finding Repeated elements in an array...

Hello everyone...
I'm trying to find repeated elements in array, and this is my code:
package rep; import java.util.Scanner; public class Rep{     public static void main(String args[]){                 int nums[] = new int[5];         int i,j,count=0;         Scanner input = new Scanner(System.in);                 System.out.print("Enter Numbers: ");          for(i=0; i<nums.length; i++)              nums[i] = input.nextInt();                   System.out.print("Repeated numbers are: ");         for(i=0; i<nums.length; i++){             for(j=0; j<nums.length; j++){                 if(nums[i] == nums[j])                     count++;             }             System.out.print(" " + nums[i] + ":" + count + ", ");               count = 0;         }                   } }

The question is i just wanna to show just repeated elements;
for e.g i enter this number {1,1,2,2,3}
the out put is:
Repeated numbers are:  1:2,  1:2,  2:2,  2:2,  3:1
i just want to show 1:2, 2:2, 3:1

what's solution?

tnx for your help...

Please explain what you are doing. Your desired output

1:2, 2:2, 3:1

doesn't show repeated elements, since 3 appears only once.

Look,
1 repeated twice
2 repeated twice
3 repeated once
I wanna the output look like above
But i got that output:
1:2, 1,:2, 2:2, 2:2, 3:1

What change my code needs to get that output?

Alex Moradi wrote:Look, . . . What change my code needs to get that output?

That isn't an explanation of what you want, simply an example. From which it appears you are trying to count values, not seek repeats.

My guess is he want's to loop his array and at the end print how many times each number occurs in the array.

Yes, that would constitute counting elements.

Is there a limit of the value that's bin entered, otherwise it's gonna be hard to do with an array? for this example let's say all numbers are between or equal to 0 and 5. If your set on using an array, cause that would be easier with a list,  I would create a counter for each value, loop the array and use a switch to increment the counter. Then at the end you can use your counters to print the occurences of each value

The reason why you getting not the desired output is cause your still in a for loop when printing it. You'll need a counter for each value, not a single counter that you set to zero each time

It is awkward to count things like that without something to record the counts. There is an example in the Java™ Tutorials, but that may be at a later stage than OP has reached.

Also I guess you entered your values on the same line seperatted with a space? Keep in  mind when the user tries to asign the values on seperatted lines,  the assignment of your array will fail, so eather warn the user not to do that or account for that mistake

Are you sure about that? I think your Scanner will be able to compensate for input on any number of lines, as long as you don't use nextLine(). Remember that the line ends between consecutive lines count as whitespace, so they are included in the default delimiter.

What if "1 1 2 3 1 1" is entered? Do you want "1:4 2:1 3:1" ?

If so, there are a couple of approaches:

Sort the array first and then traverse in order.

Use a Map<Integer,Integer> where key is number and value is count.

Maintain a parallel boolean array to track which numbers you've already accounted for.

I think we are going to have to wait for a response from OP before we can do any more.

With what you have now, create an empty Set<String>, fill it with whatever you are generating on line 20, print out elements of the set  

package files; import java.util.Scanner; import java.util.ArrayList; import java.util.Arrays; public class Sample { public static void main(String[] args) { Scanner sc=new Scanner(System.in); ArrayList<Integer> key=new ArrayList<Integer>(); ArrayList<Integer> times=new ArrayList<Integer>(); int n; n=sc.nextInt(); int[] A=new int[n]; for(int i=0;i<n;i++) A[i]=sc.nextInt(); Arrays.sort(A); int count=1; int prev=A[0]; for(int i=1;i<n;i++) { if(A[i]==prev) { count++; } else { key.add(prev); times.add(count); count=1; prev=A[i]; } } key.add(prev); times.add(count); for(int i=0;i<key.size();i++) { System.out.print(key.get(i)+":"+times.get(i)+"  "); } } }

It works fine for your problem.

Welcome to the Ranch

How do you know it works? OP hasn't told us, so we don't actually know what the problem is. It might be counting, but that looks a very complicated way to count things. Running two Lists in parallel is very error‑prone, so please explain all the details of how your code works.

