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LeetCode 1829: Maximum XOR for Each Query – Step-by-Step Guide to Optimizing the Solution


This problem involves maximizing the XOR operation across multiple queries, making it an interesting and challenging problem.

In this blog post, we’ll walk through how to approach LeetCode 1829, starting with a brute-force solution and gradually optimizing it for better performance. We’ll break down the problem, explain the brute-force method, and then introduce an optimized approach that significantly improves efficiency.

LeetCode 1829: Maximum XOR for Each Query

You are given an array of non-negative integers (nums) sorted in ascending order and an integer maximumBit. Your task is to perform n queries (where n is the size of the array) and for each query, find a non-negative integer k that maximizes the XOR result when combined with all elements of the array up to that point.

For each query:

  1. You XOR all elements of the array up to the current query.
  2. You find a k (where 0 <= k < 2^maximumBit) such that the result of XOR-ing all elements with k is maximized.

After each query, the last element is removed from the array and the process continues until you have processed all queries.

Let's take an example to understand how this works:

Example 1:

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

    For each query, we want to maximize the XOR result. Here's how we would process the queries:

  1. Query 1: With nums = [0, 1, 1, 3], we find that k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
  2. Query 2: After removing the last element, nums = [0, 1, 1], we find k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
  3. Query 3: After removing another element, nums = [0, 1], we find k = 2 since 0 XOR 1 XOR 2 = 3.
  4. Query 4: Finally, with nums = [0], we find k = 3 since 0 XOR 3 = 3.

Output: [0, 3, 2, 3]

Brute-Force Approach

The brute-force solution for this problem involves iterating over every possible value of k for each query and calculating the XOR of all the elements in the array up to that query. While this approach is easy to understand, it's not efficient enough for large inputs (with n up to 100,000).

Step-by-Step Brute Force Solution:

class Solution {
public:
    vector<int> getMaximumXor(vector<int>& nums, int maximumBit)
    {
        int n = nums.size();
        vector<int> ans(n);
        int maxK = (1 << maximumBit) - 1;  // Calculate the maximum possible value of k
       
        for (int i = n - 1; i >= 0; --i)
        {
            int currentXor = 0;
            // XOR all elements in the array up to the current index i
            for (int j = 0; j <= i; ++j)
            {
                currentXor ^= nums[j];
            }
           
            // Try all possible k values from 0 to maxK
            int bestK = 0;
            for (int k = 0; k <= maxK; ++k)
            {
                if ((currentXor ^ k) > (currentXor ^ bestK))
                {
                    bestK = k;
                }
            }
            ans[i] = bestK;
        }
       
        return ans;
    }
};

Time Complexity of the Brute Force Approach

  • Outer Loop: We process n queries, so we iterate over the array n times.
  • Inner Loop (Array Traversal): For each query, we compute the XOR of all elements in the array up to the current query, which takes O(n) time.
  • Inner Inner Loop (K values): For each query, we check all possible values of k (from 0 to 2^maximumBit - 1), which requires up to 2^maximumBit checks.

Thus, the overall time complexity is O(n^2 * 2^maximumBit), which is inefficient for large input sizes.

Optimized Approach

The brute-force solution is too slow, especially when n is large. To optimize this, we can take advantage of the cumulative XOR and process the queries in reverse order.

Key Insight for Optimization:

  1. Cumulative XOR: Instead of recalculating the XOR for each element every time, we can maintain a running cumulative XOR of all elements up to the current point.
  2. Reverse Query Processing: Since the queries require removing the last element of nums after each step, processing the queries in reverse order helps reuse previous computations.

Optimized Algorithm

class Solution {
public:
    void solve(vector<int>& nums, vector<int>& ans)
    {
        int n = nums.size();
        ans[n - 1] = nums[0]; // Initial XOR for the last element
       
        // Compute cumulative XOR in reverse order
        for (int i = n - 2; i >= 0; i--)
        {
            ans[i] = ans[i + 1] ^ nums[n - i - 1];
        }
    }

    vector<int> getMaximumXor(vector<int>& nums, int maximumBit)
    {
        vector<int> ans(nums.size());
        int k = (1 << maximumBit) - 1; // Calculate the maximum value of k
       
        // Compute cumulative XOR array
        solve(nums, ans);

        // Perform XOR operation with k for each query
        for (int i = 0; i < nums.size(); i++)
        {
            ans[i] = ans[i] ^ k;
        }
       
        return ans;
    }
};

Explanation of the Optimized Solution

  1. Cumulative XOR Calculation:

    • We initialize ans[n-1] with nums[0]. Then, we iterate backward through the array and calculate the cumulative XOR for each element.

  2. Maximizing XOR with k:

    • Once we have the cumulative XOR for each query, we calculate the maximum k by performing a bitwise XOR with k = 2^maximumBit - 1.

  3. Efficient Query Processing:

    • By computing the cumulative XOR in reverse and using the XOR result directly, we avoid recomputing the XOR for each query, resulting in a more efficient solution.

