LeetCode 2683 - Neighboring Bitwise XOR

The problem presents us with a binary array called derived of length n. This array is constructed from another binary array original of the same length using the bitwise XOR operation on adjacent elements.

leetcodemediumarraybit-manipulation

LeetCode Problem 2683

Difficulty: 🟡 Medium
Topics: Array, Bit Manipulation

Solution

Problem Understanding

The problem presents us with a binary array called derived of length n. This array is constructed from another binary array original of the same length using the bitwise XOR operation on adjacent elements. Specifically, each element derived[i] is the XOR of original[i] and original[i+1], with a wrap-around at the end where derived[n-1] = original[n-1] ⊕ original[0].

Our task is to determine if there exists at least one valid original array that could produce the given derived array. The output should be a boolean: true if such an array exists, and false otherwise.

Constraints give us n up to 10^5, meaning any brute-force approach that tries all possible binary arrays is infeasible because the number of binary arrays grows exponentially with n (2^n possibilities). Each element in derived is guaranteed to be 0 or 1, which allows us to focus purely on bitwise reasoning rather than more general integer logic.

Important edge cases include arrays of length 1 or 2, where the wrap-around behavior is critical, and arrays where the number of 1s is odd, which may prevent a valid binary original from existing.

Approaches

The brute-force approach would be to generate all possible binary arrays of length n and check if their XORs of adjacent elements match derived. This is correct because it exhaustively checks all possibilities, but it is too slow. For n = 10^5, iterating through 2^100000 arrays is impossible.

The key observation for an optimal solution is that XOR has two properties: it is associative and commutative, and x ⊕ x = 0. If we denote original as [a0, a1, ..., an-1], the XOR of all elements in derived is equivalent to a0 ⊕ a1 ⊕ a1 ⊕ a2 ⊕ ... ⊕ an-1 ⊕ a0. Because XOR cancels itself in pairs, all middle terms cancel, leaving a0 ⊕ a0 = 0. Thus, the XOR of all elements in derived must be 0 for a valid original to exist. This provides an O(n) solution that only requires iterating through the array once to compute the XOR.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n * n) O(n) Generate all possible original arrays and check XORs
Optimal O(n) O(1) Use XOR properties to check if sum of derived is 0

Algorithm Walkthrough

  1. Initialize a variable xor_sum to 0. This will hold the XOR of all elements in derived.
  2. Iterate through each element d in derived.
  3. For each element, update xor_sum with xor_sum ⊕ d. This accumulates the XOR of the entire array.
  4. After the loop, check if xor_sum equals 0.
  5. If xor_sum is 0, return true; otherwise, return false.

Why it works: XORing all elements in derived cancels out all intermediate original elements, leaving the XOR of the first element with itself. If the final XOR is 0, a valid binary array exists. If it is 1, no valid array can exist because XOR properties prevent constructing an original that satisfies all derived values.

Python Solution

from typing import List

class Solution:
    def doesValidArrayExist(self, derived: List[int]) -> bool:
        xor_sum = 0
        for d in derived:
            xor_sum ^= d
        return xor_sum == 0

The code initializes xor_sum to accumulate the XOR of all elements in derived. Iterating through the array, we update xor_sum with the XOR of each element. Finally, if xor_sum is 0, it returns true, which indicates a valid original array exists.

Go Solution

func doesValidArrayExist(derived []int) bool {
    xorSum := 0
    for _, d := range derived {
        xorSum ^= d
    }
    return xorSum == 0
}

The Go solution mirrors the Python logic. We initialize xorSum and iterate through derived, XORing each element. At the end, we check if xorSum is 0 to decide whether a valid array exists. Go handles slices efficiently, and there is no need for additional memory allocation beyond the loop.

Worked Examples

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

Step xor_sum Element xor_sum after XOR
1 0 1 1
2 1 1 0
3 0 0 0

xor_sum = 0 → return true.

Example 2: derived = [1, 1]

Step xor_sum Element xor_sum after XOR
1 0 1 1
2 1 1 0

xor_sum = 0 → return true.

Example 3: derived = [1, 0]

Step xor_sum Element xor_sum after XOR
1 0 1 1
2 1 0 1

xor_sum = 1 → return false.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Iterate through derived once to compute XOR
Space O(1) Only a single integer variable is used, independent of n

The reasoning is that XOR accumulation is a constant-time operation per element, and we do not store any additional arrays or data structures.

Test Cases

# Provided examples
assert Solution().doesValidArrayExist([1,1,0]) == True  # Example 1
assert Solution().doesValidArrayExist([1,1]) == True    # Example 2
assert Solution().doesValidArrayExist([1,0]) == False   # Example 3

# Edge cases
assert Solution().doesValidArrayExist([0]) == True      # Single element zero
assert Solution().doesValidArrayExist([1]) == False     # Single element one
assert Solution().doesValidArrayExist([0,0,0,0]) == True # All zeros
assert Solution().doesValidArrayExist([1,1,1,1]) == True # Even number of ones
assert Solution().doesValidArrayExist([1,0,1]) == False # Odd number of ones
Test Why
[1,1,0] Typical small case with valid array
[1,1] Smallest even-length array that is valid
[1,0] Smallest invalid case
[0] Single-element zero edge case
[1] Single-element one edge case
[0,0,0,0] All zeros, trivial valid array
[1,1,1,1] Even number of ones, valid XOR sum
[1,0,1] Odd number of ones, invalid XOR sum

Edge Cases

One important edge case is when derived has only one element. If that element is 0, then the original array can also be [0], but if it is 1, no valid original array exists because 1 ⊕ 1 = 0 is impossible.

Another edge case is when all elements in derived are zeros. In this case, any original array of identical elements, all zeros or all ones, satisfies the XOR condition, so the result should be true.

A third edge case is when the number of ones in derived is odd. Since XOR accumulates such that each element in original cancels out except for the first element, an odd number of ones results in a final XOR of 1, which means no valid original array exists. Handling this edge case is crucial because it is not immediately intuitive from just inspecting pairs of elements.