Find third largest no in Java

Question

I'm little confused with this simple program.I have to find third largest no in array.I have done some code but getting only second largest no problem in third largest no so please suggest me what is wrong with this solution:

class ArrayExample {
    public static void main(String[] args) {
        int secondlargest = Integer.MIN_VALUE;
        int thirdlargest = Integer.MIN_VALUE;
        int largest = Integer.MIN_VALUE;
        Scanner input = new Scanner(System.in);
        System.out.println("Enter array values: ");
        int arr[] = new int[5];
        for (int i = 0; i < arr.length; i++) {
            arr[i] = input.nextInt();
            if (largest < arr[i]) {
                secondlargest = largest;
                largest = arr[i];
            }
            if (secondlargest < arr[i] && largest != arr[i]) {
                thirdlargest = secondlargest;
                secondlargest = arr[i];
                if (thirdlargest < arr[i] && secondlargest != arr[i])
                    thirdlargest = arr[i];
            }

        }
        System.out.println("Second Largest number is: " + secondlargest
                + "\nThird largest number is=====" + thirdlargest);
    }
}

You could simply sort the array and return `array[2]` (if it is in descending order). — assylias, Sep 05 '12 at 10:45
@assylias You're right, but it's a lot less efficient. To be honest I don't think the OP cares about that though. — Aske B., Sep 05 '12 at 10:46
Is it homework? If it is - mention it, it will help us refer the question better, otherwise you are likely to get solutions based on more advanced algorithms (sorting, selection, min-heap) — amit, Sep 05 '12 at 10:48
@assylias what if the number at position 1 and 2 are the same value? — Andy Clark, Sep 05 '12 at 10:49
@AndyClark then they'd still be the 2nd and 3rd largest numbers! — Timm, Sep 05 '12 at 10:50
Sorting is typically O(nlogn) whereas you only need to scan through the array once to find the 3rd largest which is only `O(n)`. — P.P, Sep 05 '12 at 10:55
@KingsIndian: Yes and no. For constant 3 - yes, but for arbitrary k? I also doubt this question is about sorting, it seems like homework designed to teach students how to think with boolean expressions & conditions and basic loops. — amit, Sep 05 '12 at 10:58
@amit I think, in that case, we can choose the algorithm which is `min(nk,nlogn)` for an arbitrary k. i.e. sorting vs. scanning linearly. — P.P, Sep 05 '12 at 11:02
@KingsIndian: FYI, If you are after efficiency, I once addressed this question (for general/larger k) in [this thread](http://stackoverflow.com/q/10751953/572670). It can be done in O(nlogk) using a heap or in O(n) using selection algorithm. But I really think this is not what the question is really about. — amit, Sep 05 '12 at 11:05

score 12 · Accepted Answer · edited Sep 05 '12 at 10:58

I would try something like this:

if (largest < ar[i]) {
    thirdlargest = secondlargest;
    secondlargest = largest;
    largest = arr[i];
} else if (secondlargest < ar[i]) {
    thirdlargest = secondlargest;
    secondlargest = ar[i];
} else if (thirdlargest < ar[i]) {
    thirdlargest = ar[i];
}

Not tested but I think the second IF isn't needed anymore.

Code Explanation:

We are verifying that if an entered number is greater than largest then move the third, second and 1st largest values one level up. If an entered value is greater than 2nd largest and less than largest, then move 3 and 2 one level up. If entered values is greated than 3rd largest and less than 2nd largest then move 3rd largest to the entered value.

Thanks for the edit, this is what I had in mind while writing the code :) — Jens, Sep 05 '12 at 10:58

score 4 · Answer 2 · answered Sep 05 '12 at 11:13

4

Collections API. Here is an example:

