How do you solve Partition Array for Maximum XOR and AND on LeetCode?

Partition Array for Maximum XOR and AND (LeetCode #3630) 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: XOR is maximized by including high-value bits.

What is the time complexity of Partition Array for Maximum XOR and AND?

The optimal solution for Partition Array for Maximum XOR and AND runs in O(n) time and uses O(1) space. We compute total XOR in O(n) and iterate through nums once, resulting in O(n) overall complexity.

What is the brute force solution for Partition Array for Maximum XOR and AND?

The brute force approach for Partition Array for Maximum XOR and AND is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible subset for B and calculate the corresponding XOR and AND values for A and C. This exhaustive search guarantees we find the maximum sum. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Partition Array for Maximum XOR and AND?

Partition Array for Maximum XOR and AND uses the following concepts: Array, Math, Greedy, Bit Manipulation, Enumeration. The recommended patterns to study are: Bit Manipulation, Greedy.

What are the interview tips for Partition Array for Maximum XOR and AND?

Explain your thought process clearly. Consider edge cases like empty arrays. Practice bit manipulation problems.

What are common mistakes when solving Partition Array for Maximum XOR and AND?

Not considering empty subsequences. Confusing XOR and AND properties.

#3630

Partition Array for Maximum XOR and AND

Hard

ArrayMathGreedyBit ManipulationEnumerationBit ManipulationGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

By fixing B and calculating the XOR of the remaining elements, we can efficiently compute the maximum possible sum without checking all subsets.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total XOR of all elements in nums.
2Step 2: Iterate through each element, treating it as part of B, and compute AND for B and XOR for A and C.
3Step 3: Keep track of the maximum sum encountered.

solution.py10 lines

1def maxPartition(nums):
2    total_xor = 0
3    for num in nums:
4        total_xor ^= num
5    max_sum = 0
6    for num in nums:
7        andB = num
8        xorA = total_xor ^ num
9        max_sum = max(max_sum, andB + xorA)
10    return max_sum

ℹ

Complexity note: We compute total XOR in O(n) and iterate through nums once, resulting in O(n) overall complexity.

1XOR is maximized by including high-value bits.
2AND is maximized by including elements with common bits.

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