# Solving Leetcode Interviews in Seconds with AI: Find The Original Array of Prefix Xor


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2433" 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
	> You are given an integer array pref of size n. Find and return the array arr of size n that satisfies:  pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i].  Note that ^ denotes the bitwise-xor operation. It can be proven that the answer is unique.   Example 1:  Input: pref = [5,2,0,3,1] Output: [5,7,2,3,2] Explanation: From the array [5,7,2,3,2] we have the following: - pref[0] = 5. - pref[1] = 5 ^ 7 = 2. - pref[2] = 5 ^ 7 ^ 2 = 0. - pref[3] = 5 ^ 7 ^ 2 ^ 3 = 3. - pref[4] = 5 ^ 7 ^ 2 ^ 3 ^ 2 = 1.  Example 2:  Input: pref = [13] Output: [13] Explanation: We have pref[0] = arr[0] = 13.    Constraints:  1 <= pref.length <= 105 0 <= pref[i] <= 106  

	# Explanation
	Here's the solution to the problem:

*   **Key Idea:** The core idea is to leverage the properties of the XOR operation. Specifically, `a ^ a = 0` and `a ^ 0 = a`. Since `pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i]` and `pref[i-1] = arr[0] ^ arr[1] ^ ... ^ arr[i-1]`, then `pref[i] ^ pref[i-1] = arr[i]`.
*   **Iterative Calculation:** We can iterate through the `pref` array, calculating each `arr[i]` by XORing `pref[i]` with `pref[i-1]`. The first element `arr[0]` is simply `pref[0]`.
*   **In-place Modification (Optional):** To save space, it is possible to perform the array modification in-place. But in this solution to improve readability a new array for results is created.

*   **Runtime & Storage Complexity:** O(n) runtime, O(n) storage.

	
	# Code
	```python
	def find_array(pref: list[int]) -> list[int]:
    """
    Given an integer array pref of size n.
    Find and return the array arr of size n that satisfies:
    pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i].

    Note that ^ denotes the bitwise-xor operation.
    It can be proven that the answer is unique.

    Example 1:
    Input: pref = [5,2,0,3,1]
    Output: [5,7,2,3,2]

    Example 2:
    Input: pref = [13]
    Output: [13]
    """
    n = len(pref)
    arr = [0] * n
    arr[0] = pref[0]

    for i in range(1, n):
        arr[i] = pref[i] ^ pref[i-1]

    return arr
	```