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, 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

    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