same ints within an array of ints enter by command line args, how it works explanation please

[Stan Austin]

hi everyone, my first post woo!
In lecture we was given the task to allow the user, to enter any number of ints in the command line, and find if any of the numbers entered are duplicates/repeats, if so respond false, if all unique respond true.
The args to be entered are "1, 2, 3, 4, 5, 6, 7, 8, 9, 10," (True)   ...     "1, 2, 3, 4, 5, 6, 7, 1, 2, 3," (False)    ...     "0, 1, 2, 3, 4, 0" (false)
So after many attempts and some guidance i managed to get to this. I was certain it would work and still unsure why it does not.

 
public class Snap {
  public static void main(String[] args) {
    int i = 0;
    int j = 0;
    int c = 0;
    int l = args.length;
    
    for (i=0; i<l; i=i+1) {
    for (j=0; j<l; j=j+1) {
        int x1 = Integer.parseInt(args[i]);
        int x2 = Integer.parseInt(args[j]);
           
            if (x1 == x2)
                c = 1;
            else if (x1 != x2)
                c = 0;     
    }        
    }
    if (c == 1)
        System.out.println("True / Unique");
    else if (c == 0)
        System.out.println("False / Duplicate");
    
}
}

the issue is this only ever prints true regardless of the args entered

at the end the lecturer helped me come to this solution

public class Snap2 {
  public static void main(String[] args) {
    int i = 0;
    int j = 0;
        int c = 0;
        int x1 = 0;
        int x2 = 0;
        int l = args.length;
    
    for (i=0; i<l; i=i+1) {
    for (j=0; j<l; j=j+1) {
        x1 = Integer.parseInt(args[i]);
        x2 = Integer.parseInt(args[j]);
            if (x1 == x2)
                c = c + 1;     
    }        
    }
    if (c == l)
        System.out.println("True / Unique");
    else 
        System.out.println("False / Duplicate");
    
}
}

Could some one please help me understand how this works? because I don't get why it does (and why mine doesn't)

[Henry Wong]

Unfortunately, in taking a very quick look, neither yours or your instructor's versions should work.

Hint: What should happen when the third parameter is compared to the third parameter? Is that a duplicate?

Henry

[Stan Austin]

I'm not quite sure what you mean i have changed the "L"'s to "as" for clarity so they aren't mixed up with 1's
and it does work! but I cant figure it out.
Apparently, its something to do with it forming a matrix and if the numbers are all different then this would cause c to go up by 1 each time there was a match, (i guess it would match itself?)
and if there was more than one duplicate it would increase the value of c more...
Since c starts as 0, plus one to it to make it equal to the array size if there was only an equivalent number of matches. if there was to many this number would be too high i guess :/
still a bit fuzzy, however writing out (badly) in my own words, what he tried explaining to me, has made it a little clearer and i think i get the gist of what might be going on here...

although could you be a little less cryptic? i don't quite get the hint

public class Snap2 {
  public static void main(String[] args) {
    int i = 0;
    int j = 0;
        int c = 0;
        int x1 = 0;
        int x2 = 0;
        int as = args.length;
     
    for (i=0; i<l; i=i+1) {
    for (j=0; j<l; j=j+1) {
        x1 = Integer.parseInt(args[i]);
        x2 = Integer.parseInt(args[j]);
            if (x1 == x2)
                c = c + 1;     
    }        
    }
    if (c == as)
        System.out.println("True / Unique");
    else
        System.out.println("False / Duplicate");
     
}
}

[Tony Docherty]

Welcome to the Ranch.

My first thoughts were, like Henry's, that the lecturers version wouldn't work but then you helpfully changed the 'l' to 'as' which cleared up the problem of seeing 'L' as '1'.

It works because you are comparing every number with every number (including itself) and so if there are no duplicates there will be exactly 'as' matches (ie 1 match for every element in the array because every element only matches itself).

[Henry Wong]

Stef Aucky wrote:
although could you be a little less cryptic? i don't quite get the hint

Okay, your parameters are "1 2 3 4 5 6 7 8 9 10". What is the third parameter? Well... it is "3".

So, when you compare the third parameter with the third parameter, ie. comparing "3" with "3", is the result a duplicate or not?

Henry


[edit: of course, this followup hint is moot, as Tony already explained it.]

[Henry Wong]

Tony Docherty wrote:
My first thoughts were, like Henry's, that the lecturers version wouldn't work but then you helpfully changed the 'l' to 'as' which cleared up the problem of seeing 'L' as '1'.

Good point. I too, saw a one instead of a letter L.

Henry

[Stan Austin]

Tony Docherty wrote:Welcome to the Ranch.

My first thoughts were, like Henry's, that the lecturers version wouldn't work but then you helpfully changed the 'l' to 'as' which cleared up the problem of seeing 'L' as '1'.

It works because you are comparing every number with every number (including itself) and so if there are no duplicates there will be exactly 'as' matches (ie 1 match for every element in the array because every element only matches itself).

Thank you hopefully I will have a great time as part of the community here,
also Noted L's make to bad/confusing names.

