901. Online Stock Span

Online Stock Span

Problem Description

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.

Approach) Using Stack

Keep the stack mono decrease

• Step 1 100 [100, 1] stack is empty and nothing left is small than 100, push [100, 1] to stack.
• Step 2 80 [100, 1][80, 1] 100 > 80, we push[80, 1] to the stack.
• Step 3 60 [100, 1][80, 1][60, 1] same as previous step, we push [60, 1] to the stack.
• Step 4 70 [100, 1][80, 1][70, 2] 60 < 70, so we pop it from the stack and add its count to current one
• Step 5 60 [100, 1][80, 1][70, 2][60, 1] still decreasing, push to stack.
• Step 6 75 [100, 1][80, 1][75, 4] 75 > 60 and 75 > 70, pop that two elements and add their counts to current one.
• Step 7 85 [100, 1][85, 6] same as previous step, pop 80 and 75.

var StockSpanner = function() {
    this.values = [];
};

/** 
 * @param {number} price
 * @return {number}
 */
StockSpanner.prototype.next = function(price) {
    let count = 1;
    while (this.values.length > 0 && this.values[this.values.length - 1][0] <= price) {
        count += this.values.pop()[1];
    }
    this.values.push([price, count]);
    return count;
};
/** 
 * Your StockSpanner object will be instantiated and called as such:
 * var obj = new StockSpanner()
 * var param_1 = obj.next(price)
 */

Conclusion

That’s all folks! In this post, we solved LeetCode problem 901. Online Stock Span

I hope you have enjoyed this post. Feel free to share your thoughts on this.

You can find the complete source code on my GitHub repository. If you like what you learn. feel free to fork 🔪 and star ⭐ it.

In this blog, I have tried to collect & present the most important points to consider when improving Data structure and logic, feel free to add, edit, comment, or ask. For more information please reach me here
Happy coding!