Finding the largest element in an array is a fundamental operation in Java programming. It is useful across many real-world applications, including statistical calculations and algorithmic problems.
An array of numbers (integers or floats), the task is to find and print the largest element in the array.
Approach 1: Using a for Loop (Basic Method)
This is the most common way to find the largest element.
// Java Program to Find the Largest
// Element of an Array
public class LargestElement {
public static void main(String[] args) {
int[] arr = {10, 45, 2, 67, 34};
// Assume first element is the largest
int max = arr[0];
for (int i = 1; i < arr.length; i++) {
if (arr[i] > max) {
// Update max if current is greater
max = arr[i];
}
}
System.out.println("Largest Element: " + max);
}
}
Output:
Largest Element: 67Approach 2: Using for-each Loop
More concise and readable with Java’s enhanced loop.
// Java Program to Find the Largest
// Element of an Array
public class LargestElement {
public static void main(String[] args) {
int[] arr = {3, 22, 11, 56, 19};
int max = arr[0];
for (int num : arr) {
if (num > max) {
max = num;
}
}
System.out.println("Largest Element: " + max);
}
}
Approach 3: Taking Input from User Using Scanner
Let the user define the array size and elements at runtime.
// Java Program to Find the Largest
// Element of an Array
import java.util.Scanner;
public class LargestElement {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter Number of Elements: ");
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
System.out.print("Enter " + (i+1) + " Element: ");
arr[i] = sc.nextInt();
}
int max = arr[0];
for (int i = 1; i < n; i++) {
if (arr[i] > max) {
max = arr[i];
}
}
System.out.println("Largest Element: " + max);
sc.close();
}
}
Approach 4: Using Java 8 Streams (Java 8+)
Java Stream API allows for a more functional approach.
// Java Program to Find the Largest
// Element of an Array
import java.util.Arrays;
public class LargestElement {
public static void main(String[] args) {
int[] arr = {12, 99, 34, 23, 87};
// or use .orElse(Integer.MIN_VALUE)
int max = Arrays.stream(arr).max().orElseThrow();
System.out.println("Largest Element: " + max);
}
}
Approach 5: Sorting the Array
Not the most efficient way if you only want the max, but still valid.
// Java Program to Find the Largest
// Element of an Array
import java.util.Arrays;
public class LargestElement {
public static void main(String[] args) {
int[] arr = {5, 9, 1, 12, 6};
Arrays.sort(arr);
int max = arr[arr.length - 1];
System.out.println("Largest Element: " + max);
}
}
Note:
- Time complexity is O(n log n) due to sorting, which is slower than a single pass (O(n)).
Approach 6: Using Collections (If Array is Converted to List)
Useful for arrays of wrapper types like Integer[].
// Java Program to Find the Largest
// Element of an Array
import java.util.Collections;
import java.util.Arrays;
import java.util.List;
public class LargestElement {
public static void main(String[] args) {
Integer[] arr = {18, 42, 7, 26, 91};
List<Integer> list = Arrays.asList(arr);
int max = Collections.max(list);
System.out.println("Largest Element: " + max);
}
}
Edge Case Handling
1. Empty Array
int[] arr = {};
// Should handle or throw exception2. Array with Negative Numbers
int[] arr = {-4, -10, -2, -11};
// Output should still be -2 (largest)3. Single Element
int[] arr = {42};
// Output: 42Always validate input arrays before processing!