# Solving Leetcode Interviews in Seconds with AI: Decode XORed Permutation


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1734" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 < 105 n is odd. encoded.length == n - 1  

	# Explanation
	*   **Calculate the XOR of all numbers from 1 to n:** Since `perm` is a permutation of numbers from 1 to n, the XOR of all elements in `perm` is simply the XOR of numbers from 1 to n.
*   **Calculate the XOR of encoded elements at even indices:** The XOR of `encoded[0]`, `encoded[2]`, `encoded[4]`, ... is equivalent to `(perm[0] ^ perm[1]) ^ (perm[2] ^ perm[3]) ^ ...`. If we XOR this result with the XOR of all numbers from 1 to n, most of the `perm` elements will cancel out, leaving us with the last element of `perm`, which is `perm[0]`.
*   **Reconstruct the perm array:** Once `perm[0]` is known, we can easily reconstruct the rest of the `perm` array by XORing `perm[i]` with `encoded[i]` to get `perm[i+1]`.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(n)

	
	# Code
	```python
	def decode(encoded: list[int]) -> list[int]:
    """
    Decodes the original permutation array from the encoded array.

    Args:
        encoded: The encoded array where encoded[i] = perm[i] XOR perm[i+1].

    Returns:
        The original permutation array perm.
    """
    n = len(encoded) + 1
    total_xor = 0
    for i in range(1, n + 1):
        total_xor ^= i

    even_xor = 0
    for i in range(0, len(encoded), 2):
        even_xor ^= encoded[i]

    first_element = total_xor ^ even_xor
    perm = [first_element]
    for i in range(len(encoded)):
        perm.append(perm[-1] ^ encoded[i])

    return perm
	```