Sort Array by Parity in Java

Sort Array by Parity in Java

This guide explains the intuition, optimized approach, and Java implementation for sort array by parity in java, with practical tips for interviews and production coding standards.

Problem

Reorder array so all even numbers come first, then odd numbers.

Two Pointer Partition

• left moves until odd
• right moves until even
• swap and continue

Java Solution

class Solution {
    public int[] sortArrayByParity(int[] nums) {
        int left = 0, right = nums.length - 1;

        while (left < right) {
            while (left < right && nums[left] % 2 == 0) left++;
            while (left < right && nums[right] % 2 != 0) right--;

            int temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
        }

        return nums;
    }
}

Complexity

• Time: O(n)
• Space: O(1)

Output order inside even or odd groups is not required to be stable.

Key Takeaways

• Start from the brute-force idea, then derive the optimized invariant.
• Use clean pointer/index boundaries to avoid off-by-one bugs.
• Validate against edge cases (empty input, single element, duplicates, extreme values).