Introduction

In this blog post, we will explore how to solve the LeetCode problem "3158" 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

You are given an array nums, where each number in the array appears either once or twice. Return the bitwise XOR of all the numbers that appear twice in the array, or 0 if no number appears twice. Example 1: Input: nums = [1,2,1,3] Output: 1 Explanation: The only number that appears twice in nums is 1. Example 2: Input: nums = [1,2,3] Output: 0 Explanation: No number appears twice in nums. Example 3: Input: nums = [1,2,2,1] Output: 3 Explanation: Numbers 1 and 2 appeared twice. 1 XOR 2 == 3. Constraints: 1 <= nums.length <= 50 1 <= nums[i] <= 50 Each number in nums appears either once or twice.

Explanation

Here's the solution to the problem, along with explanations:

• High-level approach:

  • Use a frequency counter (dictionary or array) to count the occurrences of each number.
  • Iterate through the frequency counter.
  • If a number appears twice, XOR it into the result.

• Complexity: O(n) runtime, O(1) storage. Since constraint mentions that 1 <= nums[i] <= 50, we can use a frequency array of fixed size which leads to O(1) storage complexity, even though it depends on the range of elements.

Code

    def xor_twice(nums):
    """
    Calculates the XOR of numbers that appear twice in an array.

    Args:
        nums: A list of integers where each number appears once or twice.

    Returns:
        The bitwise XOR of all numbers that appear twice, or 0 if no number appears twice.
    """

    counts = [0] * 51  # Frequency counter, considering numbers from 1 to 50
    for num in nums:
        counts[num] += 1

    result = 0
    for i in range(1, 51):
        if counts[i] == 2:
            result ^= i

    return result