Anagram

Two words are anagrams if they contain exactly the same letters, just in a different order. listen and silent. evil and vile. astronomer and moon starer (ok, that last one has a space, but you get the idea)

How do we check? The simplest approach: sort the letters and compare

def is_anagram(word1, word2):
    return sorted(word1.lower()) == sorted(word2.lower())

print(is_anagram("listen", "silent"))
print(is_anagram("hello", "world"))
print(is_anagram("Evil", "Vile"))

Output

True
False
True

Why does this work? If two words have the same letters, sorting those letters will produce the same result. "listen" sorted → "eilnst". "silent" sorted → "eilnst". Same? Yes → anagram

We use .lower() so that uppercase and lowercase don't matter

But let's also solve it using what we learned: a frequency dictionary. Two words are anagrams if every letter appears the same number of times in both words

def is_anagram(word1, word2):
    word1 = word1.lower()
    word2 = word2.lower()

    if len(word1) != len(word2):
        return False

    freq1 = {}
    for char in word1:
        if char in freq1:
            freq1[char] += 1
        else:
            freq1[char] = 1

    freq2 = {}
    for char in word2:
        if char in freq2:
            freq2[char] += 1
        else:
            freq2[char] = 1

    return freq1 == freq2

print(is_anagram("listen", "silent"))
print(is_anagram("hello", "world"))

Output

True
False

We build a frequency dictionary for each word, then compare them. If the dictionaries are equal, the words have the same letters with the same counts → anagram

Notice the early exit: if the lengths are different, they can't be anagrams, no need to count anything

The frequency approach is actually faster than sorting for very long strings. Sorting takes roughly n × log(n) steps, while counting takes only n steps. For this exercise, both are fine, but it's good to think about these things

Mission: Passphrase Match

Two crew members each transmit a scrambled passphrase. The airlock only opens if the two passphrases are anagrams of each other: the same letters in a different order (case doesn't matter).

1. Write a function are_anagrams(a, b) that returns True if the two words are anagrams, False otherwise. Use either approach (sort the letters and compare, or a frequency dictionary). Lowercase both first so capitals don't matter.
2. Read two passphrases, then print Anagram: followed by the result.
3. On the next line, print Access granted if they match, or Access denied if they don't.

Input:

• the first passphrase
• the second passphrase

Output: two lines: the anagram check, then the access verdict.

Example

If the user types

listen
silent

the program should print

Anagram: True
Access granted

If the user types

orbit
robot

the program should print

Anagram: False
Access denied