# LeetCode 1720. Decode XORed Array

## Description

https://leetcode.com/problems/decode-xored-array/

There is a hidden integer array `arr` that consists of `n` non-negative integers.

It was encoded into another integer array `encoded` of length `n - 1`, such that `encoded[i] = arr[i] XOR arr[i + 1]`. For example, if `arr = [1,0,2,1]`, then `encoded = [1,2,3]`.

You are given the `encoded` array. You are also given an integer `first`, that is the first element of `arr`, i.e. `arr`.

Return the original array `arr`. It can be proved that the answer exists and is unique.

Example 1:

```Input: encoded = [1,2,3], first = 1
Output: [1,0,2,1]
Explanation: If arr = [1,0,2,1], then first = 1 and encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3]
```

Example 2:

```Input: encoded = [6,2,7,3], first = 4
Output: [4,2,0,7,4]
```

Constraints:

• `2 <= n <= 104`
• `encoded.length == n - 1`
• `0 <= encoded[i] <= 105`
• `0 <= first <= 105`

## Explanation

The key is to find the inverse is XOR.

If you have:

c = a^b;

You can get a or b back if you have the other value available:

a = c^b; // or b^c (order is not important)
b = c^a; // or a^c

For example if `a = 5``b = 3``c = 6` you get:

```b=0011 (3)            a=0101 (5)
c=0110 (6) XOR   or   c=0110 (6) XOR
----------            ----------
a=0101 (5)            b=0011 (3)```

## Python Solution

``````class Solution:
def decode(self, encoded: List[int], first: int) -> List[int]:
results = []
results.append(first)

i = 0
for e in encoded:
reverse_xor = e ^ results[i]
results.append(reverse_xor)
i += 1

return results``````
• Time Complexity: O(N).
• Space Complexity: O(N).