Minimum Number of Operations to Make Array XOR Equal to K

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Bit Manipulation

You are given a 0-indexed integer array nums and a positive integer k. You can apply the following operation on the array any number of times:

Choose any element of the array and flip a bit in its binary representation. Flipping a bit means changing a 0 to 1 or vice versa.

Return the minimum number of operations required to make the bitwise XOR of all elements of the final array equal to k.

Note that you can flip leading zero bits in the binary representation of elements. For example, for the number (101)₂ you can flip the fourth bit and obtain (1101)₂.

Example 1:

Input: nums = [2, 1, 3, 4], k = 1 Output: 2

Explanation: We can do the following operations:

Choose element 2 which is 3 == (011)₂, we flip the first bit and we obtain (010)₂ == 2. nums becomes [2, 1, 2, 4].
Choose element 0 which is 2 == (010)₂, we flip the third bit and we obtain (110)₂ = 6. nums becomes [6, 1, 2, 4]. The XOR of elements of the final array is (6 XOR 1 XOR 2 XOR 4) == 1 == k. It can be shown that we cannot make the XOR equal to k in less than 2 operations.

Example 2:

Input: nums = [2, 0, 2, 0], k = 0 Output: 0

Explanation: The XOR of elements of the array is (2 XOR 0 XOR 2 XOR 0) == 0 == k. So no operation is needed.

Minimum Number of Operations to Make Array XOR Equal to K

Medium

Google

1 view

Topics:

Bit Manipulation

You are given a 0-indexed integer array nums and a positive integer k. You can apply the following operation on the array any number of times:

Choose any element of the array and flip a bit in its binary representation. Flipping a bit means changing a 0 to 1 or vice versa.

Return the minimum number of operations required to make the bitwise XOR of all elements of the final array equal to k.

Note that you can flip leading zero bits in the binary representation of elements. For example, for the number (101)₂ you can flip the fourth bit and obtain (1101)₂.

Example 1:

Input: nums = [2, 1, 3, 4], k = 1 Output: 2

Explanation: We can do the following operations:

Choose element 2 which is 3 == (011)₂, we flip the first bit and we obtain (010)₂ == 2. nums becomes [2, 1, 2, 4].
Choose element 0 which is 2 == (010)₂, we flip the third bit and we obtain (110)₂ = 6. nums becomes [6, 1, 2, 4]. The XOR of elements of the final array is (6 XOR 1 XOR 2 XOR 4) == 1 == k. It can be shown that we cannot make the XOR equal to k in less than 2 operations.

Example 2:

Input: nums = [2, 0, 2, 0], k = 0 Output: 0

Explanation: The XOR of elements of the array is (2 XOR 0 XOR 2 XOR 0) == 0 == k. So no operation is needed.

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 elements within the input array, and what is the range for K?
Can the input array be empty or null? If so, what should I return?
Is K guaranteed to be non-negative?
If it's impossible to achieve XOR equal to K through the allowed operations, what value should I return?
Is the array read-only, or am I allowed to modify it directly during the operation?

Brute Force Solution

Approach

The brute force method for this problem means trying every possible combination of changes we can make to the numbers. We want to find the fewest changes needed so that combining all the numbers using a special operation (XOR) equals a target number.

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

First, consider making no changes to any number at all and see if the combined result (XOR) is already equal to the target number. If it is, we're done - zero changes needed!
If not, then go through each number one at a time. For each number, consider changing it to every other possible number and check if doing so gets the combined result closer to the target, or even equals the target. Count how many changes it took to achieve this.
Next, explore changing two numbers at a time. Try every possible pair of numbers and change each pair into every possible combination of other numbers. Again, check the combined result after each change and keep track of the number of changes made.
Continue this pattern, increasing the number of numbers we change at the same time. For example, consider changing three numbers at a time, then four, and so on.
At each step, keep track of the fewest number of changes needed to make the combined result equal to the target number. If we find a combination of changes that needs fewer changes than what we have tracked before, then update it.
Eventually, we will have tried changing all the numbers in every possible way. The minimum number of changes that we tracked along the way is our answer.

