Maximum XOR for Each Query

Maximum XOR for Each Query

What is this problem about?

The "Maximum XOR for Each Query" coding problem typically involves a given array of non-negative integers and a specified maximumBit value. For each query, you need to find a non-negative integer k such that k is less than 2^maximumBit and the bitwise XOR of all elements currently in the array, along with k, results in the maximum possible value. After each query, the last element of the array is removed. This problem challenges your ability to efficiently compute XOR sums over changing subarrays and to determine the optimal k for maximization, often hinting at the use of prefix XORs and bit manipulation.

Why is this asked in interviews?

This Maximum XOR for Each Query interview question is frequently asked by companies like Uber and Google because it tests a candidate's understanding of bitwise operations, particularly XOR, and how to optimize computations over dynamically changing arrays. It evaluates your ability to leverage the properties of XOR (e.g., A ^ A = 0, A ^ B ^ B = A) to efficiently update XOR sums without re-calculating from scratch. Furthermore, finding the optimal k to maximize the XOR sum requires a good grasp of bit manipulation strategy – knowing how to flip bits to achieve the largest possible result. It's a robust problem that highlights your analytical skills and grasp of array, bit manipulation, and prefix sum interview patterns.

Algorithmic pattern used

The "Maximum XOR for Each Query" problem is efficiently solved using Prefix XORs and Bit Manipulation.

1. Prefix XORs: Compute the prefix XORs of the given array. Let prefix_xor[i] be the XOR sum of nums[0] through nums[i-1]. The total XOR sum of the current array (which initially is the entire array) can be found using the last prefix XOR value. As elements are removed from the end, the current XOR sum effectively becomes prefix_xor[current_length].
2. Finding Optimal k: For a given current_xor_sum of the array, we want to find k < 2^maximumBit such that current_xor_sum ^ k is maximized. To maximize this, for each bit position from maximumBit - 1 down to 0:
   • If the (bit_pos)-th bit of current_xor_sum is 0, we want the (bit_pos)-th bit of k to be 1 to make the resulting XOR sum's bit 1.
   • If the (bit_pos)-th bit of current_xor_sum is 1, we want the (bit_pos)-th bit of k to be 0 to make the resulting XOR sum's bit 1. Essentially, k should be the bitwise complement of current_xor_sum within the maximumBit range. We construct k by setting bits where current_xor_sum has 0 and unsetting where current_xor_sum has 1. We also need to ensure k does not exceed 2^maximumBit - 1. This array, bit manipulation, and prefix sum interview pattern allows for an efficient solution by avoiding recalculations.

Example explanation

Suppose nums = [0, 1, 1, 3], maximumBit = 2. 2^maximumBit = 2^2 = 4. So k must be < 4 (i.e., 0, 1, 2, 3).

1. Initial Array: [0, 1, 1, 3]

   • XOR Sum: 0 ^ 1 ^ 1 ^ 3 = 3. (Binary 011_2)
   • To maximize 3 ^ k where k < 4:
     • We want k to have 0 at bit 0 and 1 at bit 1 (complement of 011_2's last two bits 11).
     • Optimal k is 0 (Binary 00_2) to get 3^0 = 3. Or 1 to get 3^1 = 2. Or 2 to get 3^2 = 1. Or 3 to get 3^3 = 0.
     • Wait, the rule is to make current_xor_sum ^ k as large as possible. current_xor_sum = 3 (011_2) Desired k for max XOR sum ^ k (within maximumBit = 2):
       • Bit 1 (MSB): current_xor_sum has 0. We want k to have 1.
       • Bit 0 (LSB): current_xor_sum has 1. We want k to have 0. So, k should be 10_2 = 2. 3 ^ 2 = 011_2 ^ 010_2 = 001_2 = 1. This is not maximal.

   Let target = (1 << maximumBit) - 1. target = 2^2 - 1 = 3 (binary 11_2). We want to find k such that (current_xor_sum ^ k) is maximized. This implies we want k to "flip" as many bits of current_xor_sum to 1 as possible. The ideal k would be ~current_xor_sum restricted to maximumBit bits. Let current_xor_sum = X. We want to maximize X ^ k. If X has bit b set, we want k to have bit b unset. If X has bit b unset, we want k to have bit b set. This means k should be X ^ ((1 << maximumBit) - 1). So, for current_xor_sum = 3 (011_2) and maximumBit = 2: k = 3 ^ ((1 << 2) - 1) = 3 ^ (4 - 1) = 3 ^ 3 = 0. Resulting XOR value: 3 ^ 0 = 3. This seems correct.

2. Remove 3 (last element): Array [0, 1, 1]

   • XOR Sum: 0 ^ 1 ^ 1 = 0. (Binary 000_2)
   • Optimal k for 0 ^ k where k < 4:
     • k = 0 ^ ((1 << 2) - 1) = 0 ^ 3 = 3.
     • Resulting XOR value: 0 ^ 3 = 3.

3. Remove 1: Array [0, 1]

   • XOR Sum: 0 ^ 1 = 1. (Binary 001_2)
   • Optimal k for 1 ^ k where k < 4:
     • k = 1 ^ ((1 << 2) - 1) = 1 ^ 3 = 2.
     • Resulting XOR value: 1 ^ 2 = 3.

The answers would be [3, 3, 3]. This example clarifies finding the optimal k for the Maximum XOR for Each Query coding problem.

Common mistakes candidates make

A common mistake in the "Maximum XOR for Each Query" coding problem is inefficiently recalculating the XOR sum of the array for each query, leading to an O(N*Q) solution which can be too slow. Another frequent error is incorrectly determining the optimal k to maximize the XOR sum; candidates might guess k values instead of systematically constructing k bit by bit or using the bitwise complement logic. Forgetting the k < 2^maximumBit constraint is also a pitfall, leading to k values outside the allowed range. Off-by-one errors in bit manipulation or prefix XOR calculations are also common.

Interview preparation tip

To ace the "Maximum XOR for Each Query" interview question, master prefix XORs and their properties for efficient range XOR queries. Develop a strong understanding of bit manipulation, particularly how to strategically set or unset bits in k to maximize an XOR result. Practice constructing target numbers bit by bit. Always write down the binary representations of numbers for small examples to trace your logic carefully. This array, bit manipulation, and prefix sum interview pattern is a strong indicator of your foundational algorithmic knowledge.