#1829

Medium

LeetCode Problem Workspace

Maximum XOR for Each Query

Find the maximum XOR for each query based on a sorted array and bit manipulation.

Problem Statement

You are given a sorted array nums of n non-negative integers and an integer maximumBit. For each query, you need to find the maximum XOR of any prefix of nums, with the constraint that the XOR result must not exceed the value (2^maximumBit - 1).

For each query, return the result of the XOR of the prefix of the array and the maximum possible value that can be achieved based on maximumBit. This needs to be repeated for every query.

Examples

Example 1

Input: nums = [0,1,1,3], maximumBit = 2

Output: [0,3,2,3]

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

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

Example details omitted.

Constraints

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

Solution Approach

Prefix XOR Calculation

For each query, compute the XOR of the entire prefix of the array up to the current index. Using the sorted property of nums allows efficient computation of each query result.

Bit Masking

Use bit manipulation to mask out results exceeding the limit defined by maximumBit. The bitwise operation ensures that the result is constrained within the valid range.

Iterative Query Processing

The array's sorted property allows processing each query in constant time, making the solution efficient with a time complexity of O(n). This method avoids recomputation of the XOR for every element in each query.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The time complexity is O(n), where n is the length of the array, since each query is processed in constant time due to the prefix XOR and bit manipulation. The space complexity is O(1) as the solution does not require any additional data structures besides the input array.

What Interviewers Usually Probe

The candidate effectively applies bit manipulation to constrain the XOR result within the specified limit.
The candidate demonstrates a good understanding of using prefix sums in the context of XOR calculations.
The candidate is able to optimize the problem solution by leveraging sorted input for efficient query processing.

Common Pitfalls or Variants

Common pitfalls

Forgetting to apply the bit mask to limit the XOR result to (2^maximumBit - 1).
Inefficiently recalculating XOR values for each query without leveraging the sorted array structure.
Misunderstanding the prefix XOR calculation and failing to compute it correctly for each query.

Follow-up variants

Increase the size of the input array and test the solution's scalability.
Use a different sorting method and compare the performance with the current approach.
Apply a different bitwise operation (e.g., AND or OR) and observe the behavior of the solution.

FAQ

What is the best way to solve Maximum XOR for Each Query?

The most efficient approach involves using bit manipulation and prefix sums to process each query in constant time, making use of the sorted array.

How do you ensure the XOR result does not exceed the limit?

Use bit masking to constrain the XOR result within the range defined by maximumBit. This ensures the result is always within the valid range.

How does GhostInterview help with solving bit manipulation problems?

GhostInterview provides a clear breakdown of bit manipulation techniques, helping you apply them correctly to solve problems like Maximum XOR for Each Query.

What is the time complexity of the solution to Maximum XOR for Each Query?

The time complexity is O(n), where n is the length of the array. Each query is processed in constant time due to efficient XOR calculation.

How does the array's sorted property help in solving the problem?

The sorted property allows for efficient computation of each query by avoiding redundant recalculations of XOR for each query, improving the solution's performance.

terminal

Solution

Solution 1: Bitwise Operation + Enumeration

First, we preprocess the XOR sum $xs$ of the array `nums`, i.e., $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n-1]$.

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class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        ans = []
        xs = reduce(xor, nums)
        for x in nums[::-1]:
            k = 0
            for i in range(maximumBit - 1, -1, -1):
                if (xs >> i & 1) == 0:
                    k |= 1 << i
            ans.append(k)
            xs ^= x
        return ans

Solution 2: Enumeration Optimization

Similar to Solution 1, we first preprocess the XOR sum $xs$ of the array `nums`, i.e., $xs=nums[0] \oplus nums[1] \oplus \cdots \oplus nums[n-1]$.

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class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        ans = []
        xs = reduce(xor, nums)
        for x in nums[::-1]:
            k = 0
            for i in range(maximumBit - 1, -1, -1):
                if (xs >> i & 1) == 0:
                    k |= 1 << i
            ans.append(k)
            xs ^= x
        return ans

Continue Practicing

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