LeetCode #2939 — MEDIUM

Maximum Xor Product

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Greedy · Bit Manipulation

Example 1

12
5
4

Example 2

6
7
5

Example 3

1
6
3

Core Insight

What unlocks the optimal approach

Iterate over bits from most significant to least significant.
For the <code>ith</code> bit, if both <code>a</code> and <code>b</code> have the same value, we can always make <code>x</code>’s <code>ith</code> bit different from <code>a</code> and <code>b</code>, so <code>a ^ x</code> and <code>b ^ x</code> both have the <code>ith</cod> bit set.
Otherwise, we can only set the <code>ith</code> bit of one of <code>a ^ x</code> or <code>b ^ x</code>. Depending on the previous bits of <code>a ^ x</code> or <code>b ^ x</code>, we should set the smaller value’s <code>ith</code> bit.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2939: Maximum Xor Product
class Solution {
    public int maximumXorProduct(long a, long b, int n) {
        final int mod = (int) 1e9 + 7;
        long ax = (a >> n) << n;
        long bx = (b >> n) << n;
        for (int i = n - 1; i >= 0; --i) {
            long x = a >> i & 1;
            long y = b >> i & 1;
            if (x == y) {
                ax |= 1L << i;
                bx |= 1L << i;
            } else if (ax < bx) {
                ax |= 1L << i;
            } else {
                bx |= 1L << i;
            }
        }
        ax %= mod;
        bx %= mod;
        return (int) (ax * bx % mod);
    }
}

// Accepted solution for LeetCode #2939: Maximum Xor Product
func maximumXorProduct(a int64, b int64, n int) int {
	const mod int64 = 1e9 + 7
	ax := (a >> n) << n
	bx := (b >> n) << n
	for i := n - 1; i >= 0; i-- {
		x, y := (a>>i)&1, (b>>i)&1
		if x == y {
			ax |= 1 << i
			bx |= 1 << i
		} else if ax < bx {
			ax |= 1 << i
		} else {
			bx |= 1 << i
		}
	}
	ax %= mod
	bx %= mod
	return int(ax * bx % mod)
}

# Accepted solution for LeetCode #2939: Maximum Xor Product
class Solution:
    def maximumXorProduct(self, a: int, b: int, n: int) -> int:
        mod = 10**9 + 7
        ax, bx = (a >> n) << n, (b >> n) << n
        for i in range(n - 1, -1, -1):
            x = a >> i & 1
            y = b >> i & 1
            if x == y:
                ax |= 1 << i
                bx |= 1 << i
            elif ax > bx:
                bx |= 1 << i
            else:
                ax |= 1 << i
        return ax * bx % mod

// Accepted solution for LeetCode #2939: Maximum Xor Product
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2939: Maximum Xor Product
// class Solution {
//     public int maximumXorProduct(long a, long b, int n) {
//         final int mod = (int) 1e9 + 7;
//         long ax = (a >> n) << n;
//         long bx = (b >> n) << n;
//         for (int i = n - 1; i >= 0; --i) {
//             long x = a >> i & 1;
//             long y = b >> i & 1;
//             if (x == y) {
//                 ax |= 1L << i;
//                 bx |= 1L << i;
//             } else if (ax < bx) {
//                 ax |= 1L << i;
//             } else {
//                 bx |= 1L << i;
//             }
//         }
//         ax %= mod;
//         bx %= mod;
//         return (int) (ax * bx % mod);
//     }
// }

// Accepted solution for LeetCode #2939: Maximum Xor Product
function maximumXorProduct(a: number, b: number, n: number): number {
    const mod = BigInt(1e9 + 7);
    let ax = (BigInt(a) >> BigInt(n)) << BigInt(n);
    let bx = (BigInt(b) >> BigInt(n)) << BigInt(n);
    for (let i = BigInt(n - 1); ~i; --i) {
        const x = (BigInt(a) >> i) & 1n;
        const y = (BigInt(b) >> i) & 1n;
        if (x === y) {
            ax |= 1n << i;
            bx |= 1n << i;
        } else if (ax < bx) {
            ax |= 1n << i;
        } else {
            bx |= 1n << i;
        }
    }
    ax %= mod;
    bx %= mod;
    return Number((ax * bx) % mod);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.