Formatted question description: https://leetcode.ca/all/1734.html

# 1734. Decode XORed Permutation

Medium

## 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 < 10^5
• n is odd.
• encoded.length == n - 1

## Solution

Use the condition that n is odd to solve this problem. First calculate totalXOR as the XOR from 1 to n. Next, calculate the XOR result of even indices of encoded as evenXOR, and the XOR result of odd indices of encoded as oddXOR. Then there is decoded = totalXOR ^ oddXOR and decoded[n - 1] = totalXOR ^ evenXOR. Finally, for i from 0 to n - 2, there is decoded[i + 1] = decoded[i] ^ encoded[i].

class Solution {
public int[] decode(int[] encoded) {
int length = encoded.length;
int n = length + 1;
int totalXOR = 0;
for (int i = 1; i <= n; i++)
totalXOR ^= i;
int[] decoded = new int[n];
int evenXOR = 0, oddXOR = 0;
for (int i = 0; i < length; i++) {
if (i % 2 == 0)
evenXOR ^= encoded[i];
else
oddXOR ^= encoded[i];
}
decoded = totalXOR ^ oddXOR;
decoded[n - 1] = totalXOR ^ evenXOR;
for (int i = 0; i < length; i++)
decoded[i + 1] = decoded[i] ^ encoded[i];
return decoded;
}
}