    List list = Arrays.asList(new Integer[] {1, 2, 29, 4, 28, 6, 27, 8});
    Collections.sort(list);
    System.out.print(list.get(list.size()-3));

answered Sep 05 '12 at 11:13

Nos

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1
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score 2 · Answer 3 · answered Jun 24 '14 at 23:56

        if(firstLargest<array[num])
        {
            thirdLargest=secondLargest;
            secondLargest=firstLargest;
            firstLargest = array[num];
        }

        else if((secondLargest<array[num])&&(array[num]!=firstLargest))
        {
            thirdLargest=secondLargest;
            secondLargest = array[num];
        }

        else if((thirdLargest<array[num])&&(array[num]!=secondLargest))
        {
            thirdLargest = array[num];
        }

score 1 · Answer 4 · answered Sep 05 '12 at 10:46

1

Use a java list, sort it. Take the third element.

java.util.Collections.sort()

answered Sep 05 '12 at 10:46

NimChimpsky

46,453
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score 1 · Answer 5 · answered Sep 05 '12 at 10:52

for (int i = 0; i < arr.length; i++)
 {
        arr[i] = input.nextInt();
        if (largest < arr[i]) {
            secondlargest = largest;
            largest = arr[i];
            continue;
        }
        if (secondlargest <= arr[i] && largest > arr[i])
        {
            thirdlargest = secondlargest;
            secondlargest = arr[i];
            continue;
        }
        if (thirdlargest <= arr[i] && secondlargest > arr[i])
        {
                thirdlargest = arr[i];
        }

 }

score 1 · Answer 6 · answered Sep 05 '12 at 10:54

1

Use Integer array and then sort it using Collections and just pick the element you need:

Code:

System.out.println("Enter array values: ");
Integer arr[] = new Integer[5];
for (int i = 0; i < arr.length; i++) {
            arr[i] = input.nextInt();
}
List<Integer> list = Arrays.asList(arr);
Collections.sort(list);
System.out.println(list);

The output is:

[0, 1, 2, 3, 6]

So, now select the 3rd larget number as list.get(list.size()-3)).

You can also reverse sort the Collection. Check it's documentation.

answered Sep 05 '12 at 10:54

mtk

13,221
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Why? If you want to sort, use `Arrays.sort()`. No need to create a wrapper (`Integer`) objects & arrays, not to convert the array to a list. – amit Sep 05 '12 at 10:55
Just wanted to give a `Collections` answer. And also, the OP get's to know how simple Collections can be. – mtk Sep 05 '12 at 10:59

score 1 · Answer 7 · answered Sep 05 '12 at 10:59

If you want that code to work, I think the problem is here:

 if (secondlargest < arr[i] && largest != arr[i]) {
                thirdlargest = secondlargest;
                secondlargest = arr[i];
                if (thirdlargest < arr[i] && secondlargest != arr[i])
                    thirdlargest = arr[i];
            }

The issue is you are setting thirdLargest to be secondLargest, which has already been identified as less than arr[i]. You are then testing if thirdLargest is less than arr[i] (which it is guaranteed to be as it it has been set to second largest within the outer condition) and then setting it to arr[i]. Try removing the

if (thirdlargest < arr[i] && secondlargest != arr[i])
                        thirdlargest = arr[i];

and if that doesn't work try adding a third separate condition to cover cases where arr[i] is less than secondGreatest but greater then thirdGreatest. (see Jens answer above), something like :

score 0 · Answer 8 · answered Sep 05 '12 at 10:46

0

Just cycle through the whole array and keep track of the three largest numbers.

Or you could sort it and then return the third element from the top.

answered Sep 05 '12 at 10:46

Infiltrator

1,611
1
16
25

2

I think the cycling is what the OP tries. – Jens Sep 05 '12 at 10:49

score 0 · Answer 9 · answered Sep 05 '12 at 10:51

Once you have the array read in, just call Arrays.sort(array) on the array (so in your case Arrays.sort(arr); ). This will sort it into ascending order, then you can just retrieve the element arr.length-2

Full description here: http://www.homeandlearn.co.uk/java/sorting_arrays.html

Example:

    Scanner input = new Scanner(System.in);
            System.out.println("Enter array values: ");
            int arr[] = new int[5];
            for (int i = 0; i < arr.length; i++) {
                arr[i] = input.nextInt();
        }
    Arrays.sort(arr);
System.out.println("Second Largest number is: " + arr[4]
                + "\nThird largest number is=====" + arr[3]);

score 0 · Answer 10 · edited Apr 20 '16 at 07:44

0

Try this code,

public static void main(String[] args) {

    int arr[] = {67, 56, 87, 42};

    for (int i = 0; i <arr.length - 1; i++) {
        if (arr[i] < arr[i + 1]) {
            int swap = arr[i];
            arr[i] = arr[i + 1];
            arr[i + 1] = swap;       
        }
    }

