Valid Anagram

Sort both strings

Sorting removes the only difference an anagram is allowed to have: character order. Reject unequal lengths, sort both strings, and compare the results.

def is_anagram(s: str, t: str) -> bool:
    if len(s) != len(t):
        return False
    return sorted(s) == sorted(t)

Time O(n log n) for the two sorts. Space O(n) for the sorted character lists. This is a strong baseline, but it does more work than a 26-letter alphabet demands.

Fixed frequency table

A frequency table stores how many times each possible value occurs. Here, index 0 stands for a, index 1 for b, and so on through index 25 for z.

Add each character from s to its slot and subtract each character from t. Equal multisets cancel perfectly, leaving 26 zeroes.

def is_anagram(s: str, t: str) -> bool:
    if len(s) != len(t):
        return False
 
    counts = [0] * 26
    for char in s:
        counts[ord(char) - ord("a")] += 1
    for char in t:
        counts[ord(char) - ord("a")] -= 1
 
    return all(count == 0 for count in counts)

The length check is more than a shortcut. It rules out extra characters before counting and lets both loops cover the same number of positions.

Time O(n), because each character is counted once. Space O(1), because the table always has exactly 26 slots rather than growing with the strings.