Hey, coding enthusiasts! 😃 Let’s dive into a unique problem that brings together bitwise manipulation and array queries. XOR operations are at the core of this challenge, and with a little understanding of binary numbers, we can quickly see how to solve it effectively. This post will not only explain the problem but will also show you the power of bitwise tricks in competitive programming. Let’s get started!

📝 Problem Statement Recap

You’re given a sorted array nums of n non-negative integers and a value called maximumBit. You need to answer n queries, each involving a different subset of the array, where the subset changes by removing the last element from the array with each query.

For each query, you need to find an integer k that maximizes the XOR result when XORing together:

• The elements of the current array subset, and
• The integer k.

Constraints

1. nums.length == n
2. 1 <= n <= 10^5
3. 1 <= maximumBit <= 20
4. nums[i] < 2^maximumBit

🛠 Key Insights to Solve the Problem

To solve this problem, we’ll need to make use of binary representations and properties of the XOR operation.

🔍 Insight #1: Maximizing XOR with k

For any XOR operation, the result is maximized when the two operands have opposite bits. To maximize the XOR result, we should XOR our cumulative XOR value with a binary number that has all bits set to 1 within the limit of maximumBit.

So, what’s the maximum possible number we can achieve with maximumBit bits?

• The largest number represented with maximumBit bits is 2^maximumBit - 1.
• This number has all bits set to 1 up to maximumBit. For example:
  • If maximumBit = 3, then 2^3 - 1 = 7, which is 111 in binary.
  • If maximumBit = 4, then 2^4 - 1 = 15, which is 1111 in binary.

Let’s call this number max_xor.

By XORing our cumulative XOR with max_xor, we “flip” all bits, maximizing the result for each query.

🔍 Insight #2: Cumulative XOR of the Array

Rather than recalculating the XOR from scratch in each query, we can use a cumulative XOR approach:

1. Start with the full array and compute the XOR of all elements.
2. For each query, instead of recalculating the XOR from scratch, we can simply remove the last element’s effect by XORing it again (since XORing a number twice negates it).
3. This gives an efficient way to track the XOR as the array shrinks for each query.

📚 Step-by-Step Solution with Binary Explanation

Let’s walk through the solution, incorporating binary representations to help clarify each step.

1. Calculate max_xor: Since we’re given maximumBit, our goal for each query will be to achieve a result that is as close as possible to max_xor, which is 2^maximumBit - 1. In binary, this value has all bits set to 1, maximizing the XOR potential for each query.

2. Compute Initial Cumulative XOR: Start by XORing all elements in the array to get the cumulative XOR value for the entire array. Let’s store this in a variable, cumulative_xor.

3. Process Each Query in Reverse Order:
   • We process the array in reverse because each query removes the last element from nums.
   • For each query:
     • Calculate the optimal k by XORing cumulative_xor with max_xor. This k will maximize the XOR result for this subset.
     • Append k to the answer list.
     • Update cumulative_xor by removing the last element’s contribution from it.
4. Reverse and Return the Result: Since we process the queries from the end of nums to the beginning, we need to reverse the answer list before returning it.

🧑‍💻 Code Implementation with Explanation

Here’s the code for this approach with detailed comments to clarify each step:

def getMaximumXor(nums, maximumBit):
    # Step 1: Calculate max_xor as the largest number within maximumBit bits
    max_xor = (1 << maximumBit) - 1  # e.g., if maximumBit = 3, max_xor = 7 (binary 111)

    # Step 2: Calculate the cumulative XOR of all elements in nums
    cumulative_xor = 0
    for num in nums:
        cumulative_xor ^= num

    # Step 3: Initialize the answer array and fill it in reverse
    answer = []
    for num in reversed(nums):
        # Optimal k for the current cumulative XOR is cumulative_xor ^ max_xor
        answer.append(cumulative_xor ^ max_xor)
        
        # Update cumulative_xor by removing the effect of the last element in reversed order
        cumulative_xor ^= num
    
    # Step 4: Since we processed queries in reverse, reverse the answer array to correct order
    return answer

🔢 Example Walkthrough in Binary

Let’s work through an example using binary representations to visualize the XOR calculations.

Example 1

Input: nums = [2, 3, 4, 7], maximumBit = 3
Output: [5, 2, 6, 5]

Step-by-Step Execution with Binary

1. Calculate max_xor:
   • maximumBit = 3
   • 2^3 - 1 = 7, which is 111 in binary.
2. Initial Cumulative XOR Calculation:
   • nums = [2, 3, 4, 7]
   • Cumulative XOR (binary):

     2:   010
     XOR 3: 011
     XOR 4: 100
     XOR 7: 111
     ----------------
     Result: 010 (binary) = 2 (decimal)
     

3. Processing Each Query:

   • 1st Query (nums = [2, 3, 4, 7]):
     • cumulative_xor = 2 (010 in binary)
     • k = cumulative_xor ^ max_xor = 2 XOR 7 = 5 (binary 101)
     • Remove 7 from cumulative XOR: cumulative_xor = 2 ^ 7 = 5
   • 2nd Query (nums = [2, 3, 4]):
     • cumulative_xor = 5 (binary 101)
     • k = cumulative_xor ^ max_xor = 5 XOR 7 = 2 (binary 010)
     • Remove 4: cumulative_xor = 5 ^ 4 = 1
   • 3rd Query (nums = [2, 3]):
     • cumulative_xor = 1 (binary 001)
     • k = cumulative_xor ^ max_xor = 1 XOR 7 = 6 (binary 110)
     • Remove 3: cumulative_xor = 1 ^ 3 = 2
   • 4th Query (nums = [2]):
     • cumulative_xor = 2 (binary 010)
     • k = cumulative_xor ^ max_xor = 2 XOR 7 = 5 (binary 101)

Final output: [5, 2, 6, 5]

🧪 Edge Cases to Consider

1. Single Element Array: If nums contains only one element, the solution should handle it without recalculating multiple XORs.
2. All Zeros in nums: If every element is 0, the cumulative_xor remains zero.
3. Large Input Size: Make sure the solution operates within O(n) complexity for large arrays.

🎉 Conclusion

This problem showcases the power of XOR in algorithm design. By leveraging cumulative XOR and binary manipulation, we efficiently solved each query while maintaining optimal performance. Bitwise operations might seem intimidating, but as we’ve seen, they can often provide elegant solutions.

Happy coding, and may you XOR your way to success! 🎉👩‍💻👨‍💻