Anagram Check

Problème

Two phrases are anagrams if they contain exactly the same letters the same number of times, ignoring case, spaces and punctuation.

Implement is_anagram(first, second).

first: The first phrase. second: The second phrase.

Return true if the phrases are anagrams.

0 ≤ length of each ≤ 10,000 (ASCII)

Exemple 1

Entrée :
first = "Listen"
second = "Silent"

Sortie attendue :
expected = True

Entrée :
std::string first = "Listen";
std::string second = "Silent";

Sortie attendue :
bool expected = true;

Entrée :
string first = "Listen";
string second = "Silent";

Sortie attendue :
bool expected = true;

Both use e, i, l, n, s, t once each.

Exemple 2

Entrée :
first = "apple"
second = "paddle"

Sortie attendue :
expected = False

Entrée :
std::string first = "apple";
std::string second = "paddle";

Sortie attendue :
bool expected = false;

Entrée :
string first = "apple";
string second = "paddle";

Sortie attendue :
bool expected = false;

The letters and counts differ.

Essayez d'abord seul, puis ouvrez les indices un par un.

Indice 1

Remove non-letters and ignore case before comparing.

Indice 2

Count character frequencies for both phrases, or sort their normalized forms.

Indice 3

Punctuation and spaces should not affect the result.

Complexité attendue

O(n) time, O(1) extra space

Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.