How do you solve Maximum Xor Product on LeetCode?

Maximum Xor Product (LeetCode #2939) 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 XOR operation can be manipulated to maximize individual bits, allowing for strategic choices in x.

What is the brute force solution for Maximum Xor Product?

The brute force approach for Maximum Xor Product is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible value of x from 0 to 2^n - 1. For each x, we calculate (a XOR x) * (b XOR x) and keep track of the maximum value found. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Xor Product?

Maximum Xor Product uses the following concepts: Math, Greedy, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Greedy Algorithms.

What are the interview tips for Maximum Xor Product?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process clearly while coding, especially when transitioning from brute force to optimal solutions. Practice bit manipulation problems, as they often appear in interviews and require a solid understanding of binary operations.

What are common mistakes when solving Maximum Xor Product?

Forgetting to apply the modulo operation, which can lead to incorrect results due to overflow. Not considering all possible values of x in the brute force approach, leading to missed maximum products.

#2939

Maximum Xor Product

Medium

MathGreedyBit ManipulationBit ManipulationGreedy Algorithms

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 approach leverages bit manipulation. By analyzing the bits of a and b, we can determine the best x that maximizes the product without needing to check every possible x. We focus on setting bits in x such that (a XOR x) and (b XOR x) yield the highest possible values.

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Algorithm

5 steps

1Step 1: Initialize max_product to 0.
2Step 2: Iterate over each bit position from the most significant to the least significant (up to n bits).
3Step 3: For each bit, determine if setting that bit in x can maximize the product based on the current bits of a and b.
4Step 4: Calculate the potential maximum product based on the chosen bits for x.
5Step 5: Return the maximum product modulo (10^9 + 7).

solution.py10 lines

1# Full working Python code
2
3def max_xor_product(a, b, n):
4    max_product = 0
5    for i in range(n - 1, -1, -1):
6        x = (1 << i)
7        product1 = (a ^ x) * (b ^ x)
8        product2 = (a ^ 0) * (b ^ 0)
9        max_product = max(max_product, product1, product2)
10    return max_product % (10**9 + 7)

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Complexity note: This complexity is efficient because we only iterate through the bits of n, making it linear with respect to the number of bits rather than the number of possible x values.

1The XOR operation can be manipulated to maximize individual bits, allowing for strategic choices in x.
2Understanding how to set bits in x based on the bits of a and b can lead to significant optimizations.

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