Anagram Check

Problema

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)

Ejemplo 1

Entrada:
first = "Listen"
second = "Silent"

Salida esperada:
expected = True

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

Salida esperada:
bool expected = true;

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

Salida esperada:
bool expected = true;

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

Ejemplo 2

Entrada:
first = "apple"
second = "paddle"

Salida esperada:
expected = False

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

Salida esperada:
bool expected = false;

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

Salida esperada:
bool expected = false;

The letters and counts differ.

Inténtalo primero por tu cuenta y luego abre las pistas una a una.

Pista 1

Remove non-letters and ignore case before comparing.

Pista 2

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

Pista 3

Punctuation and spaces should not affect the result.

Complejidad esperada

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

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