Java – Find Second Largest number in an Arrays or List ?

April 26, 2022 SJ Collection 0

In this article, we will discuss how to find second largest number in an Arrays and List

1. Finding second largest number in an Arrays

We will follow below 2 approaches to get 2nd Largest number in an Arrays

  • Using SortedSet and TreeSet
  • Using Arrays sorting

1.1 Using SortedSet & TreeSet approach

  • Create TreeSet object with SortedSet reference and pass actual Arrays (as List after converting) as constructor-argument to be sorted
  • Now, elements inside TreeSet object will be sorted according to natural-order and last element will be the largest element
  • Using last() method of SortedSet reference, we can get last element and that will be the largest element
  • Remove last element using remove() method of SortedSet reference and now the last element in the TreeSet object will be the 2nd largest number

FindSecondLargestNumberInAnArraysUsingTreeSet.java

package in.bench.resources.second.largest.number;

import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.SortedSet;
import java.util.TreeSet;

public class FindSecondLargestNumberInAnArraysUsingTreeSet {

	public static void main(String[] args) {

		// random numbers
		Integer[] numbers = {5, 9, 11, 2, 8, 21, 1};


		// Execution - start time
		LocalTime startTime = LocalTime.now();


		// sort Integer[] arrays using TreeSet - stores in ASC order
		SortedSet<Integer> sortedSet = new TreeSet<Integer>(
				Arrays.asList(numbers) // convert arrays to List
				);


		// remove last element which will be largest number in an Arrays
		sortedSet.remove(sortedSet.last());


		// now, this will be second largest number in an Arrays
		int secondLargestNumber = sortedSet.last();


		// Execution - end time
		LocalTime endTime = LocalTime.now();


		// find difference
		Duration duration = Duration.between(startTime, endTime);
		long differenceInNano = duration.getNano();


		// print sum to console
		System.out.println("Second largest number in an Arrays is - "
				+ secondLargestNumber);


		// print execution time in Nano seconds
		System.out.println("\nExecution time - "
				+ differenceInNano + " ns");
	}
}

Output:

Second largest number in an Arrays is - 11

Execution time - 2000000 ns

1.2 Using Arrays sorting approach

FindSecondLargestNumberInAnArraysUsingSortMethod.java

package in.bench.resources.second.largest.number;

import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;

public class FindSecondLargestNumberInAnArraysUsingSortMethod {

	public static void main(String[] args) {

		// random numbers
		Integer[] numbers = {5, 9, 11, 2, 8, 21, 1};


		// Execution - start time
		LocalTime startTime = LocalTime.now();


		// sort Arrays element in ascending order
		Arrays.sort(numbers);


		// 2nd last element will be second largest number in an Arrays
		int secondLargestNumber = numbers[numbers.length - 2];


		// Execution - end time
		LocalTime endTime = LocalTime.now();


		// find difference
		Duration duration = Duration.between(startTime, endTime);
		long differenceInNano = duration.getNano();


		// print sum to console
		System.out.println("Second largest number in an Arrays is - "
				+ secondLargestNumber);


		// print execution time in Nano seconds
		System.out.println("\nExecution time - "
				+ differenceInNano + " ns");
	}
}

Output:

Second largest number in an Arrays is - 11

Execution time - 1000000 ns

2. Finding second largest number in List

FindSecondLargestNumberInListUsingCollectionsSortMethod.java

package in.bench.resources.second.largest.number;

import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;

public class FindSecondLargestNumberInListUsingCollectionsSortMethod {

	public static void main(String[] args) {

		// random numbers
		List<Integer> numbers = Arrays.asList(5, 9, 11, 2, 8, 21, 1);


		// Execution - start time
		LocalTime startTime = LocalTime.now();


		// sort List element in ascending order
		Collections.sort(numbers);


		// 2nd last element will be second largest number in an Arrays
		int secondLargestNumber = numbers.get(numbers.size() - 2);


		// Execution - end time
		LocalTime endTime = LocalTime.now();


		// find difference
		Duration duration = Duration.between(startTime, endTime);
		long differenceInNano = duration.getNano();


		// print sum to console
		System.out.println("Second largest number in List is - "
				+ secondLargestNumber);


		// print execution time in Nano seconds
		System.out.println("\nExecution time - "
				+ differenceInNano + " ns");
	}
}

Output:

Second largest number in List is - 11

Execution time - 1000000 ns

3. Points to remember w.r.t execution time:

  • Execution time differs in different platforms
  • With small set of numbers, we may not find large difference in execution time
  • But with large set of numbers, difference will be significant to consider

Related Articles:

Happy Coding !!
Happy Learning !!

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