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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].

• Java
• C++
• Python

• 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;
      }
  };
  

• 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]
  
  ############
  
  # 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
  

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