Puzzle: Given an input array of n positive integers where the integers are in random order. Each number in that array can occur many times. You need to find all the distinct elements and put all those elements in an array i.e. output1. If no number is duplicated in input, then output should be {-1}.
Input Specifications: input: number of elements in input2 (n) input 2: an array of n positive integers
Output Specifications: output: an array of distinct elements which are duplicate in input2
Example 1: input : 6 input2 : {4,4,7,8,8,9} output : {4,8}
Example 2: input : {2,3,6,8,90,58,58,60} output : {58}
Example 3: input : {3,6,5,7,8,19,32} output : {-1}Solution
import java.util.HashSet;
import java.util.Set;
public class DuplicatesInArray
{
public static void main(String[] args)
{
Integer[] array = {1,2,3,4,5,6,7,8}; //input 1
int size = array.length; //input 2
Set<Integer> set = new HashSet<Integer>();
Set<Integer> duplicates = new HashSet<Integer>();
for(int i = 0; i < size ; i++)
{
if(set.add(array[i]) == false)
{
duplicates.add(array[i]);
}
}
if(duplicates.size() == 0)
{
duplicates.add(-1);
}
System.out.println(duplicates);
}
}The above program will find all duplicate elements from the array and put them into a separate set. You can find more discussion on this logic here: //howtodoinjava.com/java/interviews-questions/find-duplicate-elements-in-an-array/
Happy Learning !!
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Vassilis
List input = Arrays.asList(1, 2, 5, 2, 5, 5, 1, 6, 9); Map frequencies = input.stream() .collect(Collectors.groupingBy(n -> n, Collectors.counting())); frequencies.entrySet().stream() .filter(entry -> entry.getValue() > 1) .forEach(System.out::println);Juan David Gomez
A better approach with stream will be
Set<Integer> uniques = new HashSet<>(); Set<Integer> collect = duplicates.stream() .filter(integer -> !uniques.add(integer)) .collect(Collectors.toSet());or
Set<Integer> collect = duplicates.stream() .filter(integer -> Collections.frequency(duplicates, integer) > 1) .collect(Collectors.toSet());