How do you solve Decode XORed Permutation on LeetCode?

Decode XORed Permutation (LeetCode #1734) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: XOR is a powerful tool for solving problems involving pairs of numbers, especially when the order doesn't matter.

What is the brute force solution for Decode XORed Permutation?

The brute force approach for Decode XORed Permutation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible permutations of the numbers from 1 to n and checking which one produces the given encoded array. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Decode XORed Permutation?

Decode XORed Permutation uses the following concepts: Array, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Array.

What are the interview tips for Decode XORed Permutation?

Always think about the properties of the operations you are using, such as XOR in this case. When faced with a permutation problem, consider how you can leverage mathematical properties rather than brute-force generation. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Decode XORed Permutation?

Students often forget that XOR is both commutative and associative, which can lead to incorrect assumptions about order. Another common mistake is not properly handling the edge cases when constructing the original permutation.

#1734

Decode XORed Permutation

Medium

ArrayBit ManipulationBit ManipulationArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the properties of XOR and the fact that n is odd. By computing the XOR of all numbers from 1 to n and using the encoded array, we can derive the original permutation efficiently.

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Algorithm

4 steps

1Step 1: Compute the XOR of all numbers from 1 to n (let's call this totalXor).
2Step 2: Compute the XOR of all elements in the encoded array (let's call this encodedXor).
3Step 3: The first element of perm can be found as first = totalXor XOR encodedXor.
4Step 4: Initialize perm[0] with first and use the encoded array to fill the rest of perm using perm[i] = perm[i-1] XOR encoded[i-1].

solution.py13 lines

1def decode_optimal(encoded):
2    n = len(encoded) + 1
3    totalXor = 0
4    for i in range(1, n + 1):
5        totalXor ^= i
6    encodedXor = 0
7    for num in encoded:
8        encodedXor ^= num
9    first = totalXor ^ encodedXor
10    perm = [first]
11    for i in range(len(encoded)):
12        perm.append(perm[-1] ^ encoded[i])
13    return perm

ℹ

Complexity note: This complexity arises because we iterate through the encoded array and the range from 1 to n, leading to linear growth in time with respect to n.

1XOR is a powerful tool for solving problems involving pairs of numbers, especially when the order doesn't matter.
2The fact that n is odd guarantees that there will be a unique solution, which simplifies our approach.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.