I did kind of get there with my own explanation, is there another way of doing this without setting up a second "duplicate array" as this was the only way i could find, my lecturer modified my logic, to work , but my version looks extremely different to the other students in my class  theirs was longer and i think used a single array)

[Knute Snortum]

As long as the code is clear, shorter is often better.  But there are often a lot of ways to do the same task.  The first thing that came to my mind was to use maps, but I don't know if that would be "better".

[Henry Wong]

My first thought would be to adjust the inner loop. There is no reason to start at zero. It can start at "i+1" instead. Checking the case where j equals i is wrong; And checking the cases where j is less than i is not needed, as they were checked with previous iterations.

My second thought would be to not use a counter. When a duplicate is found, the task is done. Why continue checking? Just break out of the loops*.

Henry

* of course, this second suggestion may open a can of worms, as we want to break out of two loops.

[Knute Snortum]

Just for illustration, here's my way of solving the problem:

package net.snortum.play;

import java.util.HashMap;
import java.util.Map;

public class Unique {

	public static void main(String[] args) {
		new Unique().run(args);
	}

	private void run(String[] args) {
		Map<String, Integer> occurrences = new HashMap<>();
		
		for (String arg : args) {
			if (occurrences.containsKey(arg)) {
				occurrences.put(arg, occurrences.get(arg) + 1);
			} else {
				occurrences.put(arg, 1);
			}
		}
		
		boolean unique = true;
		
		for (int times : occurrences.values()) {
			if (times > 1) {
				unique = false;
				break;
			}
		}
		
		System.out.println(unique ? "Unique" : "Not Unique");
	}
}

Notice that it's longer (more statements).  It doesn't use arrays at all.  It will work with any input on the command line, not just integers.  It probably uses more memory.  It might be faster.  So "better"?  Not really -- except that mine uses descriptive variable names.

[Junilu Lacar]

You can stop when you find the first duplicate. Here's another way:

	public static void main(String[] args) {
		check("1, 2, 3, 4, 5, 6, 7, 8, 9, 10");
		check("1, 2, 3, 4, 5, 6, 7, 1, 2, 3");
	}

	public static void check(String input) {
		System.out.printf("%nInput: %s%n", input);
		String[] args = input.split(", ");
		int[] nums = new int[args.length];
		for (int i = 0; i < args.length; i++) {
			int n = nums[i] = Integer.parseInt(args[i]);
			for (int j = 0; j < i; j++) {
				if (n == nums[j]) {
					System.out.printf("Duplicate found: %d at index %d and %d%n", n, j, i);
					System.out.println("False");
					return;
				}
			}
		}
		System.out.println("True");
	}

Output:

Input: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
True

Input: 1, 2, 3, 4, 5, 6, 7, 1, 2, 3
Duplicate found: 1 at index 0 and 7
False

[Junilu Lacar]

Knute's solution looks a lot sexier though

[Tobias Bachert]

Knute's solution seems like a bit overkill in my opinion, as a one-liner is sufficient (at least if you can use Java8):

public class Unique {
    
    public static void main(
            String[] args) {
        ////
        System.out.println(allUnique(args) ? "Unique" : "Not Unique");
    }
    
    public static <T> boolean allUnique(
            final T[] array) {
        ////
        return Stream.of(array).distinct().count() == array.length;
    }
}

[Junilu Lacar]

Tobias, the overachiever. A cow for the one-liner!

[Junilu Lacar]

Henry Wong wrote:
Just break out of the loops*.

Henry

* of course, this second suggestion may open a can of worms, as we want to break out of two loops.

No shame in doing that in a small, controlled environment. I beg to differ with whoever says otherwise.

[Campbell Ritchie]

Stef Aucky wrote:. . . any number of ints in the command line, . . . "1, 2, 3, 4, 5, 6, 7, 8, 9, 10," . . .

That isn't how you enter command‑line arguments. You simply write

java MyApp 1 2 3 4 5 6 7 8 9 10

Just the arguments. No quote marks (unless you need to join two words into one argument), no commas, no nothing.
Don't use numbers instead of booleans.
boolean b = x == y;
or boolean b = (x == y); Round brackets optional.
And once you have that boolean, don't attempt == true or == false. Later on, find out about the ?: operator:-
System.out.println(b ? "Unique" : "Repeated"); or something like that.
Declare your loop indices inside the loop, as in the Java™ Tutorials. Don't use a separate variable for the size of the array; always use myArray.length.

I see you have got lots of other replies, and got your app sorted out.
And welcome to the Ranch

[Campbell Ritchie]

Junilu Lacar wrote:. . . No shame in doing that in a small, controlled environment. I beg to differ with whoever says otherwise.

You s‍hould have anaesthetised me first, then Unless you get into bad habits and think you can do the same in a large wild environment. You can always include a flag as part of the loop header:-

{
   boolean notYetFound = true;
   for (int i = 0; i < myArray.length && notYetFound; i++)
   {
      // ...
      notYetFound = ...;
      // ...
   }
} // {} to restrict scope of netYetFound

If you have nested loops, you would put && notYetFound in both their headers. I will ackowledge that here strict adherence to structured programming does look inelegant.