2683. Neighboring Bitwise XOR

[mah-shamim]

Topics: Array, Bit Manipulation

A 0-indexed array derived with length n is derived by computing the bitwise XOR (⊕) of adjacent values in a binary array original of length n.

Specifically, for each index i in the range [0, n - 1]:

• If i = n - 1, then derived[i] = original[i] ⊕ original[0].
• Otherwise, derived[i] = original[i] ⊕ original[i + 1].

Given an array derived, your task is to determine whether there exists a valid binary array original that could have formed derived.

Return true if such an array exists or false otherwise.

• A binary array is an array containing only 0's and 1's

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

Hint:

1. Understand that from the original element, we are using each element twice to construct the derived array
2. The xor-sum of the derived array should be 0 since there is always a duplicate occurrence of each element.

[mah-shamim]

To determine whether a valid binary array original exists that could form the given derived array, we can use the properties of XOR:

Key Observations:

1. Each derived[i] is the XOR of two adjacent elements in original:

   • For i = n - 1, derived[i] = original[n-1] ⊕ original[0].
   • Otherwise, derived[i] = original[i] ⊕ original[i+1].

2. XOR properties:

   • a ⊕ a = 0 (XOR of a number with itself is 0).
   • a ⊕ 0 = a (XOR of a number with 0 is the number itself).
   • XOR is commutative and associative.

3. To construct a valid original, the XOR of all elements in derived must equal 0. Why? Because for a circular XOR relationship to work (start and end wrap back to the beginning), the cumulative XOR of derived should cancel out.

Steps:

1. Compute the XOR of all elements in derived.
2. If the result is 0, a valid original exists; otherwise, it does not.

Algorithm:

• Compute the XOR of all elements in derived.
• If the XOR is 0, return true.
• Otherwise, return false.

Let's implement this solution in PHP: 2683. Neighboring Bitwise XOR

<?php
/**
 * @param Integer[] $derived
 * @return Boolean
 */
function doesValidArrayExist($derived) {
    $xorSum = 0;

    // Compute the XOR of all elements in derived
    foreach ($derived as $value) {
        $xorSum ^= $value;
    }

    // If the XOR sum is 0, a valid original array exists
    return $xorSum === 0;
}

// Example 1
$derived1 = [1, 1, 0];
echo doesValidArrayExist($derived1) ? 'true' : 'false'; // Output: true

// Example 2
$derived2 = [1, 1];
echo doesValidArrayExist($derived2) ? 'true' : 'false'; // Output: true

// Example 3
$derived3 = [1, 0];
echo doesValidArrayExist($derived3) ? 'true' : 'false'; // Output: false
?>

Explanation:

1. We initialize $xorSum to 0.
2. For each element in derived, we XOR it with $xorSum.
3. At the end of the loop, $xorSum will contain the XOR of all elements in derived.
4. If $xorSum is 0, return true; otherwise, return false.

Complexity:

• Time Complexity: O(n), where n is the length of derived. We only iterate through the array once.
• Space Complexity: O(1), as we only use a single variable to compute the XOR.

This approach efficiently checks for the existence of a valid original array within the given constraints.

[kovatz]

Thanks for solving the problem of "Neighboring Bitwise XOR"

[mah-shamim]

Thanks