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
Java 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
boolean isCleanPalindrome(String text) {
}Worked examples
| Call | Result |
|---|---|
isCleanPalindrome("race a car") | false |
isCleanPalindrome("A man, a plan, a canal: Panama") | true |
isCleanPalindrome("racecar") | true |
isCleanPalindrome(" ") | true |
Hint
Walk from both ends at once, skipping anything that is not a letter or digit, and compare the survivors.
Reference solution in Java
boolean isCleanPalindrome(String text) {
int i = 0, j = text.length() - 1;
while (i < j) {
char a = text.charAt(i), b = text.charAt(j);
if (!Character.isLetterOrDigit(a)) { i++; continue; }
if (!Character.isLetterOrDigit(b)) { j--; continue; }
if (Character.toLowerCase(a) != Character.toLowerCase(b)) return false;
i++; j--;
}
return true;
}