Break a list into fixed-size pieces
A bulk API takes at most a hundred records per call, so a long list has to be handed over in pieces.
- Every piece is the given size except possibly the last, which takes what is left.
- A size of zero or less gives an empty result.
- An empty input gives an empty result, not a list holding one empty piece.
chunkList(values: list<int>, perChunk: int) → list<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<std::vector<int>> chunkList(std::vector<int> values, int perChunk) {
}Worked examples
| Call | Result |
|---|---|
chunkList(std::vector<int>{1, 2, 3, 4, 5}, 2) | std::vector<std::vector<int>>{std::vector<int>{1, 2}, std::vector<int>{3, 4}, std::vector<int>{5}} |
chunkList(std::vector<int>{1, 2, 3, 4}, 2) | std::vector<std::vector<int>>{std::vector<int>{1, 2}, std::vector<int>{3, 4}} |
chunkList(std::vector<int>{1}, 5) | std::vector<std::vector<int>>{std::vector<int>{1}} |
chunkList(std::vector<int>{}, 2) | std::vector<std::vector<int>>{} |
Hint
Step the index forward by the chunk size and slice, rather than pushing one item at a time.
Reference solution in C++
std::vector<std::vector<int>> chunkList(std::vector<int> values, int perChunk) {
std::vector<std::vector<int>> result;
if (perChunk <= 0) return result;
for (size_t i = 0; i < values.size(); i += perChunk) {
size_t j = std::min(i + (size_t) perChunk, values.size());
result.push_back(std::vector<int>(values.begin() + i, values.begin() + j));
}
return result;
}