Leetcode 2317: Maximum XOR After Operations

Input: The input consists of an integer array nums where 1 <= nums.length <= 10^5 and 0 <= nums[i] <= 10^8.

Example: nums = [5, 7, 3, 10]

Constraints:

• 1 <= nums.length <= 10^5

• 0 <= nums[i] <= 10^8

Output: Return the maximum possible bitwise XOR of all elements after applying the operation any number of times.

Example: For nums = [5, 7, 3, 10], the output is 15.

Constraints:

• The result is a non-negative integer.

Goal: The goal is to calculate the maximum bitwise XOR achievable after applying the operation on the array any number of times.

Steps:

• 1. Iterate over the entire array and calculate the OR of all elements.

• 2. Return the OR result, which represents the maximum possible XOR after applying the operations.

Goal: The problem constraints are as follows:

Steps:

• 1 <= nums.length <= 10^5

• 0 <= nums[i] <= 10^8

Assumptions:

• The array nums is non-empty.

• Input: nums = [5, 7, 3, 10]

• Explanation: In this example, applying the operation with x = 5 at index 3 gives us nums = [5, 7, 3, 0]. The XOR of these elements is 15, which is the maximum possible XOR.

Approach: To achieve the maximum bitwise XOR of all elements, the operation modifies the array in a way that the result is influenced by the bitwise OR of the elements.

Observations:

• The operation essentially affects the bits of each element, and we are interested in maximizing the XOR across the entire array.

• The operation will allow us to manipulate the bits of the elements, but the final result can be determined by the OR of all elements.

Steps:

• 1. Initialize a variable mask to 0.

• 2. Iterate through the array and apply the OR operation on each element to update the mask.

• 3. Return the final value of the mask, as it represents the maximum possible XOR.

Empty Inputs:

• If the array is empty, the result should be 0.

Large Inputs:

• The algorithm should efficiently handle large arrays with up to 10^5 elements.

Special Values:

• If all elements are zero, the result will be 0.

Constraints:

• Ensure that large inputs are processed efficiently within time limits.

int maximumXOR(vector<int>& nums) {
    
    int mask = 0;
    
    for(auto it : nums) {
        mask |= it;
    }
    
    return mask;
}

1 : Method Declaration

int maximumXOR(vector<int>& nums) {

Declare the method `maximumXOR` which takes a vector of integers `nums` as input and returns an integer representing the maximum XOR achievable.

2 : Variable Initialization

    int mask = 0;

Initialize a variable `mask` to store the cumulative result of the bitwise OR operation over all elements of `nums`. Initially set to 0.

3 : Loop Initialization

    for(auto it : nums) {

Start a loop to iterate over each element `it` in the vector `nums`.

4 : Bitwise OR Operation

        mask |= it;

For each number `it` in `nums`, perform a bitwise OR operation with the current `mask` value, updating `mask` to include the bits set by `it`.

5 : Return Statement

    return mask;

Return the final value of `mask`, which holds the result of the bitwise OR of all elements in `nums`.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n), where n is the length of the array, as we iterate over the array once to compute the OR.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is O(1), as we only need a constant amount of extra space to store the mask.

Leetcode 2317: Maximum XOR After Operations

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