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
XOR.
a = b ^ c
=> a ^ b = b ^ c ^ b
=> c = a ^ b
.
class Solution:
def decode(self, encoded: List[int], first: int) -> List[int]:
ans = [first]
for e in encoded:
ans.append(ans[-1] ^ e)
return ans
class Solution {
public int[] decode(int[] encoded, int first) {
int n = encoded.length;
int[] ans = new int[n + 1];
ans[0] = first;
for (int i = 0; i < n; ++i) {
ans[i + 1] = ans[i] ^ encoded[i];
}
return ans;
}
}
class Solution {
public:
vector<int> decode(vector<int>& encoded, int first) {
vector<int> ans{{first}};
for (int i = 0; i < encoded.size(); ++i)
ans.push_back(ans[i] ^ encoded[i]);
return ans;
}
};
func decode(encoded []int, first int) []int {
ans := []int{first}
for i, e := range encoded {
ans = append(ans, ans[i]^e)
}
return ans
}