How to Find Second Largest Element in Array in Java

The Challenge

Find the second largest element in an array!

Solutions

Method 1: Using Two Variables

java

int[] numbers = {5, 2, 8, 1, 9, 3};
int largest = Integer.MIN_VALUE;
int secondLargest = Integer.MIN_VALUE;

for (int num : numbers) {
    if (num > largest) {
        secondLargest = largest;
        largest = num;
    } else if (num > secondLargest && num != largest) {
        secondLargest = num;
    }
}

System.out.println("Second Largest: " + secondLargest);
// Output: Second Largest: 8

Method 2: Using Sorting

java

import java.util.Arrays;

int[] numbers = {5, 2, 8, 1, 9, 3};
Arrays.sort(numbers);
int secondLargest = numbers[numbers.length - 2];

System.out.println("Second Largest: " + secondLargest);
// Output: Second Largest: 8

Did You Know?

The two-variable method is more efficient (O(n)) than sorting (O(n log n)) and works in a single pass!

Watch Out!

Handle duplicate values correctly! The condition num != largest ensures we don't count the same value twice.

Real-World Use Cases

Gaming: Find runner-up score
Competitions: Determine second place
Analytics: Find second-best performance

›Short versionOpen if you only have a minute.

One pass can track the largest and a distinct second.
When a new largest arrives, the old largest becomes the second.
A duplicate of the largest must not become the second.
Sorting and reading length - 2 mutates the array and treats a repeated maximum as the second value.

Second largest

Find the largest value that is strictly below the maximum.

Working approach

Need at least two distinct values. Otherwise there is no answer.
Update both trackers when a new maximum is seen.
Update only the second when the value sits strictly between the two.

Walk the sample

Values 5, 9, 9, 3, 8
9 becomes largest and 5 becomes second
The extra 9 is ignored
8 replaces 5
The result is 8

The trap

Integer.MIN_VALUE as a sentinel cannot tell "no second value" from "the second value is Integer.MIN_VALUE". Sorting then taking the second index also returns the maximum twice when that maximum is duplicated.

java

static int secondLargest(int[] values) {
    if (values.length < 2) {
        throw new IllegalArgumentException("need two elements");
    }
    int largest = values[0];
    Integer second = null;
    for (int i = 1; i < values.length; i++) {
        int value = values[i];
        if (value > largest) {
            second = largest;
            largest = value;
        } else if (value < largest && (second == null || value > second)) {
            second = value;
        }
    }
    if (second == null) {
        throw new IllegalArgumentException("no distinct second");
    }
    return second;
}

For {9, 9, 9}, what is the second largest distinct value?

There is not one. Every element equals the maximum.

numbers[numbers.length - 2] after sorting is 9, the maximum again. The condition value != largest is what keeps a duplicate from occupying the second slot, and the method still has to say when that slot was never filled.

The one-pass scan is O(n) time and O(1) space. Sorting is O(n log n) and mutates.

Easy to get wrong

Returning the duplicated maximum as if it were second.
Using Integer.MIN_VALUE as the second value without a flag that says it was seen.