Decode XORed Permutation - Problem
Decode XORed Permutation
Imagine you have a secret permutation array containing the first
The encoding process creates an array
Your Mission: Given only the encoded array, recover the original permutation! The solution is guaranteed to be unique.
Hint: Think about the properties of XOR operations and how permutations work. The fact that
Imagine you have a secret permutation array containing the first
n positive integers (where n is always odd), but it's been encoded using XOR operations! πThe encoding process creates an array
encoded of length n-1 where each element encoded[i] = perm[i] XOR perm[i+1]. For example, if the original permutation is [1,3,2], the encoded array becomes [1β3, 3β2] = [2,1].Your Mission: Given only the encoded array, recover the original permutation! The solution is guaranteed to be unique.
Hint: Think about the properties of XOR operations and how permutations work. The fact that
n is odd is a crucial clue! π‘ Input & Output
example_1.py β Simple case
$
Input:
encoded = [3, 1]
βΊ
Output:
[1, 2, 3]
π‘ Note:
The original permutation [1, 2, 3] was encoded as [1β2, 2β3] = [3, 1]. We can decode it by finding that the first element is 1, then 1β3=2, and 2β1=3.
example_2.py β Medium case
$
Input:
encoded = [6, 5, 4, 6]
βΊ
Output:
[2, 4, 1, 5, 3]
π‘ Note:
Using XOR properties: total_xor of 1β2β3β4β5 = 1, odd_indices_xor = 5β6 = 3, so first element = 1β3 = 2. Then reconstruct: 2β4β1β5β3.
example_3.py β Single element
$
Input:
encoded = [0]
βΊ
Output:
[1, 1]
π‘ Note:
Wait, this is invalid! A permutation can't have duplicate elements. The smallest valid case is n=3. This test case would be [some_valid_encoded_array] for nβ₯3.
Visualization
Tap to expand
Understanding the Visualization
1
Analyze the Pattern
Each encoded[i] = perm[i] β perm[i+1], creating a chain of XOR relationships
2
Mathematical Insight
XOR of numbers 1 to n has predictable behavior, especially when n is odd
3
Find the Key
Calculate total XOR and odd-position XOR to isolate the first element
4
Reconstruct Chain
Use the first element and XOR properties to rebuild the entire permutation
Key Takeaway
π― Key Insight: The constraint that n is odd makes the XOR of numbers 1 to n predictable. Combined with the fact that each number appears in exactly two XOR operations (except boundaries), we can mathematically isolate the first element without brute force!
Time & Space Complexity
Time Complexity
O(n)
Single pass to calculate XORs, then single pass to reconstruct permutation
β Linear Growth
Space Complexity
O(1)
Only constant extra space for calculations, output array not counted
β Linear Space
Constraints
- 3 β€ n β€ 3 * 105
- n is odd
- encoded.length == n - 1
- 1 β€ encoded[i] β€ 109
- perm is a valid permutation of [1, 2, ..., n]
π‘
Explanation
AI Ready
π‘ Suggestion
Tab
to accept
Esc
to dismiss
// Output will appear here after running code