Problem Description

You are given a derived array that was created from an original binary array through XOR operations. Your task is to determine if a valid original binary array could exist that would produce this derived array.

Here's how the derived array is created from the original array:

• Both arrays have the same length n and are 0-indexed
• For each position i in the derived array:
  • If i is at the last position (i = n - 1), then derived[i] = original[i] ⊕ original[0] (XOR of last and first elements)
  • Otherwise, derived[i] = original[i] ⊕ original[i + 1] (XOR of consecutive elements)

The key insight is that each element in the derived array is the XOR of two adjacent elements in the original array, with the last element wrapping around to XOR with the first element, forming a circular relationship.

Example of the transformation:

• If original = [a₀, a₁, a₂, a₃]
• Then derived = [a₀⊕a₁, a₁⊕a₂, a₂⊕a₃, a₃⊕a₀]

The solution leverages the mathematical property that if we XOR all elements in the derived array: b₀ ⊕ b₁ ⊕ ... ⊕ b_{n-1} = (a₀⊕a₁) ⊕ (a₁⊕a₂) ⊕ ... ⊕ (a_{n-1}⊕a₀)

Since each element in the original array appears exactly twice in this expression and x ⊕ x = 0, the entire expression equals 0. Therefore, a valid original array exists if and only if the XOR of all elements in the derived array equals 0.

The solution uses Python's reduce function with the xor operator to compute the XOR of all elements and checks if the result is 0.

Intuition

Let's think about what happens when we create the derived array. Each element in the original array participates in exactly two XOR operations:

• It gets XORed with its right neighbor to create one derived element
• It gets XORed with its left neighbor to create another derived element

For example, if we have original = [a, b, c, d], then:

• a appears in derived[0] = a ⊕ b and derived[3] = d ⊕ a
• b appears in derived[0] = a ⊕ b and derived[1] = b ⊕ c
• And so on...

Now, what happens if we XOR all the elements in the derived array together?

derived[0] ⊕ derived[1] ⊕ derived[2] ⊕ derived[3] = (a ⊕ b) ⊕ (b ⊕ c) ⊕ (c ⊕ d) ⊕ (d ⊕ a)

Using the properties of XOR (commutative and associative), we can rearrange: = a ⊕ a ⊕ b ⊕ b ⊕ c ⊕ c ⊕ d ⊕ d

Since x ⊕ x = 0 for any value x, each pair cancels out: = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0

This reveals the key insight: if a valid original array exists, the XOR of all elements in the derived array must equal 0.

But does the reverse hold true? If the XOR sum is 0, can we always construct a valid original array? Yes! We can start with any value for original[0], then work our way through:

• original[1] = derived[0] ⊕ original[0]
• original[2] = derived[1] ⊕ original[1]
• And so on...

The XOR sum being 0 guarantees that when we compute the last element and check if derived[n-1] = original[n-1] ⊕ original[0], this equation will hold true, completing the cycle consistently.

Therefore, checking if the XOR of all derived elements equals 0 is both necessary and sufficient to determine if a valid original array exists.

Solution Approach

The implementation is remarkably concise, leveraging Python's built-in functions to solve the problem efficiently.

Mathematical Foundation:

Based on our intuition, we need to verify that: b₀ ⊕ b₁ ⊕ ... ⊕ b_{n-1} = 0

Where b represents the derived array. This condition is both necessary and sufficient for a valid original array to exist.

Implementation Details:

The solution uses Python's reduce function from the functools module along with the xor operator from the operator module:

class Solution:
    def doesValidArrayExist(self, derived: List[int]) -> bool:
        return reduce(xor, derived) == 0

How reduce works here:

• reduce(xor, derived) applies the XOR operation cumulatively to all elements in the derived array
• It starts with the first two elements: derived[0] ⊕ derived[1]
• Then XORs the result with the next element: (derived[0] ⊕ derived[1]) ⊕ derived[2]
• Continues until all elements are processed

Step-by-step execution for derived = [1, 1, 0]:

1. First operation: 1 ⊕ 1 = 0
2. Second operation: 0 ⊕ 0 = 0
3. Result: 0, so return True

Time and Space Complexity:

• Time Complexity: O(n) where n is the length of the derived array, as we iterate through all elements once
• Space Complexity: O(1) as we only use a constant amount of extra space for the XOR accumulation

The elegance of this solution lies in recognizing the mathematical property that all original elements appear exactly twice in the XOR chain, causing them to cancel out completely when a valid original array exists.

