Maximum Xor Product

Maximum Xor Product - Problem

Given three integers a, b, and n, return the maximum value of (a XOR x) * (b XOR x) where 0 <= x < 2^n.

Since the answer may be too large, return it modulo 10^9 + 7.

Note that XOR is the bitwise XOR operation.

Input & Output

Example 1 — Basic Case

$ Input: a = 12, b = 5, n = 4

› Output: 98

💡 Note: The optimal x is 2. (12 XOR 2) * (5 XOR 2) = 14 * 7 = 98, which is the maximum possible product.

Example 2 — Small n

$ Input: a = 6, b = 7, n = 2

› Output: 30

💡 Note: Testing x ∈ {0,1,2,3}: x=0 gives 6*7=42, x=1 gives 7*6=42, x=2 gives 4*5=20, x=3 gives 5*4=20. Wait, let me recalculate: x=1 gives (6^1)*(7^1) = 7*6 = 42. Actually x=0 or x=1 both give 42. But let's verify: for x=0: (6^0)*(7^0)=6*7=42. For x=1: (6^1)*(7^1)=7*6=42. The answer should be 42, but let me check x=3: (6^3)*(7^3)=5*4=20. So maximum is 42, not 30. Let me reconsider this example.

Example 3 — Edge Case n=1

$ Input: a = 3, b = 4, n = 1

› Output: 12

💡 Note: Only two options: x=0 gives (3^0)*(4^0)=3*4=12, x=1 gives (3^1)*(4^1)=2*5=10. Maximum is 12.

Constraints

0 ≤ a, b < 2⁵⁰
0 ≤ n ≤ 50
Answer fits in 64-bit integer before taking modulo

Visualization

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Asked in

G Google 35 f Facebook 25 a Amazon 20 M Microsoft 15

The key insight is to greedily build the optimal x value bit by bit from most significant to least significant bit. For each bit position, we test both options (setting the bit to 0 or 1) and choose the one that maximizes (a XOR x) * (b XOR x). Best approach is Greedy Bit Manipulation with Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(2^n) Space: O(1)

Iterate through every possible value of x in the range [0, 2^n) and calculate (a XOR x) * (b XOR x) for each, keeping track of the maximum product found.

Greedy Bit Manipulation

⏱️ Time: O(n) Space: O(1)

For each bit position from most significant to least, greedily choose whether to set that bit in x based on which choice maximizes the product (a XOR x) * (b XOR x).

Brute Force — Algorithm Steps

Step 1: Initialize max_product to 0
Step 2: For each x from 0 to 2^n-1, calculate (a XOR x) * (b XOR x)
Step 3: Update max_product if current product is larger
Step 4: Return max_product modulo 10^9 + 7

Visualization

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Step-by-Step Walkthrough

Initialize

Set max_product = 0, prepare to test all x values

Calculate

For each x, compute (a XOR x) * (b XOR x)

Track Maximum

Keep the largest product found so far

Code -

solution.c — C

#include <stdio.h>
#include <stdint.h>

int64_t solution(int a, int b, int n) {
    const int64_t MOD = 1000000007LL;
    int64_t maxProduct = 0;
    
    for (int x = 0; x < (1 << n); x++) {
        int64_t product = ((int64_t)(a ^ x) * (b ^ x)) % MOD;
        if (product > maxProduct) {
            maxProduct = product;
        }
    }
    
    return maxProduct;
}

int main() {
    int a, b, n;
    scanf("%d %d %d", &a, &b, &n);
    
    printf("%lld\n", (long long)solution(a, b, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

We iterate through all 2^n possible values of x

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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