Code Implementation

def min_operations_brute_force(numbers, target):
    array_length = len(numbers)
    minimum_operations = array_length + 1

    # Check initial XOR without changes.
    current_xor = 0
    for number in numbers:
        current_xor ^= number
    if current_xor == target:
        return 0

    for i in range(1 << (array_length * 7)):
        number_of_operations = 0
        temporary_numbers = numbers[:]
        temporary_index = 0
        bitmask = i

        # Iterate through array and bitmask to determine changes
        while temporary_index < array_length:
            change_value = bitmask & 127
            bitmask >>= 7

            # If change_value > 100, consider it to mean no change.
            if change_value <= 100:
                temporary_numbers[temporary_index] = change_value
                number_of_operations += 1

            temporary_index += 1

        current_xor = 0
        for number in temporary_numbers:
            current_xor ^= number

        # Update minimum operations needed
        if current_xor == target:
            minimum_operations = min(minimum_operations, number_of_operations)

    if minimum_operations > array_length:
        return -1

    return minimum_operations

Big(O) Analysis

Time Complexity

O(2^n * n) – The provided brute force approach considers all possible subsets of the array and, for each subset, tries all possible values for the elements in that subset. There are 2^n possible subsets. For each subset, in the worst case, we might need to iterate through the elements in the array which takes O(n) time, calculating the XOR sum and comparing it to the target. Therefore, the overall time complexity is O(2^n * n).

Space Complexity

O(1) – The brute force approach primarily involves iterating through the input array and potentially modifying elements. It keeps track of the minimum number of changes made so far. The number of changes tracked is stored in a single variable. Since the additional space used by the algorithm remains constant regardless of the input array size N, the space complexity is O(1).

Optimal Solution

Approach

The key to efficiently finding the minimum operations involves recognizing that we only care about the total combined effect of all the numbers. Instead of changing numbers one by one, we focus on the overall 'XOR' result and change it directly to what we need.

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

First, figure out what the combined XOR of all the numbers in the list is.
Next, calculate what the combined XOR needs to be to equal the target value.
Compare the two XOR values and count the number of bits that are different. This number directly tells you the minimum number of changes needed.

Code Implementation

def min_operations_to_make_array_xor_equal_to_k(numbers, target):
    combined_xor_value = 0

    for number in numbers:
        combined_xor_value ^= number

    # Calculate XOR needed to reach target
    required_xor_value = combined_xor_value ^ target

    # Count differing bits
    bit_difference_count = 0

    # Iterating from 0 to 31 since constraints imply 32 bit integers
    for bit_index in range(32):
        if (required_xor_value >> bit_index) & 1:
            # Count each differing bit as an operation
            bit_difference_count += 1

    return bit_difference_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array of size n once to calculate the XOR sum. Then, a bitwise XOR operation between two integers and bit counting are performed, both of which are constant time operations. Therefore, the dominant factor is the single iteration through the array, making the time complexity O(n).

Space Complexity

O(1) – The algorithm calculates the XOR of the array elements and compares it with the target value K. It primarily uses constant space to store the intermediate XOR results and the final count of differing bits. No auxiliary data structures, such as arrays or hash maps, are created that scale with the input size N (the number of elements in the input array). Therefore, the space complexity remains constant, independent of the input array's size.

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 if the array is empty or null, as no operations are needed to make the XOR equal to K.
Array with one element	If the array has only one element, check if it's equal to K; if not, return 1, otherwise return 0.
Array with all elements equal to K	Return 0, as the XOR sum is already K and no operations are needed.
Large input array exceeding memory constraints	Ensure the algorithm uses memory efficiently, possibly using bitwise operations and avoiding excessive data structures.
K is 0 and all array elements are 0	Return 0 since the XOR is already 0.
Integer overflow during XOR calculations with large numbers	Use appropriate data types (e.g., long) to prevent integer overflows during XOR calculations.
When no valid solution exist	If the XOR of array can never be K, return an error code or -1.
Maximum size array	Ensure the solution's time complexity can handle the constraints by precomputing XOR or using bit manipulation for faster computations.