Introduction

In this blog post, we will explore how to solve the LeetCode problem "831" 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

You are given a personal information string s, representing either an email address or a phone number. Return the masked personal information using the below rules. Email address: An email address is: A name consisting of uppercase and lowercase English letters, followed by The '@' symbol, followed by The domain consisting of uppercase and lowercase English letters with a dot '.' somewhere in the middle (not the first or last character). To mask an email: The uppercase letters in the name and domain must be converted to lowercase letters. The middle letters of the name (i.e., all but the first and last letters) must be replaced by 5 asterisks "*". Phone number: A phone number is formatted as follows: The phone number contains 10-13 digits. The last 10 digits make up the local number. The remaining 0-3 digits, in the beginning, make up the country code. Separation characters from the set {'+', '-', '(', ')', ' '} separate the above digits in some way. To mask a phone number: Remove all separation characters. The masked phone number should have the form: "--XXXX" if the country code has 0 digits. "+-**-*-XXXX" if the country code has 1 digit. "+---XXXX" if the country code has 2 digits. "+---XXXX" if the country code has 3 digits. "XXXX" is the last 4 digits of the local number. Example 1: Input: s = "LeetCode@LeetCode.com" Output: "l**e@leetcode.com" Explanation: s is an email address. The name and domain are converted to lowercase, and the middle of the name is replaced by 5 asterisks. Example 2: Input: s = "AB@qq.com" Output: "a*b@qq.com" Explanation: s is an email address. The name and domain are converted to lowercase, and the middle of the name is replaced by 5 asterisks. Note that even though "ab" is 2 characters, it still must have 5 asterisks in the middle. Example 3: Input: s = "1(234)567-890" Output: "--7890" Explanation: s is a phone number. There are 10 digits, so the local number is 10 digits and the country code is 0 digits. Thus, the resulting masked number is "--7890". Constraints: s is either a valid email or a phone number. If s is an email: 8 <= s.length <= 40 s consists of uppercase and lowercase English letters and exactly one '@' symbol and '.' symbol. If s is a phone number: 10 <= s.length <= 20 s consists of digits, spaces, and the symbols '(', ')', '-', and '+'.

Explanation

Here's the breakdown of the solution:

• Identify the Input Type: Determine if the input string s is an email address or a phone number by checking for the presence of the "@" symbol.

• Mask Accordingly: Apply the appropriate masking logic based on the identified type. For emails, convert to lowercase, replace the middle of the name with asterisks, and format the domain. For phone numbers, remove separators, determine the country code length, and format the masked number with asterisks and digits.

• Runtime Complexity: O(n), where n is the length of the input string s.

• Storage Complexity: O(1).

Code

    class Solution:
    def maskPII(self, s: str) -> str:
        if '@' in s:
            # Email
            name, domain = s.split('@')
            name = name.lower()
            domain = domain.lower()
            return name[0] + "*****" + name[-1] + "@" + domain
        else:
            # Phone number
            digits = ''.join(c for c in s if c.isdigit())
            last_four = digits[-4:]
            country_code_len = len(digits) - 10
            if country_code_len == 0:
                return "***-***-" + last_four
            elif country_code_len == 1:
                return "+*-***-***-" + last_four
            elif country_code_len == 2:
                return "+**-***-***-" + last_four
            else:
                return "+***-***-***-" + last_four