Solving Leetcode Interviews in Seconds with AI: Decode XORed Array
Introduction
In this blog post, we will explore how to solve the LeetCode problem "1720" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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 a hidden integer array arr that consists of n non-negative integers. It was encoded into another integer array encoded of length n - 1, such that encoded[i] = arr[i] XOR arr[i + 1]. For example, if arr = [1,0,2,1], then encoded = [1,2,3]. You are given the encoded array. You are also given an integer first, that is the first element of arr, i.e. arr[0]. Return the original array arr. It can be proved that the answer exists and is unique. Example 1: Input: encoded = [1,2,3], first = 1 Output: [1,0,2,1] Explanation: If arr = [1,0,2,1], then first = 1 and encoded = [1 XOR 0, 0 XOR 2, 2 XOR 1] = [1,2,3] Example 2: Input: encoded = [6,2,7,3], first = 4 Output: [4,2,0,7,4] Constraints: 2 <= n <= 104 encoded.length == n - 1 0 <= encoded[i] <= 105 0 <= first <= 105
Explanation
Here's the breakdown of the approach, complexity, and the Python code:
Approach:
- Initialize the original array
arrwith the givenfirstelement. - Iterate through the
encodedarray. For each elementencoded[i], calculatearr[i+1]using the XOR property:arr[i+1] = arr[i] ^ encoded[i]. Append the calculated value toarr. - Return the constructed
arr.
- Initialize the original array
Complexity:
- Runtime Complexity: O(n), where n is the length of the
encodedarray. - Storage Complexity: O(n) to store the resulting
arrarray.
- Runtime Complexity: O(n), where n is the length of the
Code
def decode(encoded, first):
"""
Decodes the original array from the encoded array and the first element.
Args:
encoded: The encoded array.
first: The first element of the original array.
Returns:
The original array.
"""
arr = [first]
for i in range(len(encoded)):
arr.append(arr[i] ^ encoded[i])
return arr