901. Online Stock Span

Question

Write a class StockSpanner which 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 backwards) for which the price of the stock 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].

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: 
First, S = StockSpanner() is initialized.  Then:
S.next(100) is called and returns 1,
S.next(80) is called and returns 1,
S.next(60) is called and returns 1,
S.next(70) is called and returns 2,
S.next(60) is called and returns 1,
S.next(75) is called and returns 4,
S.next(85) is called and returns 6.

Note that (for example) S.next(75) returned 4, because the last 4 prices (including today’s price of 75) were less than or equal to today’s price.

Note:

1. Calls to StockSpanner.next(int price) will have 1 <= price <= 10^5.
2. There will be at most 10000 calls to StockSpanner.next per test case.
3. There will be at most 150000 calls to StockSpanner.next across all test cases.
4. The total time limit for this problem has been reduced by 75% for C++, and 50% for all other languages.

Solution

• Method 1: This question can be solved using monotic stack in decreasing order.
  • 100 [100, 1] Nothing int the stack, save value and frequency pair
  • 80 [100, 1][80, 1] 80 is smaller than 100, stack is still monotic, add value and frequency pair
  • 60 [100, 1][80, 1][60, 1] 60 is smaller than 80, stack is still monotic, add value and frequency pair
  • 70 [100, 1][80, 1][70, 2] 70 > 60, we pop 60 and add its count to 70 pair, it means for next values, if it is bigger than 70, it must bigger than 60
  • 60 [100, 1][80, 1][70, 2][60, 1] 60 is smaller than 70, stack is still monotic, add value and frequency pair
  • 75 [100, 1][80, 1][75, 4] 75 is bigger than 60 and bigger then 70, so we need to pop them and add their count to 75 pair.
  • 85 [100, 1][85, 6] 85 > 75 and 85 > 80, count of 85 = 1 +4 + 1

    class StockSpanner {
        private static class Pair{
            int value;
            int count;
            public Pair(int value, int count){
                this.value = value;
                this.count = count;
            }
        }
        private Stack<Pair> stack;
        public StockSpanner() {
            this.stack = new Stack<>();
        }
        public int next(int price) {
            int count = 1;
            if(stack.isEmpty() || stack.peek().value > price){
                stack.push(new Pair(price, 1));
            }else{
                while(!stack.isEmpty() && stack.peek().value <= price){
                    count += stack.pop().count;
                }
                stack.push(new Pair(price, count));
            }
            return count;
        }
    }
    /**
     * Your StockSpanner object will be instantiated and called as such:
     * StockSpanner obj = new StockSpanner();
     * int param_1 = obj.next(price);
     */
  

Reference

1. 花花酱 LeetCode 901. Online Stock Span - 刷题找工作 EP224