README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1700-1799/1734.Decode%20XORed%20Permutation/README_EN.md	2024	Biweekly Contest 44 Q3	Bit Manipulation Array

1734. Decode XORed Permutation

中文文档

Description

There is an integer array perm that is a permutation of the first n positive integers, where n is always odd.

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

Given the encoded array, return the original array perm. It is guaranteed that the answer exists and is unique.

 

Example 1:

Input: encoded = [3,1]
Output: [1,2,3]
Explanation: If perm = [1,2,3], then encoded = [1 XOR 2,2 XOR 3] = [3,1]

Example 2:

Input: encoded = [6,5,4,6]
Output: [2,4,1,5,3]

 

Constraints:

• 3 <= n < 105
• n is odd.
• encoded.length == n - 1

Solutions

Solution 1: Bitwise Operation

We notice that the array $perm$ is a permutation of the first $n$ positive integers, so the XOR of all elements in $perm$ is $1 \oplus 2 \oplus \cdots \oplus n$, denoted as $a$. And $encode[i]=perm[i] \oplus perm[i+1]$, if we denote the XOR of all elements $encode[0],encode[2],\cdots,encode[n-3]$ as $b$, then $perm[n-1]=a \oplus b$. Knowing the last element of $perm$, we can find all elements of $perm$ by traversing the array $encode$ in reverse order.

The time complexity is $O(n)$, where $n$ is the length of the array $perm$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

Python3

class Solution:
    def decode(self, encoded: List[int]) -> List[int]:
        n = len(encoded) + 1
        a = b = 0
        for i in range(0, n - 1, 2):
            a ^= encoded[i]
        for i in range(1, n + 1):
            b ^= i
        perm = [0] * n
        perm[-1] = a ^ b
        for i in range(n - 2, -1, -1):
            perm[i] = encoded[i] ^ perm[i + 1]
        return perm

Java

class Solution {
    public int[] decode(int[] encoded) {
        int n = encoded.length + 1;
        int a = 0, b = 0;
        for (int i = 0; i < n - 1; i += 2) {
            a ^= encoded[i];
        }
        for (int i = 1; i <= n; ++i) {
            b ^= i;
        }
        int[] perm = new int[n];
        perm[n - 1] = a ^ b;
        for (int i = n - 2; i >= 0; --i) {
            perm[i] = encoded[i] ^ perm[i + 1];
        }
        return perm;
    }
}

C++

class Solution {
public:
    vector<int> decode(vector<int>& encoded) {
        int n = encoded.size() + 1;
        int a = 0, b = 0;
        for (int i = 0; i < n - 1; i += 2) {
            a ^= encoded[i];
        }
        for (int i = 1; i <= n; ++i) {
            b ^= i;
        }
        vector<int> perm(n);
        perm[n - 1] = a ^ b;
        for (int i = n - 2; ~i; --i) {
            perm[i] = encoded[i] ^ perm[i + 1];
        }
        return perm;
    }
};

Go

func decode(encoded []int) []int {
	n := len(encoded) + 1
	a, b := 0, 0
	for i := 0; i < n-1; i += 2 {
		a ^= encoded[i]
	}
	for i := 1; i <= n; i++ {
		b ^= i
	}
	perm := make([]int, n)
	perm[n-1] = a ^ b
	for i := n - 2; i >= 0; i-- {
		perm[i] = encoded[i] ^ perm[i+1]
	}
	return perm
}