    System.out.println("third highest element is: " + arr[2]);
}

edited Apr 20 '16 at 07:44

Dovydas Šopa

2,282
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answered Apr 20 '16 at 07:07

unknown

1
1

score 0 · Answer 11 · edited Jul 28 '17 at 16:23

package algo;

public class LargestNumbers {

    public static void main(String args[]){
        int arr[] = new int[]{5,2,3,4,6};

        int largest=Integer.MIN_VALUE;;
        int secondLargest=Integer.MIN_VALUE;;
        int thirdLargest=Integer.MIN_VALUE;;


        for(int i=0;i<arr.length;i++){

              if(largest<arr[i])
                {
                    thirdLargest=secondLargest;
                    secondLargest=largest;
                    largest = arr[i];
                }

               else if((secondLargest<arr[i])&&(arr[i]!=largest))
                {
                    thirdLargest=secondLargest;
                    secondLargest = arr[i];
                }

               else if((thirdLargest<arr[i])&&(arr[i]!=secondLargest))
                {
                    thirdLargest = arr[i];
                }

        }//for


         System.out.println("Numbers are: " + largest + " "  + secondLargest
                    + "\nThird largest number is=====" + thirdLargest);
    }

}

This is really substandard input. Your formatting is broken; and just dumping code; without any further explanations; also not a good idea. — GhostCat, Jul 17 '17 at 14:08

score 0 · Answer 12 · answered Oct 02 '21 at 08:23

package arrays;

import java.util.Arrays;
public class ArraythirdHighestvalues {
    //method 1 
    public static  int highest(int []arr,int length)
    {
        int temp;
        for(int i=0;i<length;i++)
        {
            for(int j=i+1;j<length;j++)
            {
                if(arr[i]>arr[j])
                {
                    temp=arr[i];
                    arr[i]=arr[j];
                    arr[j]=temp;
                }
            }
        }
        return arr[length-2];
    }
    public static void main(String[] args) {
        int arr[]= {1,3,5,6,7,2,};
        System.out.println(highest(arr,arr.length));
        System.out.println("method 2 to find array second higest element value --");
        Arrays.parallelSort(arr);
        System.out.println(arr[arr.length-3]);
        
        
        
    }
    
}

score 0 · Answer 13 · answered Oct 02 '21 at 11:14

package algo;

public class LargestNumbers {

public static void main(String args[]){
    int arr[] = new int[]{5,2,3,4,6};

    int largest=Integer.MIN_VALUE;;
    int secondLargest=Integer.MIN_VALUE;;
    int thirdLargest=Integer.MIN_VALUE;;


    for(int i=0;i<arr.length;i++){

          if(largest<arr[i])
            {
                thirdLargest=secondLargest;
                secondLargest=largest;
                largest = arr[i];
            }

           else if((secondLargest<arr[i])&&(arr[i]!=largest))
            {
                thirdLargest=secondLargest;
                secondLargest = arr[i];
            }

           else if((thirdLargest<arr[i])&&(arr[i]!=secondLargest))
            {
                thirdLargest = arr[i];
            }

    }//for


     System.out.println("Numbers a`enter code here`re: " + largest + " "  + secondLargest
                + "\nThird largest number is=====" + thirdLargest);
}

}

Mohammed Imran · Answer 14 · 2022-04-19T13:12:37.063

0

Stream.of(23, 11, 96, 33, 89, 63, 17, 44 ,67, 55, 7, 76,42).sorted(Comparator.reverseOrder()).distinct().limit(3).skip(2).findFirst().get();

Initialize Arrays.asList(new int[]{23, 11, 9}).stream()

edited Apr 19 '22 at 13:12

answered Apr 19 '22 at 12:14

Mohammed Imran

59
1
9

Find third largest no in Java

14 Answers14

Code: