LeetCode 1720. Decode XORed Array

Description

https://leetcode.com/problems/decode-xored-array/

There is a hidden integer array arr that consists of n non-negative integers.

It was encoded into another integer array encoded of length n - 1, such that encoded[i] = arr[i] XOR arr[i + 1]. For example, if arr = [1,0,2,1], then encoded = [1,2,3].

You are given the encoded array. You are also given an integer first, that is the first element of arr, i.e. arr[0].

Return the original array arr. It can be proved that the answer exists and is unique.

Example 1:

Input: encoded = [1,2,3], first = 1
Output: [1,0,2,1]
Explanation: If arr = [1,0,2,1], then first = 1 and encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3]

Example 2:

Input: encoded = [6,2,7,3], first = 4
Output: [4,2,0,7,4]

Constraints:

  • 2 <= n <= 104
  • encoded.length == n - 1
  • 0 <= encoded[i] <= 105
  • 0 <= first <= 105

Explanation

The key is to find the inverse is XOR.

If you have:

c = a^b;

You can get a or b back if you have the other value available:

a = c^b; // or b^c (order is not important)
b = c^a; // or a^c

For example if a = 5b = 3c = 6 you get:

b=0011 (3)            a=0101 (5)
c=0110 (6) XOR   or   c=0110 (6) XOR
----------            ----------
a=0101 (5)            b=0011 (3)

Python Solution

class Solution:
    def decode(self, encoded: List[int], first: int) -> List[int]:
        results = []
        results.append(first)
        
        i = 0
        for e in encoded:
            reverse_xor = e ^ results[i]            
            results.append(reverse_xor)
            i += 1
        
        return results
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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