How long this price has held up
A trading widget shows, for each day, how many days back the price has been no higher than it is today — today included.
- Count today and then each earlier consecutive day whose price is less than or equal to today’s.
- Stop at the first earlier day priced above today.
- The first day always answers 1.
priceRun(prices: 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> priceRun(std::vector<int> prices) {
}Worked examples
| Call | Result |
|---|---|
priceRun(std::vector<int>{100, 80, 60, 70, 60, 75, 85}) | std::vector<int>{1, 1, 1, 2, 1, 4, 6} |
priceRun(std::vector<int>{10, 20, 30}) | std::vector<int>{1, 2, 3} |
priceRun(std::vector<int>{30, 20, 10}) | std::vector<int>{1, 1, 1} |
priceRun(std::vector<int>{5, 5, 5}) | std::vector<int>{1, 2, 3} |
Hint
Rather than walking backwards each day, keep a stack of earlier days that were priced higher. Popping the ones that were not gives you the run in one pass.
Reference solution in C++
std::vector<int> priceRun(std::vector<int> prices) {
std::vector<int> runs;
std::vector<int> higher;
for (int i = 0; i < static_cast<int>(prices.size()); i++) {
while (!higher.empty() && prices[higher.back()] <= prices[i]) higher.pop_back();
runs.push_back(higher.empty() ? i + 1 : i - higher.back());
higher.push_back(i);
}
return runs;
}