LeetCode in Kotlin
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 < 105nis odd.encoded.length == n - 1
Solution
class Solution {
fun decode(encoded: IntArray): IntArray {
val decoded = IntArray(encoded.size + 1)
run {
var i = 1
while (i < encoded.size) {
decoded[0] = decoded[0] xor encoded[i]
decoded[0] = decoded[0] xor i
decoded[0] = decoded[0] xor (i + 1)
i += 2
}
}
decoded[0] = decoded[0] xor decoded.size
for (i in 1 until decoded.size) {
decoded[i] = decoded[i - 1] xor encoded[i - 1]
}
return decoded
}
}