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).
Input
first: The first phrase. second: The second phrase.
Output
Return true if the phrases are anagrams.
Vincoli
0 ≤ length of each ≤ 10,000 (ASCII)
Esempi
Esempio 1
Input:
first = "Listen"
second = "Silent"
Output atteso:
expected = TrueInput:
std::string first = "Listen";
std::string second = "Silent";
Output atteso:
bool expected = true;Input:
string first = "Listen";
string second = "Silent";
Output atteso:
bool expected = true;Spiegazione
Both use e, i, l, n, s, t once each.
Esempio 2
Input:
first = "apple"
second = "paddle"
Output atteso:
expected = FalseInput:
std::string first = "apple";
std::string second = "paddle";
Output atteso:
bool expected = false;Input:
string first = "apple";
string second = "paddle";
Output atteso:
bool expected = false;Spiegazione
The letters and counts differ.
Come affrontarlo
Prova prima da solo, poi apri un suggerimento alla volta.
Suggerimento 1
Remove non-letters and ignore case before comparing.
Suggerimento 2
Count character frequencies for both phrases, or sort their normalized forms.
Suggerimento 3
Punctuation and spaces should not affect the result.
Complessità attesa
O(n) time, O(1) extra space
Riferimenti
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