Leetcode 2683: Neighboring Bitwise XOR

Input: You are given a binary array `derived`, where each value is either 0 or 1. You need to determine if there exists a binary array `original` such that applying the XOR operation on adjacent values produces `derived`.

Example: Input: derived = [0, 1, 1]

Constraints:

• 1 <= derived.length <= 10^5

• The values in the derived array are either 0 or 1.

Output: Return `true` if a valid binary array `original` exists, otherwise return `false`.

Example: Output: true

Constraints:

• The output should be a boolean value.

Goal: The goal is to check if there exists a valid `original` array that could have been transformed into the `derived` array using the bitwise XOR operation.

Steps:

• Step 1: For the given array `derived`, initialize two variables to track the XOR of elements.

• Step 2: Iterate through the `derived` array and compute the cumulative XOR to check if a valid `original` exists.

• Step 3: Return `true` if the cumulative XOR values meet the conditions for a valid `original` array, otherwise return `false`.

Goal: The input size can be as large as 100,000, so the solution must be efficient.

Steps:

• The array `derived` has a length of at least 1 and at most 100,000.

Assumptions:

• The derived array will always be valid (it will not contain values outside the range of 0 and 1).

• The problem will be solved by checking the XOR conditions across the array.

• Input: Input: derived = [0, 1, 1]

• Explanation: The `derived` array is obtained by XORing adjacent values of the `original` array. A valid `original` array that satisfies this condition is [0, 1, 0]. Therefore, the output is true.

• Input: Input: derived = [1, 0]

• Explanation: It is not possible to create a valid `original` array from this `derived` array using the XOR operation, so the output is false.

Approach: We can determine if a valid `original` array exists by computing the XOR of elements in `derived` and checking if the cumulative XOR satisfies certain conditions.

Observations:

• The XOR operation is reversible, so we can check the feasibility of creating the `original` array by comparing the XOR values across adjacent elements.

• By using XOR properties, we can efficiently determine if a valid `original` exists without explicitly constructing it.

Steps:

• Step 1: Initialize two variables to track the XOR of the first and second elements.

• Step 2: Iterate over the `derived` array, applying XOR to simulate the original array's formation.

• Step 3: Check the resulting XOR values to see if they satisfy the conditions for a valid `original` array.

• Step 4: Return `true` if the conditions are met, otherwise return `false`.

Empty Inputs:

• The input array is always non-empty as per the constraints.

Large Inputs:

• The solution needs to be efficient enough to handle arrays with up to 100,000 elements.

Special Values:

• The solution should handle cases where the `derived` array contains all 0's or all 1's.

Constraints:

• The values in `derived` are either 0 or 1, and the length of the array is within the specified constraints.

bool doesValidArrayExist(vector<int>& nums) {
    int n = nums.size();
    // same 0 0
    
    int fst = 0, scd = 1;
    for(int i = 0; i < n; i++) {
        fst ^= nums[i];
        scd ^= nums[i];
    }
    if(fst == 0 || scd == 1) return true;
    return false;
    // same 1 1
    // different 1 0
    // different 0 1        
}

1 : Function Definition

bool doesValidArrayExist(vector<int>& nums) {

The function 'doesValidArrayExist' takes a reference to a vector 'nums' and returns a boolean indicating whether a valid array exists based on the XOR conditions.

2 : Variable Initialization

    int n = nums.size();

The variable 'n' is initialized to the size of the 'nums' vector, representing the number of elements in the array.

3 : Variable Initialization

    int fst = 0, scd = 1;

Two variables, 'fst' and 'scd', are initialized to 0 and 1 respectively. These variables will be used to accumulate the XOR results as the loop iterates through 'nums'.

4 : Loop

    for(int i = 0; i < n; i++) {

A for loop is initiated to iterate through the elements of the 'nums' array from index 0 to 'n-1'.

5 : XOR Operation

        fst ^= nums[i];

The variable 'fst' is updated by applying the XOR operation between its current value and the current element of 'nums'. This operation will accumulate the XOR of all elements.

6 : XOR Operation

        scd ^= nums[i];

The variable 'scd' is updated by applying the XOR operation between its current value and the current element of 'nums'. This will accumulate a separate XOR of all elements.

7 : Condition Check

    if(fst == 0 || scd == 1) return true;

The function checks if either 'fst' is 0 or 'scd' is 1. If this condition is satisfied, it returns true, indicating that a valid array exists.

8 : Return

    return false;

If the condition is not met, the function returns false, indicating that no valid array exists.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: We iterate through the array once to check the XOR conditions, which gives a linear time complexity.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is constant as we only need a few variables to track the XOR results.

Leetcode 2683: Neighboring Bitwise XOR

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