Flashcards for:
Solving the 'Valid Anagram' Problem

1. Ensure strings are of the same length. 2. Sort both strings. 3. Compare the sorted strings.

Discussing these aspects demonstrates your understanding of algorithm efficiency and helps in handling unexpected inputs or scenarios.

True, because both strings have the same characters with the same frequency.

Using `collections.Counter`, which provides a convenient way to count character occurrences in a string.

def is_anagram(s, t): return sorted(s) == sorted(t)

from collections import Counter def is_anagram(s, t): return Counter(s) == Counter(t)

The space complexity varies; it is often O(1) if sorting is done in-place, but typically considered O(n).

False, because the strings contain different characters.

The objective is to determine if two strings `s` and `t` are anagrams, meaning `t` can be formed by rearranging the characters of `s`.

Time Complexity: O(s + t); Space Complexity: O(s + t).

Hash map approach has a time complexity of O(s + t), whereas the sorting approach has a time complexity of O(n log n).

def is_anagram(s, t): if len(s) != len(t): return False count_s, count_t = {}, {} for i in range(len(s)): count_s[s[i]] = count_s.get(s[i], 0) + 1 count_t[t[i]] = count_t.get(t[i], 0) + 1 for char in count_s: if count_s[char] != count_t.get(char, 0): return False return True

Using two hash maps where keys are the characters and values are the counts of each character in the strings.

Ensure both strings have the same length. If not, return false.

Sort both strings and compare the sorted versions to determine if they are anagrams.

The `get` method helps to avoid key errors by providing a default value, typically `0`, for missing keys.