variables for the two largest // compute the two largest for ( int index= ; index < data.length; index++) { } // write out the two largest System.out.println( ); } }

Now i know how to compute the maximum of an array, like this:

import java.io.* ;

class Largest
{

  public static void main ( String[] args ) throws IOException
  {
    int[] data = {3, 1, 5, 7, 4, 12, -3, 8, -2};
    
    // declare and initialize variable for the largest
    int largest = data[0];
    
    
    // compute the largest
    for ( int index= 0; index < data.length; index++)
    {
      if(largest < data[index]) {
         largest = data[index];
      }
    }
      
    // write out the largest
    System.out.println("The largest is: " + largest);

  }
}

However, I don't know how i can compute the second largest element (in the same for-loop as the largest element) like in the exercise. I hope someone can help me.

OrangyTang

Mar 30, 2007 at 8:30 AM

Keep track of the second largest found as before, and update it if it's greater than before but less that the largest?

Personally I'd just sort the array and take the last two element, but thats just me. [grin]

M-E

Mar 30, 2007 at 8:35 AM

Quote:
Original post by OrangyTang
Keep track of the second largest found as before, and update it if it's greater than before but less that the largest?

Personally I'd just sort the array and take the last two element, but thats just me. [grin]

Ok, but thats not the exercise [grin]

And about the first part: That's what i thought about too, but that won't work if the two largest elements are the same.

OrangyTang

Mar 30, 2007 at 8:50 AM

Thats a good point. You might also want to keep track of the indicies for the current and second largest. Then you could detect if you've stumbled across the same number twice. The logic isn't going to be pretty though.

M-E

Mar 30, 2007 at 8:58 AM

I know how to do it, but it's not (i think it's not) what the exercise wants:

class TwoLargest{  public static void main ( String[] args ) throws IOException  {    int[] data = {3, 1, 5, 7, 4, 12, -3, 8, -2};        // declare and initialize variable for the largest    int largest = data[0];    int secLargest = 0;    int skip = 0;        // compute the largest    for ( int index= 0; index < data.length; index++)    {      if(largest < data[index]) {         largest = data[index];         skip = index;      }    }        for ( int index= 0; index < data.length; index++)    {      if(secLargest < data[index] && index != skip) {         secLargest = data[index];      }    }          // write out the largest    System.out.println("The largest is: " + largest + " and the second largest is: " + secLargest);  }}

But this program has a bug too, because when all elements are negative the second largest will be zero. If you initialize secLargest like largest you will get a problem if the first element of the array is the largest.

giveblood

Mar 30, 2007 at 9:02 AM

The second largest works the same way, only you don't even consider it unless its less than the largest.

import java.io.* ;class TwoLargest{  public static void main ( String[] args ) throws IOException  {    int[] data = {3, 1, 5, 7, 4, 12, -3, 8, -2};    // declare and initialize variable for the largest    int largest = data[0];    int second = data[0];    // compute the largest    for ( int index= 0; index < data.length; index++)    {		if(data[index]>=largest)		{			largest=data[index];		}		else if (data[index]>second)		{			second=data[index];		}	}    // write out the largest    System.out.println("The largest is: " + largest);    System.out.println("The second is: " + second);  }}

OrangyTang

Mar 30, 2007 at 9:06 AM

If the first element of the array is the largest, your second largest won't be calculated correctly.

Easier would be to initialise the largest and second largest to Integer.MIN_VALUE instead.

Stani R

Mar 30, 2007 at 9:06 AM

I'll just add that I find this exercise to be a waste of your time. Like OrangyTang said, you should be sorting the array, which is what this is aiming at in the end anyway. You could start by implementing selection sort and insertion sort, then move on to more complex algorithms such as quicksort or heapsort. You will get a lot more mileage out of those :D

giveblood

Mar 30, 2007 at 9:07 AM

I didn't think about if the first value was the largest. You could get around it by making second a very small number, but that's not a good idea. I can't think of a good way to do this without having to pass through the away twice, or adding extra statements that specifically check to see if so and so is the largest

M-E

Mar 30, 2007 at 9:08 AM

Yeah, that works for this exercise, thanks. But it still doesn't work for all versions of the problem, because if the second largest element already passed between data[0] and largest it will not compute the second largest.

anyway, thanks all. I will just skip this exercise then...

Stani R

Mar 30, 2007 at 9:14 AM

I didn't spend too much time on it, but I think the following loop should catch both. If there's a problem with it, it's that it won't detect the case where the first element is the largest. For that, you need to do something else instead of initializing both to the first element right away. That's why you should just sort the array - you won't have to deal with special cases, and can just take the last two elements to answer the problem.

    for ( int index= 0; index < data.length; index++)    {    	if( data[index] > largest) {    		second = largest;    		largest = data[index];    	}    	else if (data[index] > second)    		second = data[index];    		}    }

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