Example Walkthrough

Let's walk through a concrete example to see how the solution works.

Example: derived = [1, 1, 0]

First, let's understand what we're checking. We need to verify if there exists some original binary array that could produce this derived array through the XOR transformation.

Step 1: Apply the XOR operation to all elements

We compute: 1 ⊕ 1 ⊕ 0

Breaking it down:

• First: 1 ⊕ 1 = 0
• Then: 0 ⊕ 0 = 0
• Final result: 0

Since the result equals 0, the function returns True.

Step 2: Verify by reconstructing a possible original array

To convince ourselves this works, let's construct an original array:

• Start with original[0] = 0 (we can choose any value)
• original[1] = derived[0] ⊕ original[0] = 1 ⊕ 0 = 1
• original[2] = derived[1] ⊕ original[1] = 1 ⊕ 1 = 0

So original = [0, 1, 0]

Step 3: Confirm this original array produces our derived array

• 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 ✓

Perfect! Our original array [0, 1, 0] indeed produces the derived array [1, 1, 0].

Counter-example: derived = [1, 0]

Let's check: 1 ⊕ 0 = 1

Since the result is 1 (not 0), the function returns False.

If we try to construct an original array:

• Start with original[0] = 0
• original[1] = derived[0] ⊕ original[0] = 1 ⊕ 0 = 1
• Check: derived[1] should equal original[1] ⊕ original[0] = 1 ⊕ 0 = 1
• But derived[1] = 0, not 1! This contradiction confirms no valid original array exists.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the derived array. The reduce function with the xor operator iterates through all elements of the array exactly once, performing a constant-time XOR operation for each element.

The space complexity is O(1). The reduce function only maintains a single accumulator value throughout the computation, using constant extra space regardless of the input size. No additional data structures are created that scale with the input.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Empty Array Handling

The reduce function will raise a TypeError when called on an empty list without an initial value.

Problem Example:

derived = []
reduce(xor, derived)  # Raises TypeError: reduce() of empty sequence with no initial value

Solution: Add a check for empty arrays or provide an initial value to reduce:

def doesValidArrayExist(self, derived: List[int]) -> bool:
    if not derived:
        return True  # Empty array is considered valid
    return reduce(xor, derived) == 0

Or use an initial value:

def doesValidArrayExist(self, derived: List[int]) -> bool:
    return reduce(xor, derived, 0) == 0

2. Misunderstanding XOR Properties

Developers might incorrectly assume that XOR operations require special handling for binary values or that the result needs to be masked.

Incorrect Assumption:

# Wrong: Trying to ensure binary result
return (reduce(xor, derived) & 1) == 0

Correct Understanding: XOR naturally handles any integer values, and the result is already in the correct form for comparison.

3. Manual XOR Implementation Instead of Using Built-ins

Writing custom loops when Python's reduce provides a cleaner solution:

Less Optimal:

def doesValidArrayExist(self, derived: List[int]) -> bool:
    result = 0
    for num in derived:
        result ^= num
    return result == 0

While this works and has the same complexity, using reduce is more Pythonic and concise.

4. Forgetting to Import Required Modules

The solution requires imports from both functools and operator:

Common Mistake:

# Missing imports will cause NameError
class Solution:
    def doesValidArrayExist(self, derived: List[int]) -> bool:
        return reduce(xor, derived) == 0  # NameError: name 'reduce' is not defined

Complete Solution:

from functools import reduce
from operator import xor

class Solution:
    def doesValidArrayExist(self, derived: List[int]) -> bool:
        return reduce(xor, derived) == 0

5. Edge Case: Single Element Array

While the current solution handles this correctly (a single element XORed with nothing returns itself), developers might overthink this case:

Unnecessary Complexity:

def doesValidArrayExist(self, derived: List[int]) -> bool:
    if len(derived) == 1:
        return derived[0] == 0  # Correct but redundant
    return reduce(xor, derived) == 0

The original solution already handles this case properly since reduce on a single-element list returns that element.