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Formatted question description: https://leetcode.ca/all/1734.html

# 1734. Decode XORed Permutation

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].

• 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;
}
}

############

class Solution {
public int[] decode(int[] encoded) {
int n = encoded.length + 1;
int[] ans = new int[n];
int a = 0;
int b = 0;
for (int i = 0; i < n - 1; i += 2) {
a ^= encoded[i];
}
for (int i = 0; i < n + 1; ++i) {
b ^= i;
}
ans[n - 1] = a ^ b;
for (int i = n - 2; i >= 0; --i) {
ans[i] = ans[i + 1] ^ encoded[i];
}
return ans;
}
}

• // 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

• func decode(encoded []int) []int {
n := len(encoded) + 1
ans := make([]int, n)
a, b := 0, 0
for i := 0; i < n-1; i += 2 {
a ^= encoded[i]
}
for i := 0; i < n+1; i++ {
b ^= i
}
ans[n-1] = a ^ b
for i := n - 2; i >= 0; i-- {
ans[i] = ans[i+1] ^ encoded[i]
}
return ans
}