2317. Maximum XOR After Operations

Problem Description

In this problem, we are given an array of integers nums, and our objective is to maximize the bitwise XOR of all elements of the array after performing a certain operation any number of times. The operation defined allows us to:

1. Choose any non-negative integer x.
2. Select any index i.
3. Update nums[i] to nums[i] AND (nums[i] XOR x).

Here AND and XOR are bitwise operations. Bitwise AND results in a bit being 1 if both bits at the same position are 1, otherwise it's 0. Bitwise XOR results in a bit being 1 if the bits at the same position differ, otherwise it's 0.

The main challenge is to figure out what x and i we should choose to maximize the final XOR of all elements.

Intuition

To solve this problem, we should understand the properties of the XOR and AND operations. An important fact to consider is the behavior of AND when combined with XOR: for any integer n, n AND (n XOR x) equals n. Armed with this knowledge, we can simplify the problem significantly.

Since applying the given operation does not change the value of any element in the array (because for any given i and x, nums[i] AND (nums[i] XOR x) will always equal nums[i]), the elements of nums can be left as they are. Thus, the operation has no effect on the outcome of maximizing the bitwise XOR of all array elements. This leads us to the conclusion that the maximum possible bitwise XOR can be obtained by simply computing the bitwise XOR of all elements in the array without modifying them.

To implement the solution, we use a reduce function from Python's functools module with or_, which is a function that performs the bitwise OR operation (imported from the operator module). By performing a bitwise OR between all numbers in nums, we effectively combine all the bits set in any number in the array. Since XOR of identical bits is 0 and 1 otherwise, the bitwise OR accumulates all the 1 bits across all the numbers, yielding the maximum XOR value achievable.

Learn more about Math patterns.

Solution Approach

The solution employs the reduce function, a Python technique that cumulatively applies an operation to all elements of an iterable. The specific operation used here is the bitwise OR, provided by the or_ function from the operator module.

Here's a step-by-step breakdown of the algorithm used in the solution:

1. Import the reduce function from the functools module and the or_ function from the operator module (this import is not shown in the given code, but is required for or_ to be used).
2. Initialize the maximumXOR method, which takes nums, the list of integers.
3. Within the method, call reduce with two arguments:
   • The first argument is or_, the function that will be applied to the elements of nums.
   • The second argument is the nums list itself.
4. reduce then applies the or_ function cumulatively to the items of nums, from left to right. This means that it starts with the first two elements a and b of nums, computes a | b, then applies the or_ operation to this result and the next element, and so on until all elements have been included.
5. Since the bitwise OR accumulates all bits set in any number in nums, the final result represents the maximum possible XOR that can be obtained, as each bit position in the final result is set to 1 if it's set in at least one number in the array.
6. The result of the reduce operation is the maximum XOR value, which is returned by the maximumXOR method.

In summary, the given solution cleverly sidesteps the need to perform any complex calculations by realizing that the maximum XOR is simply the accumulation of all the unique bits set in the original nums list, and it uses reduce with bitwise OR to calculate this result efficiently.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Suppose we have the following array of integers nums:

1nums = [3, 10, 5]

Based on the provided content, our goal is to maximize the bitwise XOR of all elements in the array by performing defined operations. However, since we know that the operations won't change any value (as n AND (n XOR x) equals n), the process simplifies to finding the XOR of all numbers as they are.

The binary representations of the numbers in nums are as follows:

13  ->  011
210 -> 1010
35  ->  101

Now let's compute the bitwise OR (not XOR, as incorrectly mentioned in the content, but it's intended to be the OR operation based on the explanation) of all elements in nums:

• Start with the first two numbers: 3 (011) | 10 (1010) = 11 (1011).

• Now, take the previous OR result and apply it to the next number: 11 (1011) | 5 (0101) = 15 (1111).

The result of 15 (1111) in binary indicates that all bit positions have been set to 1. This is the maximum value we can obtain from a bitwise XOR of any subsequence of these numbers because all possible 1 bits are included in the result.

Thus, using the solution approach, we would simply call reduce with the or_ function:

1from functools import reduce
2from operator import or_
3
4nums = [3, 10, 5]
5max_xor_value = reduce(or_, nums)
6print(max_xor_value)  # Output should be 15

Indeed, the maximum possible bitwise XOR value for this array is 15, which is the result we obtained by applying the bitwise OR operation to all elements of nums using the reduce function. The computation of the result agrees perfectly with our manual calculation, illustrating the correctness and efficiency of the solution approach.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(N), where N is the number of elements in the list nums. This is because the function reduce applies the or_ operator (|) sequentially to the elements of the list, and applying the or_ operator takes constant time O(1) for each pair of integers.

The space complexity of the function is O(1), as it reduces all the elements to a single integer without using any additional space that grows with the input size.

Learn more about how to find time and space complexity quickly using problem constraints.