Leetcode 1829: Maximum XOR for Each Query

Input: You are given two inputs: a sorted array `nums` of non-negative integers and an integer `maximumBit`.

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

Constraints:

• 1 <= nums.length <= 10^5

• 1 <= maximumBit <= 20

• 0 <= nums[i] < 2^maximumBit

Output: Return an array where each element is the best `k` for each query. `k` maximizes the XOR result of the array `nums` and `k`.

Example: [0, 3, 2, 3]

Constraints:

• The returned array should contain one result for each query.

Goal: To maximize the XOR of the entire array and `k`, where `k` is less than `2^maximumBit`.

Steps:

• Iterate through the array and compute the XOR of all elements.

• Determine the best `k` that maximizes the XOR result for each query.

• Update the array after each query by removing the last element.

Goal: Constraints to ensure the correctness of the problem.

Steps:

• The array `nums` is sorted in ascending order.

• All elements in `nums` and the value of `maximumBit` are within the given bounds.

Assumptions:

• The input array `nums` is non-empty.

• The value of `maximumBit` is within the specified bounds.

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

• Explanation: Each query asks to find the optimal value of `k` to maximize the XOR of all elements in the current array with `k`, considering the constraint that `k` must be less than `2^maximumBit`.

Approach: We will compute the XOR of all elements in the array, and for each query, find the optimal `k` that maximizes the result.

Observations:

• XOR operations are sensitive to bitwise manipulation, and the optimal `k` can be determined by checking the result of XORing all array elements with `k`.

• The main challenge is to efficiently compute the result for each query while modifying the array after each query.

Steps:

• Compute the XOR of all elements in `nums`.

• Find the maximum `k` that maximizes the XOR result for each query.

• Update the array by removing the last element after each query.

Empty Inputs:

• The array `nums` should not be empty.

Large Inputs:

• Ensure the algorithm efficiently handles large arrays with up to `10^5` elements.

Special Values:

• Consider cases where all elements in the array are zero or the array has a single element.

Constraints:

• Make sure to handle all constraints such as the sorted nature of the array and the bounds for `maximumBit`.

vector<int> getMaximumXor(vector<int>& nums, int bit) {
    int ui = 0;
    for(int x : nums)
        ui ^= x;
    int msk = (1 << bit) - 1;
    
    
    

    
    int res = ui ^ msk;
    vector<int> ans;
    ans.push_back(res);
    while(nums.size() > 1) {
        
        res ^= nums.back();
        nums.pop_back();
        ans.push_back(res);
        
    }
    return ans;
}

1 : Function Definition

vector<int> getMaximumXor(vector<int>& nums, int bit) {

Defines the function `getMaximumXor`, which takes a vector of integers `nums` and an integer `bit` as input and returns a vector of integers.

2 : Variable Initialization

    int ui = 0;

Initializes the variable `ui` to 0, which will be used to store the XOR of all the elements in the `nums` vector.

3 : Loop Through Numbers

    for(int x : nums)

Starts a loop to iterate through each number `x` in the `nums` vector.

4 : XOR Operation

        ui ^= x;

Performs the XOR operation between `ui` and `x`, updating `ui` with the result. This effectively calculates the XOR of all numbers in the vector.

5 : Mask Calculation

    int msk = (1 << bit) - 1;

Calculates a mask `msk` by shifting 1 left by `bit` positions and subtracting 1. This mask is used to keep only the relevant bits.

6 : XOR Masking

    int res = ui ^ msk;

Applies the mask `msk` to `ui` by performing an XOR operation. This gives the final result after masking.

7 : Vector Initialization

    vector<int> ans;

Initializes an empty vector `ans` to store the results of the XOR calculations.

8 : Store Initial Result

    ans.push_back(res);

Adds the initial XOR result `res` to the `ans` vector.

9 : While Loop Start

    while(nums.size() > 1) {

Starts a while loop that continues as long as there is more than one element left in the `nums` vector.

10 : Reverse XOR Operation

        res ^= nums.back();

Performs the XOR operation with the last element in `nums` (using `nums.back()`) and updates `res`.

11 : Remove Last Element

        nums.pop_back();

Removes the last element from the `nums` vector after performing the XOR operation.

12 : Store Updated Result

        ans.push_back(res);

Adds the updated result `res` to the `ans` vector after each XOR operation.

13 : Return Statement

    return ans;

Returns the `ans` vector, which contains the results of the XOR operations.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: Since we are performing XOR operations and modifying the array after each query, the time complexity for each query is O(n).

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) due to the storage of the result array and intermediate computations.

Leetcode 1829: Maximum XOR for Each Query

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