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
Go 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
func isCleanPalindrome(text string) bool {
}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 Go
func isCleanPalindrome(text string) bool {
alnum := func(c byte) bool {
return (c >= '0' && c <= '9') || (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')
}
i, j := 0, len(text)-1
for i < j {
a, b := text[i], 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
}