Time Complexity of the Optimized Approach

  • Cumulative XOR Calculation: This takes O(n) time as we only need to traverse the array once.

  • Final XOR Calculation with k: This also takes O(n) time.

  • Therefore, the overall time complexity is O(n), which is a significant improvement over the brute-force approach.

  • Space Complexity: We store the results in the ans array, so the space complexity is O(n).

Why the Optimized Approach Works

The optimized solution is far more efficient because:

  1. It avoids recalculating the XOR for the entire array on each query by using cumulative XOR.
  2. The solution processes the queries in reverse order, leveraging previously computed results, thus reducing the time complexity to O(n), compared to the brute-force method's O(n^2 * 2^maximumBit).

This makes the optimized approach ideal for handling large arrays (up to 100,000 elements) as specified in the problem's constraints.

FAQs  (LeetCode 1829: Maximum XOR for Each Query)

1. What is LeetCode 1829 about?
LeetCode 1829 involves maximizing XOR results across multiple queries on an array. Given nums and maximumBit, you must find the best k for each query to maximize XOR(nums[0..i] ^ k).

2. What does "maximize XOR" mean in this problem?
It means finding a value k (where 0 ≤ k < 2^maximumBit) such that XOR(nums[0..i] ^ k) produces the largest possible result.

3. Why is XOR important in this problem?
XOR is a bitwise operation that flips bits, making it useful for maximizing binary representations. The goal is to exploit XOR properties to find optimal k efficiently.

4. What is the brute-force approach for this problem?

The brute-force method checks every possible k for each query, computes the XOR, and picks the k that maximizes the result. However, it’s inefficient (O(n² * 2^maximumBit)).

5. Why is the brute-force solution inefficient?

It recalculates XOR for every subarray in each query, leading to O(n²) time complexity, which is too slow for large inputs (e.g., n = 100,000).

6. How does the optimized solution improve efficiency?

By using cumulative XOR and processing queries in reverse, it reduces time complexity to O(n)—ideal for large inputs.

7. What is cumulative XOR?

Cumulative XOR is a running XOR value stored for each prefix of the array. It avoids recomputation by reusing prior results.

8. Why process queries in reverse order?

Since each query removes the last element, reverse processing lets us reuse XOR computations from previous steps, saving time.

9. What is the role of maximumBit in this problem?

It defines the upper bound for k (i.e., k < 2^maximumBit). For maximumBit = 2, k can be 0, 1, 2, 3.

10. How do you compute the optimal k efficiently?

The optimal k is always (1 << maximumBit) - 1 (e.g., maximumBit = 2 → k = 3). XOR-ing with this value flips all bits in the cumulative XOR, maximizing the result.

11. What is the time complexity of the optimized solution?

O(n)—linear time, as it computes cumulative XOR in one pass and processes queries in reverse.

12. What is the space complexity of the optimized approach?

O(n) for storing the result array. No additional space is needed beyond input and output.

13. Can this problem be solved with prefix XOR arrays?

Yes! The optimized solution implicitly uses prefix XOR by tracking cumulative XOR values in reverse order.

14. What are the key bitwise tricks used here?

  • XOR properties: a ^ a = 0, a ^ 0 = a.
  • Bitmasking: (1 << maximumBit) - 1 generates a number with maximumBit set bits.

15. How does XOR maximize the result in this problem?

XOR-ing with k = (1 << maximumBit) - 1 flips all bits of the cumulative XOR, producing the maximum possible value.

16. What’s an example walkthrough of the optimized solution?

For nums = [0, 1, 1, 3], maximumBit = 2:
  • Cumulative XOR: [0, 1, 0, 3] (reverse order).
  • Optimal k = 3 (since 2^2 - 1 = 3).
  • Result: [0^3, 1^3, 0^3, 3^3] = [3, 2, 3, 0] → Reversed to [0, 3, 2, 3].

17. How does this problem relate to real-world applications?

It’s useful in:
  • Cryptography (XOR-based encryption).
  • Error detection (checksums).
  • Data compression (bitwise optimizations).

18. What are similar LeetCode problems to practice?

  • 421. Maximum XOR of Two Numbers in an Array
  • 1318. Minimum Flips to Make a OR b Equal to c
  • 1442. Count Triplets That Can Form Two Arrays of Equal XOR

19. Why is this problem important for coding interviews?

It tests:
  • Bit manipulation skills.
  • Optimization techniques (brute-force → cumulative operations).
  • Reverse-query processing for efficiency.

20. How can I further optimize this solution?

  • In-place computation: Avoid extra arrays.
  • Early termination: Stop if k reaches maximum early.
  • Bitmask caching: Precompute k values.

LeetCode 1829 is a great problem for practicing bit manipulation and learning how to optimize solutions using cumulative operations. By starting with a brute-force approach and gradually moving to an optimized solution, we can efficiently solve the problem with a time complexity of O(n).

The key to solving this problem efficiently lies in the observation that you can compute the XOR values incrementally using cumulative XOR and reduce the overall computational complexity significantly.

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