Rotate a list
A carousel shows the same items starting from a different one each time it advances.
- Rotating left by one moves the first item to the end.
- A shift larger than the list wraps around; a negative shift rotates the other way.
- An empty list rotates to an empty list.
rotateLeft(values: list<int>, by: int) → list<int>
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::vector<int> rotateLeft(std::vector<int> values, int by) {
}Worked examples
| Call | Result |
|---|---|
rotateLeft(std::vector<int>{1, 2, 3, 4}, 1) | std::vector<int>{2, 3, 4, 1} |
rotateLeft(std::vector<int>{1, 2, 3, 4}, 5) | std::vector<int>{2, 3, 4, 1} |
rotateLeft(std::vector<int>{1, 2, 3, 4}, -1) | std::vector<int>{4, 1, 2, 3} |
rotateLeft(std::vector<int>{1, 2, 3}, 0) | std::vector<int>{1, 2, 3} |
Hint
Reduce the shift with a modulo first, and remember that the modulo of a negative number is negative in most of these languages.
Reference solution in C++
std::vector<int> rotateLeft(std::vector<int> values, int by) {
int n = (int) values.size();
std::vector<int> result;
if (n == 0) return result;
int k = ((by % n) + n) % n;
for (int i = 0; i < n; i++) result.push_back(values[(i + k) % n]);
return result;
}