1734. Decode XORed Permutation
Medium
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 < 10^5nis odd.encoded.length == n - 1
func decode(encoded []int) []int {
// perm[i] = perm[i – 1] ^ encoded[i – 1]
// 1. p[0] ^ p[1] ^ … ^ p[n-1] = 1 ^ 2 ^ … ^ n
// 2. encoded[1] ^ encode[3] ^ … ^ encoded[n-2] = (p[1] ^ p[2]) ^ (p[3] ^ p[4]) ^ … ^ (p[n-2] ^ p[n-1])
// 1. xor 2. = p[0]
// now we can calculate perm[i] = perm[i – 1] ^ encoded[i – 1] from i = 1
sum := 0
for i := 1; i <= len(encoded) + 1; i++ {
sum ^= i
}
nohead := 0
for i := 1; i < len(encoded); i+=2 {
nohead ^= encoded[i]
}
head := nohead ^ sum
ans := make([]int, 0)
ans = append(ans, head)
for i := 1; i < len(encoded) + 1; i++ {
ans = append(ans, ans[i-1]^encoded[i-1])
}
return ans
}
Last modified 4mo ago