Introduction

In this blog post, we will explore how to solve the LeetCode problem "136" 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 a non-empty array of integers nums, every element appears twice except for one. Find that single one. You must implement a solution with a linear runtime complexity and use only constant extra space. Example 1: Input: nums = [2,2,1] Output: 1 Example 2: Input: nums = [4,1,2,1,2] Output: 4 Example 3: Input: nums = [1] Output: 1 Constraints: 1 <= nums.length <= 3 104 -3 104 <= nums[i] <= 3 * 104 Each element in the array appears twice except for one element which appears only once.

Explanation

Here's the solution to find the single number in an array where every other number appears twice, optimized for linear runtime and constant space:

• Leverage XOR: The XOR (exclusive OR) operation has the property that a ^ a = 0 and a ^ 0 = a. Therefore, XORing all elements in the array will cancel out the pairs, leaving only the single number.
• Iterate and XOR: Iterate through the array, applying the XOR operation cumulatively.

• Return the Result: The final result of the XOR operation will be the single number.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def singleNumber(nums):
    """
    Finds the single number in an array where every other number appears twice.

    Args:
        nums: A list of integers.

    Returns:
        The single number that appears only once.
    """
    result = 0
    for num in nums:
        result ^= num
    return result