Are two strings anagrams of each other?

Challenge

Decide whether two strings are anagrams of each other.

Task

Implement int is_anagram(const char *a, const char *b) that returns 1 if a and b contain exactly the same multiset of bytes (one is a rearrangement of the other), else 0. The comparison is case-sensitive and considers all bytes.

Input

a, b: two NUL-terminated strings (either may be empty).

Output

int: 1 if the strings are anagrams, 0 otherwise.

Example

"listen", "silent"   ->   1
"abc", "cab"         ->   1
"aa", "a"            ->   0   (length mismatch)
"", ""               ->   1
"Tea", "Ate"         ->   0   (case-sensitive: 'T' != 't')
"abcd", "abce"       ->   0

Edge cases

• Two empty strings are anagrams (returns 1).
• Strings of different length can never be anagrams.

Rules

• O(n + m): use a 256-entry byte-frequency table rather than sorting.

Why this matters

A frequency-count table (a poor-man's hash table) is the cleanest way to compare permutations of a fixed alphabet. The technique generalises directly to counting requests-per-IP, votes-per-candidate, words-per-document, and so on.

Input format

a, b: two NUL-terminated strings (either may be empty).

Output format

int: 1 if the strings are anagrams, else 0.

Constraints

O(n + m). Use a 256-byte frequency table.

Starter code

int is_anagram(const char *a, const char *b) { /* TODO */ return 0; }

Common mistakes

Sorting the strings — O(n log n) when O(n) suffices. Forgetting to compare lengths first (cheaper short-circuit).

Edge cases to handle

Empty strings (both empty → anagram); different lengths; identical strings.

Complexity

O(n).

Background lessons

• Hashing — turning keys into table indices
• C strings

Up next

• Longest substring without repeating characters