README_EN.md

1734. Decode XORed Permutation

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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

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(n + 1):
            b ^= i
        ans = [a ^ b]
        for e in encoded[::-1]:
            ans.append(ans[-1] ^ e)
        return ans[::-1]

Java

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

C++

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

Go

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

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