How do you solve Maximum XOR for Each Query on LeetCode?

Maximum XOR for Each Query (LeetCode #1829) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The maximum possible XOR value is always 2^(maximumBit) - 1.

What is the time complexity of Maximum XOR for Each Query?

The optimal solution for Maximum XOR for Each Query runs in O(n) time and uses O(n) space. This is efficient because we only traverse the nums array once, calculating the cumulative XOR and the result in a single pass.

What is the brute force solution for Maximum XOR for Each Query?

The brute force approach for Maximum XOR for Each Query is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will compute the XOR of all elements in the array and then try every possible value of k (from 0 to 2^maximumBit - 1) to find the maximum XOR result. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum XOR for Each Query?

Maximum XOR for Each Query uses the following concepts: Array, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Bit Manipulation, Prefix Sum.

What are the interview tips for Maximum XOR for Each Query?

Always consider the properties of the operations involved (like XOR) to simplify your approach. Think about cumulative results and how they can save computation time. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Maximum XOR for Each Query?

Not recognizing that the maximum XOR value is fixed based on maximumBit. Overcomplicating the solution by trying to brute-force every possible k instead of leveraging properties of XOR.

#1829

Maximum XOR for Each Query

Medium

ArrayBit ManipulationPrefix SumBit ManipulationPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Instead of checking every possible k, we can leverage the property of XOR and the maximum possible value we can achieve. The maximum XOR value for any prefix is 2^maximumBit - 1, and we can directly compute the result for each query based on the cumulative XOR.

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Algorithm

3 steps

1Step 1: Initialize totalXor to 0 and create an empty result array.
2Step 2: Iterate through nums in reverse order, updating totalXor with each element.
3Step 3: For each query, calculate k as (2^maximumBit - 1) ^ totalXor and store it in the result array.

solution.py8 lines

1def maximizeXor(nums, maximumBit):
2    total_xor = 0
3    result = []
4    max_k = (1 << maximumBit) - 1
5    for num in reversed(nums):
6        total_xor ^= num
7        result.append(total_xor ^ max_k)
8    return result[::-1]

ℹ

Complexity note: This is efficient because we only traverse the nums array once, calculating the cumulative XOR and the result in a single pass.

1The maximum possible XOR value is always 2^(maximumBit) - 1.
2Using cumulative XOR allows us to compute results efficiently without iterating through all possible k values.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.