Cut a description to length
A listing card has room for a fixed number of characters, and a description cut mid-word looks broken.
- Text already within the limit is returned untouched.
- Otherwise take the first `limit` characters and append three dots.
- If that slice would land mid-word, fall back to the last space inside it — but a slice that already ends on a word boundary keeps its last word.
- With no space to fall back to, cut hard at the limit and still append the dots.
- A limit of zero or less gives an empty string.
truncateWords(text: string, limit: int) → string
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
std::string truncateWords(std::string text, int limit) {
}Worked examples
| Call | Result |
|---|---|
truncateWords(std::string("The quick brown fox"), 10) | std::string("The quick...") |
truncateWords(std::string("Supercalifragilistic"), 5) | std::string("Super...") |
truncateWords(std::string("Short"), 10) | std::string("Short") |
truncateWords(std::string("a b c d e f"), 5) | std::string("a b c...") |
Hint
Look at the character sitting at `limit`: if it is a space, the slice is already clean and needs no trimming back.
Reference solution in C++
std::string truncateWords(std::string text, int limit) {
if (limit <= 0) return "";
if ((int) text.size() <= limit) return text;
string cut = text.substr(0, limit);
if (text[limit] != ' ') {
size_t sp = cut.rfind(' ');
if (sp != string::npos && sp > 0) cut = cut.substr(0, sp);
}
while (!cut.empty() && isspace(static_cast<unsigned char>(cut.back()))) cut.pop_back();
return cut + "...";
}