Count different values in array in Java

Question

I\'m writing a code where I have an int[a] and the method should return the number of unique values. Example: {1} = 0 different values, {3,3,3} = 0 different values, {1,2} = 2 d

Answer 1

Try this simple code snippet.

public static int differentValuesUnsorted(int[] a)
{
    ArrayList<Integer> list=new ArrayList<Integer>();   //import java.util.*;
    for(int i:numbers)                                  //Iterate through all the elements
      if(!list.contains(i))                             //checking for duplicate element
        list.add(i);                                    //Add to list if unique
    return list.size();
}

Answer 2

Use a set to remove duplicates

 public static int differentValuesUnsorted(int[] a) {
     if (a.length < 2) {
         return 0;
     }

     Set<Integer> uniques = new HashSet<>(a);
     return singleUnique.size();
 }

Answer 3

First create distinct value array, It can simply create using HashSet.

Then alreadyPresent.size() will provide number of different values. But for the case such as -{3,3,3} = 0 (array contains same elements); output of alreadyPresent.size() is 1. For that use this simple filter

if(alreadyPresent.size() == 1)){
    return 0;
}

Following code will give the count of different values.

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class Demo {

  public static void main(String[] args)
  {
       int array[] = {9,9,5,2,3};
       System.out.println(differentValuesUnsorted(array));
  }

  public static int differentValuesUnsorted(int[] array)
  {

     Set<Integer> alreadyPresent = new HashSet<Integer>();

     for (int nextElem : array) {
         alreadyPresent.add(nextElem);
     }

     if(alreadyPresent.size() == 1){
         return 0;
     }

     return alreadyPresent.size();

  }
}