2683

2683. Neighboring Bitwise XOR (Medium)

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

Related Topics:
Array, Bit Manipulation

Solution 1.

// OJ: https://leetcode.com/problems/neighboring-bitwise-xor
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool doesValidArrayExist(vector<int>& A) {
        int N = A.size(), prev = 0;
        for (int i = 0; i < N - 1; ++i) {
            prev = A[i] ^ prev;
        }
        return prev ^ 0 == A.back();
    }
};