In this article, we will discuss how to find second smallest number in an Arrays and List
1. Finding Second smallest number in an Arrays
We will follow below 2 approaches to get 2nd Smallest 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 anonymous Comparator object with logic for descending-order sorting of integer numbers
- And add actual Arrays (as List after converting) to newly created TreeSet object which will sort integer numbers in descending order
- Now, elements inside TreeSet object will be sorted according to descending-order and last element will be the smallest element
- Using last() method of SortedSet reference, we can get last element and that will be the smallest element
- Remove last element using remove() method of SortedSet reference and now last element in the TreeSet object will be the 2nd smallest number
FindSecondSmallestNumberInAnArraysUsingTreeSet.java
package in.bench.resources.second.smallest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.Comparator;
import java.util.SortedSet;
import java.util.TreeSet;
public class FindSecondSmallestNumberInAnArraysUsingTreeSet {
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 DESC order
SortedSet<Integer> sortedSet = new TreeSet<Integer>(
new Comparator<Integer>() {
@Override
public int compare(Integer i1, Integer i2) {
return Integer.compare(i2, i1);
}
}
);
// add int[] arrays elements to TreeSet
sortedSet.addAll(Arrays.asList(numbers));
// remove last element which will be largest number in an Arrays
sortedSet.remove(sortedSet.last());
int secondSmallestNumber = 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 smallest number in an Arrays is - "
+ secondSmallestNumber);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Second smallest number in an Arrays is - 2
Execution time - 2000000 ns
1.2 Using Arrays sorting approach
- First step is to sort Arrays of integer numbers using Arrays.sort() method in descending order
- Arrays.sort() method accepts 2 input-arguments
- 1st argument is the actual Arrays to be sorted
- 2nd argument is the anonymous Comparator object with logic for descending-order sorting of integer numbers
- After sorting, get 2nd smallest number by passing index of second-last element to integer Arrays
FindSecondSmallestNumberInAnArraysUsingSortMethod.java
package in.bench.resources.second.smallest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.Comparator;
public class FindSecondSmallestNumberInAnArraysUsingSortMethod {
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 elements in descending order
Arrays.sort(numbers, new Comparator<Integer>() {
@Override
public int compare(Integer i1, Integer i2) {
return Integer.compare(i2, i1);
}
});
// 2nd last element will be second smallest number in an Arrays
int secondSmallestNumber = 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 smallest number in an Arrays is - "
+ secondSmallestNumber);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Second smallest number in an Arrays is - 2
Execution time - 0 ns
2. Finding second smallest number in List
- First step is to sort List of integer numbers using Collections.sort() method in descending order
- Collections.sort() method accepts 2 input-arguments
- 1st argument is the actual List to be sorted
- 2nd argument is the anonymous Comparator object with logic for descending-order sorting of integer numbers
- After sorting, get 2nd smallest number by passing index of second-last element to integer List
FindSecondSmallestNumberInListUsingCollectionsSortMethod.java
package in.bench.resources.second.smallest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
public class FindSecondSmallestNumberInListUsingCollectionsSortMethod {
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 descending order
Collections.sort(numbers, new Comparator<Integer>() {
@Override
public int compare(Integer i1, Integer i2) {
return Integer.compare(i2, i1);
}
});
// 2nd last element will be second smallest number in an Arrays
int secondSmallestNumber = 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 smallest number in List is - "
+ secondSmallestNumber);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Second smallest number in List is - 2
Execution time - 0 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
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Happy Coding !!
Happy Learning !!
