Introduction

In this blog post, we will explore how to solve the LeetCode problem "2939" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given three integers a, b, and n, return the maximum value of (a XOR x) (b XOR x) where 0 <= x < 2n. Since the answer may be too large, return it modulo 109 + 7. Note that XOR is the bitwise XOR operation. Example 1: Input: a = 12, b = 5, n = 4 Output: 98 Explanation: For x = 2, (a XOR x) = 14 and (b XOR x) = 7. Hence, (a XOR x) (b XOR x) = 98. It can be shown that 98 is the maximum value of (a XOR x) (b XOR x) for all 0 <= x < 2n. Example 2: Input: a = 6, b = 7 , n = 5 Output: 930 Explanation: For x = 25, (a XOR x) = 31 and (b XOR x) = 30. Hence, (a XOR x) (b XOR x) = 930. It can be shown that 930 is the maximum value of (a XOR x) (b XOR x) for all 0 <= x < 2n. Example 3: Input: a = 1, b = 6, n = 3 Output: 12 Explanation: For x = 5, (a XOR x) = 4 and (b XOR x) = 3. Hence, (a XOR x) (b XOR x) = 12. It can be shown that 12 is the maximum value of (a XOR x) * (b XOR x) for all 0 <= x < 2n. Constraints: 0 <= a, b < 250 0 <= n <= 50

Explanation

Here's a breakdown of the solution strategy, followed by the code:

• Maximize Individual XORs: The core idea is to maximize (a XOR x) and (b XOR x) independently as much as possible within the constraint 0 <= x < 2^n. We iterate through the bits from the most significant bit (n-1) down to 0.

• Greedy Bit Setting: For each bit position, we check if setting the corresponding bit in x would cause a and b to have different bits at that position after the XOR operation. If they are the same originally, setting the bit in x makes them different, which generally leads to a larger result. If they are different, we don't set the bit, as keeping them different could result in a smaller product.

• Modulo Arithmetic: The problem requires taking the modulo at the end due to the potentially large result.

• Runtime & Storage Complexity: O(n) time complexity, O(1) space complexity.

Code

    def solve():
    a, b, n = map(int, input().split())

    x = 0
    for i in range(n - 1, -1, -1):
        bit_mask = 1 << i

        if (a & bit_mask) != (b & bit_mask):
            continue
        else:
            x |= bit_mask

    result = (a ^ x) * (b ^ x)
    return result % (10**9 + 7)

print(solve())