# Solving Leetcode Interviews in Seconds with AI: Single Number II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "137" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 where every element appears three times except for one, which appears exactly once. Find the single element and return it. You must implement a solution with a linear runtime complexity and use only constant extra space.   Example 1: Input: nums = [2,2,3,2] Output: 3 Example 2: Input: nums = [0,1,0,1,0,1,99] Output: 99    Constraints:  1 <= nums.length <= 3 * 104 -231 <= nums[i] <= 231 - 1 Each element in nums appears exactly three times except for one element which appears once.  

	# Explanation
	Here's the solution to find the single number in an array where every other number appears three times, adhering to linear time complexity and constant space constraints:

*   **Bit Manipulation:** The core idea is to use bit manipulation to track the count of each bit position (0 to 31) across all numbers in the array.
*   **Modulo 3:** Since each number appears three times (except the single number), the count of each bit position will be a multiple of 3 if only the repeating numbers are considered. Therefore, taking the count modulo 3 will give us the corresponding bit of the single number.
*   **Reconstruct Number:** We then reconstruct the single number from its constituent bits.

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

	
	# Code
	```python
	def singleNumber(nums):
    """
    Finds the single number in an array where every other number appears three times.

    Args:
        nums: A list of integers.

    Returns:
        The single number.
    """

    result = 0
    for i in range(32):  # Iterate through each bit position (0 to 31)
        bit_sum = 0
        for num in nums:
            bit_sum += (num >> i) & 1  # Extract the i-th bit of the number
        
        bit_sum %= 3  # Take modulo 3 to find the remainder

        result |= (bit_sum << i) # Set the i-th bit of the result

    # Handle negative numbers (two's complement)
    if result >= 2**31:
        result -= 2**32

    return result
	```