minte9

<?php

/**
 * Define $output on multiple $lines
 * Use Herodoc syntax.
 *
 * output =
 *      She said "This is John's test"
 *      on multiple rows
 */

minte9

<?php

/**
 * Define getMaxValue() method
 * Use static syntax
 */
class MyClass {

    const MAX_VALUE = 100;

minte9

<?php

/**
 * array_map()
 *
 * Round up array values
 * Use array_map() with built-in ceil functions
 * The result array $mapped will be [2, 3, 4]
 */

minte9

<?php

/**
 * Define class constructor and ...
 * Display the constructor method name
 */
$obj = new A();

class A
{

minte9

<?php

/**
 * Compose $m octal value using myFunc() recursively
 * Use reminder and quontient values
 *
 * 97 (decimal) = 141 (octal)
 * 97 % 8 => reminder 1
 * 12 % 8 => reminder 4
 * 1  % 8 => reminder 1

minte9

<?php

/**
 * Given the users $role and access $rights ...
 * Check if Mike has admin rights and downgrade him
 * Wrong approach:
 *  if ($is_admin && $can_edit_articles ...
 */
 $rights = (object) ["read"=>1, "write"=>2, "readwrite"=>16, "admin"=> 32];
 $role   = (object) ["jim"=>96, "mike"=>32];

minte9

<?php

/**
 * Replace
 * {b} abc {/b} with:
 * <b> abc </b>
 * Use preg_replace() function
 */
$str = "{b} abc {/b}";

minte9

# Write a function convert() that returns 
# days, hours, minutes, seconds since UTC
# The time module provides a function, also named time, 
# that returns the current GMT in “the epoch”

import time
today = time.time()

# SOLUTION

minte9

# A Caesar cypher is a weak form on encryption
# It involves "rotating" each letter by a number (shift it through the alphabet)
# A rotated by 3 is D; Z rotated by 1 is A
# In a SF movie the computer is called HAL, which is IBM rotated by -1

# Write a function rotate_word()
# Use built-in functions ord (char to code_number), chr (codes to char)

# SOLUTION

minte9

# Get the words with three consecutive double letters
# https://github.com/AllenDowney/ThinkPython2/blob/master/code/words.txt

def has3_2consecutive(word):
    i = 0; count = 0
    while i < len(word) - 1:
        if word[i] == word[i+1]:
            count = count + 1
            if count == 3:
                return True