What about my noobish solution?

import java.util.*; public class Main {    public static void main(String[] args) {        Integer[] myArray = {10, 10, 3, 4, 5, 3, 3, 4, 10};        printRepeatedElementsInArray(myArray);            }    static void printRepeatedElementsInArray(Integer[] a) {        List<Integer> list = Arrays.asList(a);        Map<Integer, Integer> map = new HashMap<>();        Collections.sort(list);        for (Integer i : list) {            map.put(i, Collections.frequency(list, i));        }        for (Integer i : map.keySet()) {            System.out.println(i + ":" + map.get(i));        }    } }

       for (Integer i : list) {            map.put(i, Collections.frequency(list, i));        }The use of frequency() is not particularly efficient in solving this problem. It will be called for each occurrence of '10' for example, coming up with the same count each time. You would only need to do this once for each unique value.

Tnx to all for the answers.

I have an array with 5 elements.

I want to count repeated element, for e.g user input this set {1,1,1,2,3}

The program must say in your set
Number 1 repeated 3 times and ignor numbers that not repeated in the set!
I don’t want print 2,3! //is it okay till now!?

An other e.g user input this set:{2,2,0,4,4}
The out put should be :
2:2 , 4:2;
0 not repeated and no need to going print !
I don’t want to use hash table & map
Just using an array and loops.

I hope understand my mean.
Tnx alot🤨

What you are asking is possible but I don't understand why you would limit yourself that way. The simplest and most straightforward solution is to use a map to create a histogram.

Without a map, you will need to set up a different array to hold the duplicate counts. You will also need to have some logic that will skip elements that have been determined to be duplicates and somehow flag them so you can avoid recounting them.

Alex Moradi wrote:. . . Number 1 repeated 3 times and ignor numbers that not repeated in the set!  . . .

That means, 3 appears 3×. I am afraid I am going into pedantic mode. You don't have a set there, beause sets don't contain duplicates; as a set, that would read {1, 2, 3}. What you are doing is counting elements equal to each other and only printing which are repeated.
Did you see the link to the Map interface in the Java™ Tutorials I posted a couple of days ago? Here it is again.

2 solutions: one with java streams and second with arrays and loops.

import java.util.*; import java.util.concurrent.*; import java.util.stream.*; public class RepeatedNumbers2 {    private static final int TOTAL_NUMBERS = 30;    public static void main(String[] args) {        int[] repeatedNumbers = generateRepeatedNumbers();        printRepeatedNumbers(repeatedNumbers);        printRepeatedNumbersWithArraysAndLoops(repeatedNumbers);    }    private static int[] generateRepeatedNumbers() {        final int MAX_NUMBER = TOTAL_NUMBERS * 2 / 3;        return IntStream.generate(() -> ThreadLocalRandom.current().nextInt(1, MAX_NUMBER + 1))            .limit(TOTAL_NUMBERS)            .toArray();    }    private static void printRepeatedNumbers(int... numbers) {        Map<Integer, Integer> mapOfNumbersToFrequency = IntStream.of(numbers)            .collect(HashMap::new,                (map, value) -> map.merge(value, 1, (count1, count2) -> count1 + count2),                (map1, map2) -> map2.forEach(                    (key, value) -> map1.merge(key, value, (value1, value2) -> value1+value2))            );        mapOfNumbersToFrequency.forEach((key, value) -> {                if (value > 1) System.out.print(key + ":" + value + ", ");            });        System.out.println();    }   private static void printRepeatedNumbersWithArraysAndLoops(int... numbers) {        int[] numbers2 = new int[numbers.length];        int[] count = new int[numbers.length];        int subsetLength = 0;        for (int x : numbers) {            boolean found = false;            for (int i=0; i<subsetLength; i++) {                if (numbers2[i] == x) {                    count[i]++;                    found = true;                }            }            if (!found) {                numbers2[subsetLength] = x;                count[subsetLength] = 1;                subsetLength++;            }        }        for (int i=0; i<subsetLength; i++) {            if (count[i] > 1)                System.out.print(numbers2[i] + ":" + count[i] + ", ");        }        System.out.println();    }     }