Introduction

In this blog post, we will explore how to solve the LeetCode problem "1829" 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 a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times: Find a non-negative integer k < 2maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the ith query. Remove the last element from the current array nums. Return an array answer, where answer[i] is the answer to the ith query. Example 1: Input: nums = [0,1,1,3], maximumBit = 2 Output: [0,3,2,3] Explanation: The queries are answered as follows: 1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3. 2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3. 3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3. 4th query: nums = [0], k = 3 since 0 XOR 3 = 3. Example 2: Input: nums = [2,3,4,7], maximumBit = 3 Output: [5,2,6,5] Explanation: The queries are answered as follows: 1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7. 2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7. 3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7. 4th query: nums = [2], k = 5 since 2 XOR 5 = 7. Example 3: Input: nums = [0,1,2,2,5,7], maximumBit = 3 Output: [4,3,6,4,6,7] Constraints: nums.length == n 1 <= n <= 105 1 <= maximumBit <= 20 0 <= nums[i] < 2maximumBit nums​​​ is sorted in ascending order.

Explanation

Here's the breakdown of the solution:

• XOR Prefix: Calculate the XOR of all elements in the current nums array. This allows us to efficiently find the k that maximizes the XOR sum in each query.
• Maximizing XOR: To maximize the XOR sum, k should be chosen such that XOR sum ^ k = (2^maximumBit - 1). Therefore, k = XOR sum ^ (2^maximumBit - 1).

• Iterative Removal: After each query, remove the last element from nums to simulate the problem statement.

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

Code

    def get_max_xor(nums, maximumBit):
    n = len(nums)
    answer = []
    xor_sum = 0
    for num in nums:
        xor_sum ^= num

    for _ in range(n):
        k = xor_sum ^ ((1 << maximumBit) - 1)
        answer.append(k)
        xor_sum ^= nums[-1]
        nums.pop()

    return answer