LeetCode Q 901 - Online Stock Span

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:

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

Solution

Solution 1: Brute Force (Time Limit Exceeds)

Code

Deque<Integer> stack;
public StockSpanner() {
  this.stack = new ArrayDeque<>();
}

public int next(int price) {
  int count = 1;
  Deque<Integer> temp = new ArrayDeque<>();

  while (!this.stack.isEmpty() && this.stack.peek() > price) {
    temp.push(this.stack.pop()); count++;
  }

  while (!temp.isEmpty()) {
    this.stack.push(temp.pop());
  }

  this.stack.push(price);

  return count;
}

Solution 2:

We use the monotonous stack, stroing int arrays int[]. Each array has two elements, the first one arr[0] denotes the price, and the second one int[1] denotes the span of stock’s price. The stack stores these arrays according to the price. The top array of the stack has minimum price.

In the previous solution, for each price, we needed to check all the elements in the stack in the worst case. So, that solution costs O(n) time for each next operation. If we call it k times, the worst time complexity is O(k^2).

Now the improved solution makes use of the previous calculation result, when we encounter a new price, we don’t need to go back checking all the previous stored prices. For example, the input prices is [100, 80, 60, 70, 60, 75, 85]. If we already known that for price 80, its span is 1, and for price75, its span is 4. Then the span of 85 is 1 + 4 + 1 = 6. Then if we call for next method k times, the time complexity is O(k).

Code

Deque<int[]> stack;
public StockSpanner() {
  this.stack = new ArrayDeque<>();
}

public int next(int price) {
  int count = 1;

  while (!this.stack.isEmpty() && this.stack.peek() <= price) {
    count += this.stack.pop()[1];
  }

  this.stack.push(new int[] {price, count});
  
  return count;
}

Reprint policy