Find Maximum Value of an Array in Java

What you’ll learn

  • How linear scan keeps a running maximum.
  • How to write a reusable findMax method in Java.
  • Why this works for negatives and duplicates too.

Prerequisites

Java arrays, loops, conditions, and console output.

  • You can iterate an array with for and access values by index.
  • You understand that maximum means the greatest numeric value in the list.

The idea

Keep one current winner (max). Scan left to right and replace it whenever you see a bigger value. One pass is enough.

Live preview

Using sample array: 14, 7, 25, 31, 10, 42.

Live result
Press "Find maximum".

Algorithm

Initialize

Set max to the first element.

Scan

Loop from index 1 to the end.

Update

If current value is larger, replace max.

Return

Return final max.

📜 Pseudocode

Pseudocode
function findMax(arr):
    max = arr[0]
    for i from 1 to arr.length - 1:
        if arr[i] > max:
            max = arr[i]
    return max
1

Find maximum (reference program)

java
public class Main {
    static int findMax(int[] arr) {
        int maxVal = arr[0];
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] > maxVal) {
                maxVal = arr[i];
            }
        }
        return maxVal;
    }

    public static void main(String[] args) {
        int[] array = {14, 7, 25, 31, 10, 42};
        int maxValue = findMax(array);
        System.out.println("Maximum value in the array: " + maxValue);
    }
}
📤 Output
Maximum value in the array: 42
2

All elements are negative

java
public class Main {
    static int findMax(int[] arr) {
        int maxVal = arr[0];
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] > maxVal) {
                maxVal = arr[i];
            }
        }
        return maxVal;
    }

    public static void main(String[] args) {
        int[] negatives = {-9, -3, -1, -7};
        System.out.println("Maximum (least negative): " + findMax(negatives));
    }
}
📤 Output
Maximum (least negative): -1

Optimization tips

Keep one pass only; sorting is unnecessary when you just need maximum.

Initialize from first element to avoid sentinel pitfalls.

❓ FAQ

It picks an initial candidate from actual data. After one pass, max stores the largest value seen.
There is no maximum in an empty array. Check arr.length first and handle that case.
Yes. The same comparison logic works, and the largest (least negative) value is returned.
The algorithm returns the maximum value itself; duplicates do not change the answer.
No for this task. Sorting is usually slower than a single O(n) scan when you only need max.
One pass over n elements: O(n) time and O(1) extra space.

Input/output examples

Input arrayOutput
[14, 7, 25, 31, 10, 42]42
[-9, -3, -1, -7]-1

Edge cases

Empty

No first element

Define behavior for empty arrays: throw, sentinel, or OptionalInt.

Single

One element

That element is the maximum.

Duplicate

Same max repeated

Result remains the same maximum value.

⏱️ Time and space complexity

TaskTimeExtra space
Find maximum in array of size nO(n)O(1)

Summary

  • Track one running maximum while scanning once.
  • The approach naturally handles negatives and duplicates.
  • Complexity is O(n) time and O(1) extra space.
Did you know?

Finding the maximum in an unsorted array needs at least n − 1 comparisons in the worst case. A single left-to-right scan achieves O(n) time and O(1) extra space.

About the author

Mari Selvan M P
Mari Selvan M P 🔗

Developer, cloud engineer, and technical writer

  • Experience 12 years building web and cloud systems
  • Focus Full Stack Development, AWS, and Developer Education

I write practical tutorials so students and working developers can learn by doing—from databases and APIs to deployment on AWS.

AWS Certified Developer – Associate (Credly)

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