How to Find Largest Element in Array in Java

The Challenge

Find the maximum (largest) element in an array!

Solutions

Method 1: Using Loop

java

int[] numbers = {5, 2, 8, 1, 9};
int max = numbers[0];

for (int i = 1; i < numbers.length; i++) {
    if (numbers[i] > max) {
        max = numbers[i];
    }
}

System.out.println("Largest: " + max);
// Output: Largest: 9

Method 2: Using Arrays.sort()

java

import java.util.Arrays;

int[] numbers = {5, 2, 8, 1, 9};
Arrays.sort(numbers);
int max = numbers[numbers.length - 1];

System.out.println("Largest: " + max);
// Output: Largest: 9

Method 3: Using Java 8 Streams

java

import java.util.Arrays;

int[] numbers = {5, 2, 8, 1, 9};
int max = Arrays.stream(numbers).max().getAsInt();

System.out.println("Largest: " + max);
// Output: Largest: 9

Did You Know?

You can find both min and max in a single loop by tracking both values simultaneously!

Pro Tip

Find both min and max efficiently:

java

int min = numbers[0], max = numbers[0];
for (int num : numbers) {
    if (num < min) min = num;
    if (num > max) max = num;
}

Real-World Use Cases

E-commerce: Find highest price
Weather: Find maximum temperature
Gaming: Find high score
Analytics: Find peak value in dataset

›Short versionOpen if you only have a minute.

One pass, starting from the first element.
int max = 0 fails when every element is negative: the largest of {-5, -1, -8} is -1, and 0 is not in the array.
Sorting to read the last index is O(n log n) and mutates the array.
An empty array has no maximum.

Largest element

Find the highest value by scanning the array once.

Working approach

Reject an empty array.
Set max to values[0].
Replace it when a later element is greater.

Walk the sample

Values 4, 9, 2
max starts at 4
9 replaces it
2 does not
The result is 9

The trap

A zero initializer looks harmless on the sample {4, 9, 2} and returns 0 for an array of negative scores.

java

static int largest(int[] values) {
    if (values.length == 0) {
        throw new IllegalArgumentException("empty array");
    }
    int max = values[0];
    for (int i = 1; i < values.length; i++) {
        if (values[i] > max) {
            max = values[i];
        }
    }
    return max;
}

What does a max that starts at 0 return for {-5, -1, -8}?

0, if you only update when a value is greater than the current max. None of those numbers is greater than 0.

The largest element is -1. Start from values[0]. A sort would also find it, and would reorder the array to do it.

Time O(n), space O(1).

Easy to get wrong

Initializing max to 0.
Sorting when a linear scan is enough.