README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1700-1799/1720.Decode%20XORed%20Array/README_EN.md	1284	Weekly Contest 223 Q1	Bit Manipulation Array

1720. Decode XORed Array

中文文档

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

Solutions

Solution 1: Bit Manipulation

Based on the problem description, we have:

$$ \textit{encoded}[i] = \textit{arr}[i] \oplus \textit{arr}[i + 1] $$

If we XOR both sides of the equation with $\textit{arr}[i]$, we get:

$$ \textit{arr}[i] \oplus \textit{arr}[i] \oplus \textit{arr}[i + 1] = \textit{arr}[i] \oplus \textit{encoded}[i] $$

Which simplifies to:

$$ \textit{arr}[i + 1] = \textit{arr}[i] \oplus \textit{encoded}[i] $$

Following the derivation above, we can start with $\textit{first}$ and sequentially calculate every element of the array $\textit{arr}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array.

Python3

class Solution:
    def decode(self, encoded: List[int], first: int) -> List[int]:
        ans = [first]
        for x in encoded:
            ans.append(ans[-1] ^ x)
        return ans

Java

class Solution {
    public int[] decode(int[] encoded, int first) {
        int n = encoded.length;
        int[] ans = new int[n + 1];
        ans[0] = first;
        for (int i = 0; i < n; ++i) {
            ans[i + 1] = ans[i] ^ encoded[i];
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> decode(vector<int>& encoded, int first) {
        vector<int> ans = {{first}};
        for (int x : encoded) {
            ans.push_back(ans.back() ^ x);
        }
        return ans;
    }
};

Go

func decode(encoded []int, first int) []int {
	ans := []int{first}
	for i, x := range encoded {
		ans = append(ans, ans[i]^x)
	}
	return ans
}

TypeScript

function decode(encoded: number[], first: number): number[] {
    const ans: number[] = [first];
    for (const x of encoded) {
        ans.push(ans.at(-1)! ^ x);
    }
    return ans;
}