https://www.w3schools.com/Html/default.asp
Understand a bit about DOM tree: https://www.w3schools.com/Js/js_htmldom.asp
Understand a bit about URI and its components: https://en.wikipedia.org/wiki/Uniform_Resource_Identifier
| rabbitmqctl add_user test test | |
| rabbitmqctl set_user_tags test administrator | |
| rabbitmqctl set_permissions -p / test ".*" ".*" ".*" |
| import logging | |
| import tempfile | |
| import subprocess | |
| class Result(object): | |
| def __init__(self, rc, stdout, stderr, cmd): | |
| self.rc = rc | |
| self.stdout = stdout | |
| self.stderr = stderr |
| """ | |
| 1.5 Implement a method to perform basic string compression using the counts | |
| of repeated characters. For example, the string aabcccccaaa would become | |
| a2b1c5a3. If the "compressed" string would not become smaller than the original | |
| string, your method should return the original string. | |
| """ | |
| # Complexity: time O(n), space O(n) | |
| def compress_string(string): | |
| if not string: | |
| return |
| """ | |
| 1.4 Write a method to replace all spaces in a string with '%20'. You may assume that | |
| the string has sufficient space at the end of the string to hold the additional | |
| characters, and that you are given the "true" length of the string. (Note: if implementing | |
| in Java, please use a character array so that you can perform this operation | |
| in place.) | |
| EXAMPLE | |
| Input: "Mr John Smith " | |
| Output: "Mr%20Dohn%20Smith" |
| """ | |
| 1.3 Given two strings, write a method to decide if one is a permutation of the other. | |
| tips: | |
| assumption - case sensitive, whitespace matters | |
| """ | |
| # sort string and compare | |
| # Complexity: time O(nlogn), space O(1) | |
| def is_permutation(str1, str2): | |
| if not str1 or not str2 or len(str1) != len(str2): | |
| return False |
| """ | |
| 1.2 Implement a function void reverse(char* str) in C or C++ which reverses a null terminated string. | |
| """ | |
| # In python, str is immutable | |
| # Complexity: time O(n), space O(n) | |
| def reverse_string(string): | |
| if not string: | |
| return | |
| res = "" | |
| for i in range(1, len(string) + 1): |
| """ | |
| 1.1 Implement an algorithm to determine if a string has all unique characters. What if you cannot use additional data structures. | |
| tips: | |
| encodings - anscii, latin1, unicode | |
| corner case - empty string's return value | |
| """ | |
| # assume ascii case | |
| # Complexity: time O(n), space O(1) | |
| def unique_chars(string): | |
| if not string or len(string) > 128: |