Read a grid by columns
A matrix is stored row by row, but a transpose report reads it column by column.
- Read columns from left to right, and within a column top to bottom.
- A ragged row contributes only the columns it actually has.
columnMajor(grid: list<list<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> columnMajor(std::vector<std::vector<int>> grid) {
}Worked examples
| Call | Result |
|---|---|
columnMajor(std::vector<std::vector<int>>{std::vector<int>{1, 2}, std::vector<int>{3, 4}}) | std::vector<int>{1, 3, 2, 4} |
columnMajor(std::vector<std::vector<int>>{std::vector<int>{1}, std::vector<int>{2, 3}}) | std::vector<int>{1, 2, 3} |
columnMajor(std::vector<std::vector<int>>{std::vector<int>{1, 2, 3}, std::vector<int>{4, 5, 6}}) | std::vector<int>{1, 4, 2, 5, 3, 6} |
columnMajor(std::vector<std::vector<int>>{std::vector<int>{5}, std::vector<int>{6}, std::vector<int>{7}}) | std::vector<int>{5, 6, 7} |
Hint
For each column position, walk every row and take the value when that row is long enough.
Reference solution in C++
std::vector<int> columnMajor(std::vector<std::vector<int>> grid) {
std::vector<int> result;
int cols = 0;
for (const auto& row : grid) if ((int) row.size() > cols) cols = (int) row.size();
for (int c = 0; c < cols; c++) {
for (const auto& row : grid) if (c < (int) row.size()) result.push_back(row[c]);
}
return result;
}