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

1734. Decode XORed Permutation

Level

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[0] = 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[0] = totalXOR ^ oddXOR;
        decoded[n - 1] = totalXOR ^ evenXOR;
        for (int i = 0; i < length; i++)
            decoded[i + 1] = decoded[i] ^ encoded[i];
        return decoded;
    }
}

// OJ: https://leetcode.com/problems/decode-xored-permutation/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> decode(vector<int>& A) {
        int N = A.size() + 1;
        vector<int> ans(N);
        for (int i = 0; i <= 16; ++i) {
            int target = 0, bit = 0, cnt = 0;
            for (int j = 1; j <= N; ++j) target += j >> i & 1;
            for (int n : A) {
                bit ^= n >> i & 1;
                cnt += bit;
            }
            ans[0] |= (cnt != target) << i;
            for (int j = 1; j < N; ++j) {
                int x = A[j - 1] >> i & 1, prev = ans[j - 1] >> i & 1, b = x ^ prev;
                ans[j] |= b << i;
            }
        }
        return ans;
    }
};

# 1734. Decode XORed Permutation
# https://leetcode.com/problems/decode-xored-permutation/

class Solution:
    def decode(self, encoded: List[int]):
        curr = start = 0
        
        for i in range(1, len(encoded)+2):
            start ^= i
        
        for x in encoded:
            curr ^= x
            start ^= curr
        
        res = [start]
        for x in encoded:
            res.append(res[-1] ^ x)
        
        return res

1734 - Decode XORed Permutation

1734. Decode XORed Permutation

Level

Description

Solution

All Problems

All Solutions