Decode XORed Permutation - Problem
Decode XORed Permutation

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! πŸ”

The 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 [1βŠ•3, 3βŠ•2] = [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! πŸ’‘

Input & Output

example_1.py β€” Simple case
$ Input: encoded = [3, 1]
β€Ί Output: [1, 2, 3]
πŸ’‘ Note: The original permutation [1, 2, 3] was encoded as [1βŠ•2, 2βŠ•3] = [3, 1]. We can decode it by finding that the first element is 1, then 1βŠ•3=2, and 2βŠ•1=3.
example_2.py β€” Medium case
$ Input: encoded = [6, 5, 4, 6]
β€Ί Output: [2, 4, 1, 5, 3]
πŸ’‘ Note: Using XOR properties: total_xor of 1βŠ•2βŠ•3βŠ•4βŠ•5 = 1, odd_indices_xor = 5βŠ•6 = 3, so first element = 1βŠ•3 = 2. Then reconstruct: 2β†’4β†’1β†’5β†’3.
example_3.py β€” Single element
$ Input: encoded = [0]
β€Ί Output: [1, 1]
πŸ’‘ 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
Decoding XOR PermutationOriginal Permutation (Hidden)?????Encoded Array (Given)6546XOR CalculationsTotal XOR (1βŠ•2βŠ•3βŠ•4βŠ•5) = 1Odd positions XOR (encoded[1]βŠ•encoded[3]) = 5βŠ•6 = 3First element = 1βŠ•3 = 2Reconstructed Permutation24153πŸ” Key Insights1. XOR Properties:β€’ a βŠ• b βŠ• b = a (self-canceling)β€’ XOR of 1 to n follows pattern2. Encoding Pattern:β€’ Each perm[i] appears twiceβ€’ Except first and last elements3. Odd Constraint:β€’ Makes XOR pattern predictableβ€’ Enables direct calculationTime: O(n) | Space: O(1)
Understanding the Visualization
1
Analyze the Pattern
Each encoded[i] = perm[i] βŠ• 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
🎯 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

n
2n
βœ“ Linear Growth
Space Complexity
O(1)

Only constant extra space for calculations, output array not counted

n
2n
βœ“ Linear Space

Constraints

  • 3 ≀ n ≀ 3 * 105
  • n is odd
  • encoded.length == n - 1
  • 1 ≀ encoded[i] ≀ 109
  • perm is a valid permutation of [1, 2, ..., n]
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