Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 9

Description

Online Stock Span

Design an algorithm that collects daily price quotes for some stock and returns the span of that stock's price for the current day.

The span of the stock's price today is defined as the maximum number of consecutive days (starting from today and going backward) for which the stock price was less than or equal to today's price.

• For example, if the price of a stock over the next 7 days were [100,80,60,70,60,75,85], then the stock spans would be [1,1,1,2,1,4,6].

Implement the StockSpanner class:

• StockSpanner() Initializes the object of the class.
• int next(int price) Returns the span of the stock's price given that today's price is price.

Example 1:

Input
["StockSpanner", "next", "next", "next", "next", "next", "next", "next"]
[[], [100], [80], [60], [70], [60], [75], [85]]
Output
[null, 1, 1, 1, 2, 1, 4, 6]

Explanation
StockSpanner stockSpanner = new StockSpanner();
stockSpanner.next(100); // return 1
stockSpanner.next(80);  // return 1
stockSpanner.next(60);  // return 1
stockSpanner.next(70);  // return 2
stockSpanner.next(60);  // return 1
stockSpanner.next(75);  // return 4, because the last 4 prices (including today's price of 75) were less than or equal to today's price.
stockSpanner.next(85);  // return 6

Constraints:

• 1 <= price <= 105
• At most 104 calls will be made to next.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class StockSpanner {
  vector<pair<int, int>> container;
public:
  StockSpanner() { }
    
  int next(int price) {
    if(container.empty() || container.back().first > price) {
      container.push_back({price, 1});
    } else {
      container.back().first = price;
      container.back().second += 1;
      while(container.size() > 1 && container.back().first >= container[container.size() - 2].first) {
        // cout << container.back().first << ' ' << container.back().second << endl;
        container[container.size() - 2].first = container.back().first;
        container[container.size() - 2].second += container.back().second;
        container.pop_back();
      }
    }
    return container.back().second;
  }
};

// Accepted
// 99/99 cases passed (318 ms)
// Your runtime beats 77.79 % of cpp submissions
// Your memory usage beats 98.88 % of cpp submissions (84.1 MB)