Mirror text, ignoring the clutter
An audit tool checks whether a phrase reads the same forwards and backwards once the formatting — spaces, punctuation, digits — is stripped away.
- Only letters and digits count; everything else is ignored.
- The comparison is case-insensitive: “A” and “a” are the same.
- A phrase with nothing countable left is a palindrome.
isCleanPalindrome(text: string) → bool
C++ needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
bool isCleanPalindrome(std::string text) {
}Worked examples
| Call | Result |
|---|---|
isCleanPalindrome(std::string("race a car")) | false |
isCleanPalindrome(std::string("A man, a plan, a canal: Panama")) | true |
isCleanPalindrome(std::string("racecar")) | true |
isCleanPalindrome(std::string(" ")) | true |
Hint
Walk from both ends at once, skipping anything that is not a letter or digit, and compare the survivors.
Reference solution in C++
bool isCleanPalindrome(std::string text) {
auto alnum = [](char c) {
return (c >= '0' && c <= '9') || (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z');
};
int i = 0, j = (int) text.size() - 1;
while (i < j) {
char a = text[i], b = text[j];
if (!alnum(a)) { i++; continue; }
if (!alnum(b)) { j--; continue; }
if (a >= 'A' && a <= 'Z') a += 32;
if (b >= 'A' && b <= 'Z') b += 32;
if (a != b) return false;
i++; j--;
}
return true;
}