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Count Odd Elements in an Array in Java

beginner· Arrays · Array

Problem

An odd number is any integer that leaves a remainder of 1 (or -1) when divided by 2.

Given an array of integers, count how many of its elements are odd.

Input

[4, 7, 10, 15, 22, 33, 8]

Output

Odd elements: 3

Java Program

Java

public class CountOdd {
    public static void main(String[] args) {
        int[] arr = {4, 7, 10, 15, 22, 33, 8};
        int count = 0;
        for (int num : arr) {
            if (num % 2 != 0) count++; // not divisible by 2
        }
        System.out.println("Odd elements: " + count);
    }
}

▶ Run Code

Output

Odd elements: 3

Core Logic

A single pass through the array, checking each number's remainder when divided by 2, tallies every odd value.

How It Works

1A for-each loop visits each element of the array in turn.
2num % 2 != 0 checks whether the current number does NOT divide evenly by 2.
3A match increments the count variable; anything else is skipped.
4After the loop, count holds the total number of odd elements.

For [4, 7, 10, 15, 22, 33, 8], the scan finds 7, 15, and 33 — three odd elements.

💡

Key Point: Using != 0 rather than == 1 matters for negative odd numbers — in Java, -7 % 2 evaluates to -1, not 1, so checking for exactly 1 would miss it.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: Each element is visited once, and only a single running counter is kept regardless of array size.

Key Concepts

modulo operatorfor-each loopcounter variable

Approach 2: Java 8

Java

import java.util.Arrays;

public class CountOddStream {
    public static void main(String[] args) {
        int[] arr = {4, 7, 10, 15, 22, 33, 8};
        // filter() keeps only odd values; count() reduces to a single total
        long count = Arrays.stream(arr).filter(num -> num % 2 != 0).count();
        System.out.println("Odd elements: " + count);
    }
}

▶ Run Code

Output

Odd elements: 3

Core Logic

The same divisibility check can filter a stream of the array's values down to just the odd ones, then count what's left.

How It Works

1Arrays.stream(arr) converts the int[] into an IntStream.
2.filter(num -> num % 2 != 0) keeps only the values NOT divisible by 2.
3.count() reduces the filtered stream down to a single long total.

Filtering [4, 7, 10, 15, 22, 33, 8] keeps 7, 15, 33, so count() returns 3.

💡

Key Point: Flipping == 0 to != 0 is the only change needed to turn the even-counting stream into this one — the rest of the pipeline stays identical.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: The stream still visits every element once, and count() reduces straight down to a single long without collecting anything.

Key Concepts

StreamIntStreamfilter()count()

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