# 1720. Decode XORed Array

## LeetCode [1720. Decode XORed Array](https://leetcode-cn.com/problems/decode-xored-array/)

### Description

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[0]`.

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 <= 10^4`
* `encoded.length == n - 1`
* `0 <= encoded[i] <= 10^5`
* `0 <= first <= 10^5`

### Tags

Bit Manipulation

### Solution

![](https://3979976701-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-MWCWmbVhbhatBCAMIGH%2Fuploads%2Fgit-blob-6225f8dd0a3cf88d52c059fdf132a071706c488a%2Fimage%20\(32\).png?alt=media)

Since `encoded[i] = arr[i] XOR arr[i + 1] (i≥-1)`, then `arr[i+1] = encoded[i] XOR arr[i]` because **XOR is reversible**. Thus, we iterate on `i` from beginning to end, and set `arr[i+1] = encoded[i] XOR arr[i]`.

### Complexity

* Time complexity: $$O(n)$$
* Space complexity: $$O(n)$$

### Code

```go
func decode(encoded []int, first int) []int {
	ans := make([]int, len(encoded)+1)
	ans[0] = first
	for i := 1; i < len(ans); i++ {
		ans[i] = ans[i-1] ^ encoded[i-1]
	}
	return ans
}
```

## Reference

1. <https://www.zhihu.com/question/28693672>
