Welcome to Subscribe On Youtube
1720. Decode XORed Array
Description
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 <= 104encoded.length == n - 10 <= encoded[i] <= 1050 <= first <= 105
Solutions
XOR.
a = b ^ c => a ^ b = b ^ c ^ b => c = a ^ b.
-
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; } }; -
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 -
func decode(encoded []int, first int) []int { ans := []int{first} for i, e := range encoded { ans = append(ans, ans[i]^e) } return ans } -
function decode(encoded: number[], first: number): number[] { const ans: number[] = [first]; for (const x of encoded) { ans.push(ans.at(-1)! ^ x); } return ans; }