In this article, we will discuss how to find largest number in an Arrays and List
1. Finding Largest number in an Arrays
We will follow below 3 approaches to get Largest number in an Arrays
- Using SortedSet and TreeSet
- Using Arrays sorting
- Using Iterative approach
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
FindLargestNumberInAnArraysUsingTreeSet.java
package in.bench.resources.largest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.SortedSet;
import java.util.TreeSet;
public class FindLargestNumberInAnArraysUsingTreeSet {
public static void main(String[] args) {
// random numbers
Integer[] numbers = {5, 9, 11, 2, 8, 21, 1};
// print to console
System.out.println("Original Integer Arrays - "
+ Arrays.toString(numbers));
// 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
);
// this will be largest number in an Arrays
int largestNumber = 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("\nLargest number in an Arrays is - "
+ largestNumber);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Original Integer Arrays - [5, 9, 11, 2, 8, 21, 1]
Largest number in an Arrays is - 21
Execution time - 0 ns
1.2 Using Arrays sorting approach
- First step is to sort Arrays of integer numbers using Arrays.sort() method in ascending order
- After sorting, get largest number by passing index of last element to integer Arrays
FindLargestNumberInAnArraysUsingSortMethod.java
package in.bench.resources.largest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
public class FindLargestNumberInAnArraysUsingSortMethod {
public static void main(String[] args) {
// random numbers
Integer[] numbers = {5, 9, 11, 2, 8, 21, 1};
// print to console
System.out.println("Original Integer Arrays - "
+ Arrays.toString(numbers));
// Execution - start time
LocalTime startTime = LocalTime.now();
// sort Arrays element in ascending order
Arrays.sort(numbers);
// last element will be second largest number in an Arrays
int largestNumber = numbers[numbers.length - 1];
// 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("\nLargest number in an Arrays is - "
+ largestNumber);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Original Integer Arrays - [5, 9, 11, 2, 8, 21, 1]
Largest number in an Arrays is - 21
Execution time - 0 ns
1.3 Using Iterative approach
- First assume 1st element of an Arrays is the largest number
- Then start iterating Arrays one-by-one using for-loop and compare assumed largest number with other iterating numbers and accordingly set largest number
FindLargestNumberInAnArraysUsingIterativeApproach.java
package in.bench.resources.largest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
public class FindLargestNumberInAnArraysUsingIterativeApproach {
public static void main(String[] args) {
// random numbers
Integer[] numbers = {5, 9, 11, 2, 8, 21, 1};
// print to console
System.out.println("Original Integer Arrays - "
+ Arrays.toString(numbers));
// Execution - start time
LocalTime startTime = LocalTime.now();
// assume largest number
int max = numbers[0];
// iterate and find largest number
for(int index = 0; index < numbers.length; index++) {
if(numbers[index] > max) {
max = numbers[index];
}
}
// 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("\nLargest number in an Arrays is - "
+ max);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Original Integer Arrays - [5, 9, 11, 2, 8, 21, 1]
Largest number in an Arrays is - 21
Execution time - 0 ns
2. Finding Largest number in List
We will follow below 2 approaches to get Largest number in a List or ArrayList
- Using Collection sorting
- Using Iterative approach
2.1 Using Collection sorting approach
- First step is to sort List of integer numbers using Collections.sort() method in ascending order
- After sorting, get Largest number by passing index of last element to integer List
FindLargestNumberInListUsingCollectionsSortMethod.java
package in.bench.resources.largest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class FindLargestNumberInListUsingCollectionsSortMethod {
public static void main(String[] args) {
// random numbers
List<Integer> numbers = Arrays.asList(5, 9, 11, 2, 8, 21, 1);
// print to console
System.out.println("Original Integer List - "
+ numbers);
// Execution - start time
LocalTime startTime = LocalTime.now();
// sort List element in ascending order
Collections.sort(numbers);
// last element will be largest number in List
int largestNumber = numbers.get(numbers.size() - 1);
// 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("\nLargest number in List is - "
+ largestNumber);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Original Integer List - [5, 9, 11, 2, 8, 21, 1]
Largest number in List is - 21
Execution time - 0 ns
2.2 Using Iterative approach
- First assume 1st element of List is the largest number
- Then start iterating List one-by-one using for-loop and compare assumed largest number with other iterating numbers and accordingly set largest number
FindLargestNumberInListUsingIterativeApproach.java
package in.bench.resources.largest.number;
import java.time.Duration;
import java.time.LocalTime;
import java.util.Arrays;
import java.util.List;
public class FindLargestNumberInListUsingIterativeApproach {
public static void main(String[] args) {
// random numbers
List<Integer> numbers = Arrays.asList(5, 9, 11, 2, 8, 21, 1);
// print to console
System.out.println("Original Integer List - "
+ numbers);
// Execution - start time
LocalTime startTime = LocalTime.now();
// assume largest number
int max = numbers.get(0);
// iterate and find largest number
for(int index = 0; index < numbers.size(); index++) {
if(numbers.get(index) > max) {
max = numbers.get(index);
}
}
// 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("\nLargest number in List is - "
+ max);
// print execution time in Nano seconds
System.out.println("\nExecution time - "
+ differenceInNano + " ns");
}
}
Output:
Original Integer List - [5, 9, 11, 2, 8, 21, 1]
Largest number in List is - 21
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 !!
