How do you solve Maximum XOR After Operations on LeetCode?

Maximum XOR After Operations (LeetCode #2317) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The ability to unset bits allows for flexibility in maximizing the XOR result.

What is the brute force solution for Maximum XOR After Operations ?

The brute force approach for Maximum XOR After Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible combination of applying the operation on each element with every possible value of x. This helps us understand the potential outcomes but is inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum XOR After Operations ?

Maximum XOR After Operations uses the following concepts: Array, Math, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Greedy.

What are the interview tips for Maximum XOR After Operations ?

Always start with a brute force approach to understand the problem better. Look for patterns in the problem, especially with bit manipulation. Explain your thought process clearly, as interviewers value understanding over just arriving at the correct answer.

What are common mistakes when solving Maximum XOR After Operations ?

Not considering the implications of unsetting bits and how it affects the final XOR. Overcomplicating the brute force approach by trying to optimize it without understanding the problem.

#2317

Maximum XOR After Operations

Medium

ArrayMathBit ManipulationBit ManipulationGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we can unset any bit in nums[i] by choosing appropriate values of x. This allows us to focus on maximizing the bits that can remain set in the final XOR result.

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Algorithm

4 steps

1Step 1: Initialize a variable to store the maximum XOR result.
2Step 2: Iterate through each number in the array and calculate the maximum possible XOR by considering the bits that can be set.
3Step 3: Use a bitmask to keep track of the bits that can be set based on the numbers in the array.
4Step 4: For each bit position, check if it can be set in the final result by ensuring at least one number can contribute that bit.

solution.py14 lines

1# Full working Python code
2
3def maxXOR(nums):
4    max_xor = 0
5    for i in range(31, -1, -1):
6        temp = max_xor | (1 << i)
7        found = False
8        for num in nums:
9            if (num & temp) == temp:
10                found = True
11                break
12        if found:
13            max_xor = temp
14    return max_xor

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Complexity note: This complexity is linear because we only iterate through the array a fixed number of times (once for each bit position) rather than checking every possible combination.

1The ability to unset bits allows for flexibility in maximizing the XOR result.
2Focusing on the highest bits first helps in constructing the maximum possible XOR efficiently.

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