1

Anyway, I have this extremely simple java program to find out if 2 objects in an array are the same. How ever when I run it when a set of unique objects it always returns 1 error, If I add more objects that are the same it counts as normal.

This is the code;

int[] array= new int[] {1,245,324,523};

    int size = array.length;
    int error=0;
    System.out.println(error);

    int i = 0;
    for(i=0; i<size; i++){

        if(array[0] == array[i]){
            error= error +1;
        }
        System.out.println(error);
    }
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2
  • 4
    This always print at least 1, since array[0] == array[0]. Is there a question here? Commented Sep 30, 2011 at 0:04
  • @user972183: what you are thinking of if(array[0]==array[i])? Commented Sep 30, 2011 at 0:25

8 Answers 8

Reset to default
5

The 1 error is because you're comparing array[0] with array[0], which is of course equal to itself.

If you want to find all pairwise duplicates, you will need to do a double loop:

for(int i=0;i<size;i++){
    for(int j=i+1;j<size;j++){
        if(array[i] == array[j]){
            if(i!=j){
                error = error + 1;
            }
        }
    }
}

You'll notice a few things from this code:

  • j starts at i+1, not at 0.
  • error is only incremented when i!=j

The first is because you're taking turns with each element in your array to compare with every other element. By the time its turn comes around (in the outer loop), it's already been compared to the elements before it, and should not be compared with them again.

The second is because you'll end up comparing an element to itself for each outer loop. You don't want to count it as an error.

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7 Comments

I think in your double loop j should always start at i+1.
i<size-1 in the first for condition.
that second if i!=j is unnecessary
yeah, er. there's a slight problem, because the index would start at size for the last outer iteration. Surely our user can figure out what to do about it.
All i'm aiming to do is find out if all the elements are distint or not, I think it would still work with out the outside loop how ever the counter would have the wrong number of errors. But if theyre are no errors it is impossible for errors to be higher than one?
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1

You're starting i at 0. Therefore the first test is if(array[0] == array[0]) ;)

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0

You're always going to get at least one error because array[0] == array[i] will be true on the first iteration, when i = 0.

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0

In your code at 

int i=0;
for(i=0;i<size;i++){
    if(array[0]==array[i]){    //this condition runs true only for the first time. as i=0 here
        error=error+1;
    }
    System.out.println(error); //now here you had put the println(error) outside the if()-condition therefore it will be printed repeatedly value of error which is 1
}
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0

Try a doubly nested for loop. Something like this

for (int i=0;i<size-1;i++){
  for (int j=i+1; j<size; j++) {
    error += (array[i] == array[j]) ? 1 : 0;
  }
}
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0

You also need to "think Java". Use Array.equals for the comparison. See the documentation here and some examples here

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0

If you are going to make use of any collection, you can easiy findout the duplicates in the array

Example:

 public static void main(String[] args) {
    boolean containsDuplicate =false;
    int[] intArray = new int[] {1,245,324,1,523};
    List<Integer> myObj = new ArrayList<Integer>();
    for(int id : intArray){
        if(myObj.contains(id)){
            containsDuplicate = true;
            System.out.println("Duplicate");
        }else{
            myObj.add(id);
        }
    }

}
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0

Following code is fine but it returns 11. Not sure if it is what was expected. 

public static void main(String[] args){
    int[] array = {23, 23, 0, 43, 545, 12, -55, 43, 12, 12, -999, -87, 12, 0, 0};
    int error = 0;
    for(int i=0;i<array.length;i++){
        for(int j=i+1;j<array.length;j++){
            if(array[i] == array[j]){
                if(i!=j){
                    error = error + 1;
                }
            }
        }
    }
    System.out.println(error);
}

I propose little bit complicated but hopefully more advanced solution.

package myjavaprogram;
import java.util.Arrays;

public class TestClass1 {
    public static void main(String[] args){
        int[] array = {23, 23, 0, 43, 545, 12, -55, 43, 12, 12, -999, -87, 12, 0, 0};
        //int[] array = {23, -22, 0, 43, 545, 12, -55, 43, 12, 0, -999, -87, 12};
        //int[] array = {23, -22, 0, 23};
        //int[] array = {23, -22, 23};
        calculate_duplicates(array);        
    }

    private static void calculate_duplicates(int[] array) {
        calculateUniqueNumbers(array);
    }

    private static void calculateUniqueNumbers(int[] array) {
        Pair[] pairs = new Pair[array.length];
        initializePairsAtrray(pairs, array);
        printPairsAtrray(pairs);

        System.out.println("array.length="+array.length);
        System.out.println("--------------------");

        // update pairs array taking in account duplicates duplicates
        for(int i = 0; i < array.length; i++) {
            System.out.println("array[i]="+array[i] + " i="+i);
            for(int j = i+1; j < array.length; j++) {
                System.out.println("array[j]="+array[j]+" j="+j);

                if(array[i] == array[j] && pairs[j].useDuringCount == true) {
                    pairs[i].occurance_num++;

                    pairs[j].occurance_num = 0;
                    pairs[j].useDuringCount = false;
                }

                if(array[i] == 0) {
                    pairs[i].occurance_num = 0;                    
                }
                if(array[j] == 0) {
                    pairs[j].occurance_num = 0;
                    pairs[j].useDuringCount = false;
                }                
            }
            pairs[i].useDuringCount = false;
            System.out.println("--------------------");
        }                
        printPairsAtrray(pairs);

        // calculate general number of duplicates (numbers whick are repeated 
        // in initial array)
        System.out.println("Duplicates in array:"+
                calculateDuplicatesNumber(pairs));        
    }

    private static void initializePairsAtrray(Pair[] pairs, int[] array) {
        for(int i=0;i<pairs.length;i++) {
            Pair p = new Pair();            
            p.occurance_num = 1;
            p.value = array[i];
            p.useDuringCount = true;
            pairs[i] = p;
        }
    }

    private static void printPairsAtrray(Pair[] pairs) {
        System.out.println("--------------------");
        for(int i=0;i<pairs.length;i++) {
            System.out.println("pairs["+i+"].occurance_num="+pairs[i].occurance_num);
            System.out.println("pairs["+i+"].value="+pairs[i].value);
            System.out.println("pairs["+i+"].useDuringCount="+pairs[i].useDuringCount);
            System.out.println("--------------------");
        }
    }

    private static int calculateDuplicatesNumber(Pair[] pairs) {
        System.out.println("-------------------- Duplicates:");
        int duplicates_num = 0;
        for(int i=0;i<pairs.length;i++) {
            if(pairs[i].occurance_num > 1) {
                duplicates_num++;
                System.out.println("number: "+pairs[i].value+" occurance_num " + pairs[i].occurance_num);
            }
        }
        return duplicates_num;
    }        
}    

class Pair {
    int value;
    int occurance_num;
    boolean useDuringCount;
}

Mykola

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2 Comments

The accepted answer, IMO, appears to do this simple task a lot more efficiently. No reason to add complexity to something solved with a much simpler and viable solution
Add additional array. Next cycle through each elements of the original array. For current element, iterate over all elements of the array to the right of the current. If found equal, this element put in an additional array. You can make a 2nd array for additional calculations, or just use two-dimensional array.

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