Introduction

In this blog post, we will explore how to solve the LeetCode problem "260" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer array nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order. You must write an algorithm that runs in linear runtime complexity and uses only constant extra space. Example 1: Input: nums = [1,2,1,3,2,5] Output: [3,5] Explanation: [5, 3] is also a valid answer. Example 2: Input: nums = [-1,0] Output: [-1,0] Example 3: Input: nums = [0,1] Output: [1,0] Constraints: 2 <= nums.length <= 3 * 104 -231 <= nums[i] <= 231 - 1 Each integer in nums will appear twice, only two integers will appear once.

Explanation

Here's a breakdown of the solution and the code:

• XOR Operation: The core idea is to use the XOR (exclusive OR) operation. XORing all elements in the array cancels out the pairs, leaving us with the XOR of the two unique numbers.
• Finding the Differentiating Bit: We find the rightmost set bit (the bit that is 1) in the XOR result. This bit indicates a position where the two unique numbers differ.

• Grouping and XORing: We divide the numbers into two groups based on whether they have that differentiating bit set or not. XORing the numbers in each group gives us the two unique numbers.

• Time & Space Complexity: O(n) time complexity, O(1) space complexity.

Code

    def singleNumber(nums):
    """
    Finds the two elements that appear only once in an array where all other elements appear exactly twice.

    Args:
        nums: A list of integers.

    Returns:
        A list containing the two unique elements.
    """

    # 1. XOR all elements to get XOR of the two unique numbers.
    xor_sum = 0
    for num in nums:
        xor_sum ^= num

    # 2. Find the rightmost set bit in xor_sum. This bit differentiates the two unique numbers.
    rightmost_bit = xor_sum & -xor_sum

    # 3. Divide the numbers into two groups based on whether they have the rightmost_bit set or not.
    num1 = 0
    num2 = 0
    for num in nums:
        if num & rightmost_bit:
            num1 ^= num
        else:
            num2 ^= num

    return [num1, num2]