Maximum XOR After Operations

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ArraysBit ManipulationGreedy Algorithms

You are given a 0-indexed integer array nums. In one operation, select any non-negative integer x and an index i, then update nums[i] to be equal to nums[i] AND (nums[i] XOR x).

Note that AND is the bitwise AND operation and XOR is the bitwise XOR operation.

Return the maximum possible bitwise XOR of all elements of nums after applying the operation any number of times.

Example 1:

Input: nums = [3,2,4,6]
Output: 7
Explanation: Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
It can be shown that 7 is the maximum possible bitwise XOR.
Note that other operations may be used to achieve a bitwise XOR of 7.

Example 2:

Input: nums = [1,2,3,9,2]
Output: 11
Explanation: Apply the operation zero times.
The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
It can be shown that 11 is the maximum possible bitwise XOR.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁸

Try coding it in LeetCode

Maximum XOR After Operations

Medium

Asked by:

8 views

Topics:

ArraysBit ManipulationGreedy Algorithms

You are given a 0-indexed integer array nums. In one operation, select any non-negative integer x and an index i, then update nums[i] to be equal to nums[i] AND (nums[i] XOR x).

Note that AND is the bitwise AND operation and XOR is the bitwise XOR operation.

Return the maximum possible bitwise XOR of all elements of nums after applying the operation any number of times.

Example 1:

Input: nums = [3,2,4,6]
Output: 7
Explanation: Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
It can be shown that 7 is the maximum possible bitwise XOR.
Note that other operations may be used to achieve a bitwise XOR of 7.

Example 2:

Input: nums = [1,2,3,9,2]
Output: 11
Explanation: Apply the operation zero times.
The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
It can be shown that 11 is the maximum possible bitwise XOR.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁸

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the numbers within the input array?
Can the input array contain negative numbers or zeros?
Are there any size constraints on the input array's length?
What type of operations are allowed? Are we limited to only XOR, or are other bitwise operations permissible?
Is the goal to maximize the XOR of all elements in the array *after* applying some number of allowed operations to each individual element, or is there some other objective?

Brute Force Solution

Approach

The brute force approach to this problem means we will try every possible combination of operations on the numbers we're given. For each combination, we will calculate the XOR sum and, in the end, pick the largest one we found.

Here's how the algorithm would work step-by-step:

Consider the first number. We can either modify it using the allowed operation or leave it as is. Let's explore both options.
For the second number, again, we have the choice to modify it or not, for each of the choices we made about the first number. This leads to exploring even more combinations.
We continue this process for all the numbers. Each number we encounter doubles the number of combinations we need to check because we either modify it or we don't.
After going through all the numbers and making a decision on each, we calculate the XOR sum of the numbers in that specific combination.
We store the XOR sum we just calculated, and compare against the largest value we have seen so far.
Once all possible combinations of modifications are explored, we will have the maximum possible XOR sum among all the possibilities.
Return the largest XOR sum we found. That will be the result.

Code Implementation

def maximum_xor_after_operations_brute_force(numbers):
    maximum_xor_sum = 0

    def calculate_xor_sum(current_numbers):
        xor_sum = 0
        for number in current_numbers:
            xor_sum ^= number
        return xor_sum

    def generate_combinations(index, current_numbers):
        nonlocal maximum_xor_sum

        # Base case: If we have processed all numbers,
        # calculate the XOR sum and update the maximum.
        if index == len(numbers):
            xor_sum = calculate_xor_sum(current_numbers)
            maximum_xor_sum = max(maximum_xor_sum, xor_sum)
            return

        # Option 1: Don't modify the current number
        generate_combinations(index + 1, current_numbers + [numbers[index]])

        # Option 2: Modify the current number
        # By setting it to 2 * the original value.

        modified_numbers = current_numbers + [2 * numbers[index]]

        generate_combinations(index + 1, modified_numbers)

    # Start generating combinations from the first number.
    generate_combinations(0, [])
    return maximum_xor_sum

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm iterates through all possible combinations of modifying or not modifying each of the n numbers in the input array. For each number, there are two choices: either apply the operation or don't. Therefore, the total number of combinations that need to be explored grows exponentially with the input size n. This leads to 2^n possible combinations, making the time complexity O(2^n).

Space Complexity

O(1) – The described brute force approach explores all possible combinations using recursion. While the description doesn't explicitly mention storing combination results, it implicitly tracks the 'largest value seen so far'. This requires a single variable. Therefore, the space used is dominated by a single variable to store the maximum XOR sum found, which takes constant space, independent of the number of input numbers N.

Optimal Solution

Approach

The problem asks us to maximize the XOR value we can get from a group of numbers by performing certain operations. The key insight is that we can set any bit in any number to 1. So, our goal is to turn every bit in the result to 1 if possible.

Here's how the algorithm would work step-by-step:

Consider each bit position separately, starting from the most significant (leftmost) bit.
Check if it's possible to make that bit a 1 in the final XOR result. This means checking if any of the numbers could potentially contribute a 1 to that bit after the operations.
Because we can always set any bit to 1, determine if at least one of the numbers, after some operations, has a 1 in that specific bit position.
If at least one number can contribute a 1 to that bit, the corresponding bit in the final XOR result should be 1. If none can, it should be 0.
Continue this process for all bit positions.
Once you have determined which bits can be 1, calculate the resulting maximum XOR value.

Code Implementation

def maximum_xor_after_operations(numbers):
    maximum_xor = 0

    # Iterate through each bit position to see if it can be set.
    for bit_position in range(32):
        # Represent current bit with mask
        bit_mask = 1 << bit_position
        found_one = False

        # Check if any number can contribute a '1' at this position.
        for number in numbers:
            if (number & bit_mask) != 0:
                found_one = True
                break

        # If a '1' can be present at this bit, include this bit in the result.
        if found_one:
            maximum_xor |= bit_mask

    # Return the maximum XOR value that can be obtained.
    return maximum_xor

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each number in the input array once. For each number, it performs a bitwise OR operation, accumulating the result. Since each number is processed exactly once, the time complexity is directly proportional to the number of elements in the input array. Therefore, the time complexity is O(n), where n is the number of elements in the input array.

Space Complexity

O(1) – The algorithm iterates through the bits of the numbers in place and does not use any auxiliary data structures that scale with the input size N (number of numbers). The operations involve only examining each bit and updating a final result which is stored in a constant amount of space. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 as no operations can be performed and the initial XOR sum is 0.
Array with a single element	The maximum XOR sum is simply the value of that single element.
All elements in the array are 0	The maximum XOR sum remains 0 as any operation on 0 yields 0.
All elements in the array are identical and non-zero	The maximum XOR sum will be the value of that element since operations can make all elements equal to this value.
Array contains a mix of very large and very small integers	The bitwise OR operation will propagate the set bits of the largest numbers to all other elements.
Integer overflow when performing the OR operation	Use a data type that can accommodate the maximum possible value of the bitwise OR across all numbers, like long in Java/C++ or equivalent in other languages.
Array contains negative numbers	Treat them as their 2's complement representation, allowing for bitwise operations to work correctly.
Very large input array size leading to potential memory issues	The solution operates on each element independently, so it scales linearly with the input size and memory usage will be determined by the size of the input array, which is unavoidable given the problem definition.