How do you solve Maximum XOR of Subsequences on LeetCode?

Maximum XOR of Subsequences (LeetCode #3681) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max)) time complexity and O(n) space complexity. Key insight: XOR is maximized when bits differ.

What is the time complexity of Maximum XOR of Subsequences?

The optimal solution for Maximum XOR of Subsequences runs in O(n log(max)) time and uses O(n) space. Building the basis takes O(n log(max)), where max is the largest number in nums. Space is used to store the basis.

What is the brute force solution for Maximum XOR of Subsequences?

The brute force approach for Maximum XOR of Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible subsequences and calculate their XOR values. Then, find the maximum XOR of any two subsequences. The optimal Optimal Solution approach improves this to O(n log(max)).

What algorithm or data structure is used to solve Maximum XOR of Subsequences?

Maximum XOR of Subsequences uses the following concepts: Array, Math, Greedy, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum XOR of Subsequences?

Explain your thought process clearly. Practice bit manipulation problems. Understand the properties of XOR.

What are common mistakes when solving Maximum XOR of Subsequences?

Forgetting to handle empty subsequences. Not considering overlapping subsequences.

#3681

Maximum XOR of Subsequences

Hard

ArrayMathGreedyBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log(max))Space O(n)

Use a linear basis to represent the numbers in terms of their binary representation, allowing efficient calculation of maximum XOR.

⚙️

Algorithm

3 steps

1Step 1: Build a basis for the binary representation of the numbers.
2Step 2: Use the basis to compute the maximum XOR possible.
3Step 3: Return the maximum XOR value derived from the basis.

solution.py11 lines

1def maxXOR(nums):
2    basis = []
3    for num in nums:
4        for b in basis:
5            num = min(num, num ^ b)
6        if num > 0:
7            basis.append(num)
8    max_xor = 0
9    for b in basis:
10        max_xor = max(max_xor, max_xor ^ b)
11    return max_xor

ℹ

Complexity note: Building the basis takes O(n log(max)), where max is the largest number in nums. Space is used to store the basis.

1XOR is maximized when bits differ.
2Building a basis helps efficiently compute maximum XOR.

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