Anagram Check

Problem

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)

Example 1

Input:
first = "Listen"
second = "Silent"

Expected output:
expected = True

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

Expected output:
bool expected = true;

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

Expected output:
bool expected = true;

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

Example 2

Input:
first = "apple"
second = "paddle"

Expected output:
expected = False

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

Expected output:
bool expected = false;

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

Expected output:
bool expected = false;

The letters and counts differ.

Try on your own first, then open one hint at a time.

Hint 1

Remove non-letters and ignore case before comparing.

Hint 2

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

Hint 3

Punctuation and spaces should not affect the result.

Expected complexity

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

This challenge is adapted for CodeUtility. Follow the statement and constraints above.