Neighboring Bitwise XOR | Approache | Dry Runs | Leetcode 2683 | CodeWithShaaz

🔍 How It Works:
XOR Property Insight:
XOR of a number with itself is 0: x ⊕ x = 0
XOR of a number with 0 is the number itself: x ⊕ 0 = x
Key Observation:
If the XOR of all elements in derived is 0, a valid original array exists.
Otherwise, it's impossible to construct such an array.
Why?
XOR is cyclic and associative, so for adjacent XORs to loop back, their total XOR must cancel out (i.e., be zero).
📊 Example:
Input: derived = [1, 1, 0]
Output: true
Explanation: A valid original array is [0, 1, 0]
Step-by-Step:
XOR of all elements → 1 ⊕ 1 ⊕ 0 = 0 ✅
Since the result is 0, a valid array exists.
✨ Key Concepts Used:
Bitwise XOR (^) properties
Associative and Commutative behavior of XOR
Simple iteration for cumulative XOR
🚀 Java Code:
class Solution {
public boolean doesValidArrayExist(int[] derived) {
int xor = 0;
for (int x : derived) {
xor ^= x; // Cumulative XOR of all elements
}
return xor == 0; // If XOR is 0, valid array exists
}
}
⚡ Complexity Analysis:
Time Complexity: O(n) 🔥 (Single pass through the array)
Space Complexity: O(1) 💡 (No extra space used)
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