Moved 1 to 5 to Python and learned a ton.

python/lib_misc.py

@@ -0,0 +1,78 @@
+def get_digits_reversed(n):
+    """
+    Returns a list of digits for n.
+    """
+    digits, rest = [], n
+    while rest != 0:
+        digits.append(rest % 10)
+        rest //= 10
+    return digits
+
+
+def is_palindrome_string(n):
+    """
+    Checks whether n is a palindrome number
+    with a string based algorithm.
+
+    :param n: number to check
+    :returns: boolean
+
+    """
+    s = str(n)
+    return s == s[::-1]
+
+
+def is_palindrome_integer(n):
+    """
+    Checks whether n is a palindrome number
+    with an integer based algorithm.
+
+    :param n: number to check
+    :returns: boolean
+    """
+    digits = get_digits_reversed(n)
+    digits_count = len(digits)
+    for i in range(digits_count // 2):
+        if digits[i] != digits[digits_count - i - 1]:
+            return False
+    return True
+
+
+def time_is_palindrom():
+    """ I just want to check which option is faster. """
+    from timeit import timeit
+    print(timeit(lambda: is_palindrome_integer(82445254428), number=100000))
+    print(timeit(lambda: is_palindrome_string(82445254428), number=100000))
+    # > 0.42330633100027626
+    # > 0.07605635199979588
+
+
+is_palindrome = is_palindrome_string
+
+
+def get_item_counts(l):
+    """
+    Counts how often each element is part of the list.
+
+    :param l: a list
+    :returns: a dictionary
+    """
+    d = {}
+    for e in l:
+        try:
+            d[e] += 1
+        except KeyError:
+            d[e] = 1
+    return d
+
+
+def product(l):
+    """
+    Calculates the product of all items in the list.
+
+    :param l: a list
+    :returns: a number that is the product of all elements in the list
+    """
+    from functools import reduce
+    import operator
+    return reduce(operator.mul, l, 1)