How to Find Smallest Element in Array in Java

The Challenge

Find the minimum (smallest) element in an array!

Solutions

Method 1: Using Loop

java

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

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

System.out.println("Smallest: " + min);
// Output: Smallest: 1

Method 2: Using Arrays.sort()

java

import java.util.Arrays;

int[] numbers = {5, 2, 8, 1, 9};
Arrays.sort(numbers);
int min = numbers[0];

System.out.println("Smallest: " + min);
// Output: Smallest: 1

Method 3: Using Java 8 Streams

java

import java.util.Arrays;

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

System.out.println("Smallest: " + min);
// Output: Smallest: 1

Did You Know?

The loop method is most efficient with O(n) time complexity. Sorting is O(n log n) but gives you a sorted array as a bonus!

Pro Tip

Use Integer.MAX_VALUE as initial min value to handle all cases safely!

Real-World Use Cases

E-commerce: Find lowest price
Weather: Find minimum temperature
Gaming: Find lowest score
Analytics: Find minimum value in dataset

›Short versionOpen if you only have a minute.

One pass is enough. Sorting is O(n log n) and reorders the array.
Start min at the first element, not at 0.
A min that starts at 0 stays 0 when every element is positive, so the result is not in the array.
An empty array has no first element. Refuse it.

Smallest element

Find the lowest value by scanning the array once.

Working approach

Reject a null or empty array.
Set min to values[0].
Replace min whenever a later element is smaller.

Walk the sample

Values 8, 3, 5
min starts at 8
3 replaces it
5 does not
The result is 3

The trap

int min = 0 never updates on {3, 8, 5}, because nothing is less than 0. The method returns a value that is not in the array.

java

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

Why does int min = 0 fail for {3, 8, 5}?

Every element is greater than 0, so the condition never fires and min stays 0. 0 is not in the array.

An all-negative array happens to update a 0 initializer, because those values are smaller than 0. The all-positive array does not. Start from the first element and both cases are covered.

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

Easy to get wrong

Initializing min to 0.
Sorting the array to read index 0.