通过按降序对 TreeSet 进行排序来查找数组的第三大元素

Question

I am trying to find the third largest element of an array by sorting it through treeset in a decreasing order, but some of the test cases fail for certain input values and most of it pass for certain input values.

My code:

// { Driver Code Starts
import java.util.Scanner;
import java.util.*;
import java.io.*;

class ThirdLargestElement
{
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t>0)
        {
            long n =sc.nextLong();
            Long arr[] = new Long[(int)n];
            for(long i=0;i<n;i++)
                arr[(int)i] = sc.nextLong();
            GfG g = new GfG();
            System.out.println(g.thirdLargest(arr));
        t--;
        }
    }
}// } Driver Code Ends
class GfG
{
    long thirdLargest(Long a[])
    {
    // Your code here
        if(a.length<3)
            return -1;
        else{
            TreeSet<Long> ts=new TreeSet<Long>(new myComparator());
            for(long i:a)
                ts.add(i);
            ArrayList<Long> al=new ArrayList<Long>(ts);
            return al.get(2);
        }
    }
}
class myComparator implements Comparator{
    public int compare(Object obj1,Object obj2){
        Long a=(Long) obj1;
        Long b=(Long) obj2;
        if(a<b)
            return 1;
        else if(a>b)
            return -1;
        else 
            return 0;
    }
}

The testcase it failed:

Link to the question where you can run the code

Please explain why this code failed to pass the given test case.

Answer 1

You can directly add elements to TreeSet and get third largest number using stream,

while(t>0)
{
    long n =sc.nextLong();
    TreeSet<Long> ts=new TreeSet<>(Comparator.comparingLong(Long::longValue).reversed());
    for(long i=0;i<n;i++)
        ts.add(sc.nextLong());
    long thirdLast = ts.stream()
            .limit(3)
            .skip(2)
            .mapToLong(e->e)
            .findAny().orElse(0l);

    System.out.println(thirdLast);
    t--;
}

Answer 2

Try This one

class GfG {
    long thirdLargest(Long a[]) {
        Arrays.sort(a);
        List<Long> numbers = Arrays.asList(a);
        Collections.reverse(numbers);
        return numbers.size() >= 3 ? numbers.get(2) : -1;
    }
}

Answer 3

The question states:

the function thirdLargest... takes two argument. The first argument is the array a[] and the second argument is the size of the array (n).

Although the question states that the array is...

an array of distinct elements

the test cases indicate that we are dealing with an array of integers.

Personally, I don't see the need for the second method parameter, because in java an array is an object and has a length member. So my implementation below only takes one parameter, namely an array of int . Maybe the people at geeksforgeeks.org simply converted a question that was originally for the C language to java, since, in C, it is difficult to determine the size of any array.

Each element in a TreeSet must be an object, so we need to convert the elements in the int array to Integer objects. Autoboxing will do this automatically, n.netheless my code below contains an explicit conversion. So in the method I create a TreeSet . Since class Integer implements interface Comparable , the default TreeSet constructor is sufficient. I add all the elements of the int array to the TreeSet , then obtain a descending iterator and then iterate to the third element returned by the iterator, which is the value that the method needs to return.

    int thirdLargest(int[] arr) {
        int third = -1;
        if (arr != null  &&  arr.length > 2) {
            TreeSet<Integer> set = new TreeSet<Integer>();
            for (int elem : arr) {
                set.add(Integer.valueOf(elem));
            }
            Iterator<Integer> iter = set.descendingIterator();
            if (iter.hasNext()) {
                iter.next();
                if (iter.hasNext()) {
                    iter.next();
                    if (iter.hasNext()) {
                        third = iter.next().intValue();
                    }
                }
            }
        }
        return third;
    }

Of-course, if you want to ignore the conditions imposed by the original question, you could get the third largest element using the stream API

IntStream.of(2, 4, 1, 3, 5)
         .boxed()
         .sorted(Collections.reverseOrder())
         .collect(Collectors.toList()).get(2)