Neighboring Bitwise XOR

Problem Statement

In this problem, we are given a derived array of length ( n ), created by applying the bitwise XOR operation between adjacent elements of a binary array named original. The original array consists only of 0s and 1s. The derived array is formed such that for each position ( i ) in original:

• If ( i ) is the last index (i.e., ( i = n - 1 )), the value at derived[i] is calculated as original[i] ⊕ original[0].
• For all other indices, derived[i] is derived as original[i] ⊕ original[i + 1].

The challenge is to ascertain whether it is possible to reconstruct a valid original binary array that could logically result in the given derived array. The result should be true if such an array original exists, or false otherwise.

Examples

Example 1

Input:

derived = [1,1,0]

Output:

true

Explanation:

A valid original array that gives derived is [0,1,0].
derived[0] = original[0] ⊕ original[1] = 0 ⊕ 1 = 1
derived[1] = original[1] ⊕ original[2] = 1 ⊕ 0 = 1
derived[2] = original[2] ⊕ original[0] = 0 ⊕ 0 = 0

Example 2

Input:

derived = [1,1]

Output:

true

Explanation:

A valid original array that gives derived is [0,1].
derived[0] = original[0] ⊕ original[1] = 1
derived[1] = original[1] ⊕ original[0] = 1

Example 3

Input:

derived = [1,0]

Output:

false

Explanation:

There is no valid original array that gives derived.

Constraints

• n == derived.length
• 1 <= n <= 105
• The values in derived are either 0's or 1's

Approach and Intuition

The solution to the problem primarily hinges on understanding the properties of XOR operations and the constraints created by the cyclic relationship in a derived array. Here's the intuition:

1. XOR Properties:

   • XOR of a number with itself is 0. (i.e., ( x ⊕ x = 0 ))
   • XOR of a number with 0 is the number itself. (i.e., ( x ⊕ 0 = x ))
   • XOR is both commutative and associative. (i.e., ( a ⊕ b = b ⊕ a ) and ( (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c) ))

2. Decoding the Derived Array:

   • The generation of derived from original can be illustrated with two adjacent values in derived. This relationship allows potentially back-calculating values in original.

3. Checking for Logical Consistency:

   • Begin with an arbitrary value in original, typically original[0]. Due to the properties of XOR, one can attempt to deduce original[1] from derived[0], and so forth.
   • Continue this derivation around the array cyclically. The key is the check when returning to the start (original[0]); the derived values must remain consistent with initial assumptions.

4. Edge Cases:

   • For arrays with a single element, the derived array element is essentially original[0] ⊕ original[0] which is always 0 regardless of the values of original[0].
   • In cases where the array length is 2, handle them separately since they directly imply each other and wrap around immediately.

By following the above steps and reasoning through the cyclic nature of the pairing in original and derived, one can determine if reconstructing original from derived is feasible. The examples provided give clear practical instances of both feasible and infeasible scenarios:

• If derived = [1,1,0], reconstructing original as [0,1,0] validly produces the derived array.
• If derived = [1,0], it is impossible to create a logical original, thereby returning false.

Solutions

class Solution {
public:
    bool isValidArray(vector<int>& numbers) {
        int total = accumulate(numbers.begin(), numbers.end(), 0);
        return total % 2 == 0;
    }
};

public class Solution {
    
    public boolean verifyEvenSum(int[] data) {
        int total = 0;
        for (int value : data) {
            total += value;
        }
        return total % 2 == 0;
    }
}

class Solution:
    def isArrayEven(self, values: List[int]) -> bool:
        return sum(values) % 2 == 0