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 = 5
, b = 3
, c = 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).