Maximum XOR for Each Query

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ArraysBit Manipulation

You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times:

Find a non-negative integer k < 2^maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the i^th query.
Remove the last element from the current array nums.

Return an array answer, where answer[i] is the answer to the i^th query.

Example 1:

Input: nums = [0,1,1,3], maximumBit = 2
Output: [0,3,2,3]
Explanation: The queries are answered as follows:
1^st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
2^nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
3^rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
4^th query: nums = [0], k = 3 since 0 XOR 3 = 3.

Example 2:

Input: nums = [2,3,4,7], maximumBit = 3
Output: [5,2,6,5]
Explanation: The queries are answered as follows:
1^st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
2^nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
3^rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
4^th query: nums = [2], k = 5 since 2 XOR 5 = 7.

Example 3:

Input: nums = [0,1,2,2,5,7], maximumBit = 3
Output: [4,3,6,4,6,7]

Constraints:

nums.length == n
1 <= n <= 10⁵
1 <= maximumBit <= 20
0 <= nums[i] < 2^maximumBit
nums is sorted in ascending order.

Try coding it in LeetCode

Maximum XOR for Each Query

Medium

Asked by:

9 views

Topics:

ArraysBit Manipulation

You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times:

Find a non-negative integer k < 2^maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the i^th query.
Remove the last element from the current array nums.

Return an array answer, where answer[i] is the answer to the i^th query.

Example 1:

Input: nums = [0,1,1,3], maximumBit = 2
Output: [0,3,2,3]
Explanation: The queries are answered as follows:
1^st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
2^nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
3^rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
4^th query: nums = [0], k = 3 since 0 XOR 3 = 3.

Example 2:

Input: nums = [2,3,4,7], maximumBit = 3
Output: [5,2,6,5]
Explanation: The queries are answered as follows:
1^st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
2^nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
3^rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
4^th query: nums = [2], k = 5 since 2 XOR 5 = 7.

Example 3:

Input: nums = [0,1,2,2,5,7], maximumBit = 3
Output: [4,3,6,4,6,7]

Constraints:

nums.length == n
1 <= n <= 10⁵
1 <= maximumBit <= 20
0 <= nums[i] < 2^maximumBit
nums is sorted in ascending order.

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 are the constraints on the size of the `nums` array and the value of `maximumBit`?
Can `nums` contain zero values, and what are the constraints on the values within `nums`?
What should I return if `nums` is initially empty?
Could you provide an example where the answer for a query is zero?
Is it safe to assume that `maximumBit` will always be a non-negative integer?

Brute Force Solution

Approach

The problem asks us to find, for each number we're given, another number that, when combined with it in a special way (XOR), gives us the biggest possible result. The brute force method involves trying every single possibility to find the best match.

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

For the very first number we're given, consider all other possible numbers we could combine it with.
Perform the special combination (XOR) with each of these numbers.
Compare all the results from these combinations, and keep track of the biggest one you find.
That biggest result is the answer for the first number.
Now, move on to the next number we were given and repeat the entire process.
Keep doing this for every single number we're given, until we've found the biggest possible combination for each one.

Code Implementation

def maximum_xor_for_each_query_brute_force(numbers, maximum_bit):
    maximum_possible_xor = (1 << maximum_bit) - 1
    results = []

    for current_number in numbers:
        maximum_xor = 0

        # Trying out all possible values to maximize XOR
        for possible_number in range(maximum_possible_xor + 1):

            xor_result = current_number ^ possible_number

            # Updating the max XOR result if needed.
            maximum_xor = max(maximum_xor, xor_result)

        results.append(maximum_xor)

    return results

Big(O) Analysis

Time Complexity

O(n²) – The outer loop iterates through each of the n numbers in the input array. For each number, the inner part of the algorithm iterates through all the other numbers in the array to find the maximum XOR. This means for each of the n elements, it performs approximately n XOR operations and comparisons. Therefore, the total number of operations is proportional to n * n, resulting in a time complexity of O(n²).

Space Complexity

O(1) – The provided solution iterates through the input numbers and calculates the XOR for each pair to find the maximum. It only stores a temporary variable to hold the current maximum XOR result, and indices for iterating through the input. This does not depend on the number of inputs, N. Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The key is to realize we don't need to try every number for each query. We can precompute the XOR of all numbers in the initial set, then efficiently determine the maximum XOR value for each query using this precomputed result and some bit manipulation.

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

First, calculate the XOR sum of all the numbers originally given to us.
Then, for each query, determine the maximum possible number based on the number of bits given (think of it as creating the largest possible number with that many bits, consisting of all 1s).
Finally, find the XOR of the precomputed XOR sum and the maximum possible number we just calculated; this will give you the answer to the query.

Code Implementation

def maximum_xor_for_each_query(numbers, maximum_bit):
    xor_sum = 0
    for number in numbers:
        xor_sum ^= number

    results = []
    for _ in range(len(numbers)):

        # Calculate the maximum possible number with maximum_bit.
        maximum_possible_number = (1 << maximum_bit) - 1

        # XOR with maximum value to get the maximum XOR.
        maximum_xor_value = xor_sum ^ maximum_possible_number

        results.append(maximum_xor_value)

        # Since the input array is not actually used
        # and the results are the same for each number
        # we can break early
        break
    return results

Big(O) Analysis

Time Complexity

O(n + q) – First, we iterate through the input array of size n once to calculate the initial XOR sum. Then, for each of the q queries, we perform a constant amount of bit manipulation operations to determine the maximum possible number and calculate the XOR result. Therefore, the time complexity is dominated by the initial XOR calculation (O(n)) plus the q constant-time operations for each query (O(q)). Combining these gives a total time complexity of O(n + q).

Space Complexity

O(1) – The algorithm calculates a precomputed XOR sum, stores the maximum possible number for each query, and then XORs these values. These operations use a fixed number of integer variables, irrespective of the size of the input array or the number of queries. Thus, the auxiliary space used is constant and does not scale with the input size N, where N is the number of elements in the initial set. Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
nums is null or empty	Return an empty list immediately as there are no elements to process.
maximumBit is 0	Return an array of zeros, since any k would have to be 0, and XORing with 0 will minimize the result at 0.
nums contains only zeros	The maximum XOR will be (2^maximumBit - 1) for each query, since k can be (2^maximumBit - 1).
nums contains values close to 2^maximumBit	Ensure the XOR calculation doesn't overflow and that the chosen 'k' remains within the defined range [0, 2^maximumBit - 1].
Maximum allowed input size for nums (e.g., 10^5)	The solution should be efficient, avoiding excessive memory allocation or nested loops that would result in TLE (Time Limit Exceeded).
All elements in nums are identical	Calculate the maximum XOR value between the repeated element and k for each query and consider the iterative removal of elements.
Large maximumBit value causing potential integer overflow during 2^maximumBit calculation	Use appropriate data types (e.g., long) or bitwise operations to prevent integer overflow when calculating the upper bound of k.
maximumBit is greater than the maximum number of bits representable by numbers in `nums`	The solution should still compute correct values, effectively padding the smaller numbers with leading zeros when calculating XOR, therefore the k would need to be chosen such that all bits up to maximumBit are flipped.