Move from brute-force thinking to an efficient approach using stack strategy.
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 in one day is the maximum number of consecutive days (starting from that day and going backward) for which the stock price was less than or equal to the price of that day.
[7,2,1,2] and the price of the stock today is 2, then the span of today is 4 because starting from today, the price of the stock was less than or equal 2 for 4 consecutive days.[7,34,1,2] and the price of the stock today is 8, then the span of today is 3 because starting from today, the price of the stock was less than or equal 8 for 3 consecutive days.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 <= 105104 calls will be made to next.Problem summary: 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 in one day is the maximum number of consecutive days (starting from that day and going backward) for which the stock price was less than or equal to the price of that day. For example, if the prices of the stock in the last four days is [7,2,1,2] and the price of the stock today is 2, then the span of today is 4 because starting from today, the price of the stock was less than or equal 2 for 4 consecutive days. Also, if the prices of the stock in the last four days is [7,34,1,2] and the price of the stock today is 8, then the span of today is 3 because starting from today, the price of the stock was less than or equal 8 for 3 consecutive days. Implement the StockSpanner class: StockSpanner() Initializes the object of the class.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack · Design
["StockSpanner","next","next","next","next","next","next","next"] [[],[100],[80],[60],[70],[60],[75],[85]]
daily-temperatures)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #901: Online Stock Span
class StockSpanner {
private Deque<int[]> stk = new ArrayDeque<>();
public StockSpanner() {
}
public int next(int price) {
int cnt = 1;
while (!stk.isEmpty() && stk.peek()[0] <= price) {
cnt += stk.pop()[1];
}
stk.push(new int[] {price, cnt});
return cnt;
}
}
/**
* Your StockSpanner object will be instantiated and called as such:
* StockSpanner obj = new StockSpanner();
* int param_1 = obj.next(price);
*/
// Accepted solution for LeetCode #901: Online Stock Span
type StockSpanner struct {
stk []pair
}
func Constructor() StockSpanner {
return StockSpanner{[]pair{}}
}
func (this *StockSpanner) Next(price int) int {
cnt := 1
for len(this.stk) > 0 && this.stk[len(this.stk)-1].price <= price {
cnt += this.stk[len(this.stk)-1].cnt
this.stk = this.stk[:len(this.stk)-1]
}
this.stk = append(this.stk, pair{price, cnt})
return cnt
}
type pair struct{ price, cnt int }
/**
* Your StockSpanner object will be instantiated and called as such:
* obj := Constructor();
* param_1 := obj.Next(price);
*/
# Accepted solution for LeetCode #901: Online Stock Span
class StockSpanner:
def __init__(self):
self.stk = []
def next(self, price: int) -> int:
cnt = 1
while self.stk and self.stk[-1][0] <= price:
cnt += self.stk.pop()[1]
self.stk.append((price, cnt))
return cnt
# Your StockSpanner object will be instantiated and called as such:
# obj = StockSpanner()
# param_1 = obj.next(price)
// Accepted solution for LeetCode #901: Online Stock Span
use std::collections::VecDeque;
struct StockSpanner {
stk: VecDeque<(i32, i32)>,
}
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl StockSpanner {
fn new() -> Self {
Self {
stk: vec![(i32::MAX, -1)].into_iter().collect(),
}
}
fn next(&mut self, price: i32) -> i32 {
let mut cnt = 1;
while self.stk.back().unwrap().0 <= price {
cnt += self.stk.pop_back().unwrap().1;
}
self.stk.push_back((price, cnt));
cnt
}
}
// Accepted solution for LeetCode #901: Online Stock Span
class StockSpanner {
private stk: number[][];
constructor() {
this.stk = [];
}
next(price: number): number {
let cnt = 1;
while (this.stk.length && this.stk.at(-1)[0] <= price) {
cnt += this.stk.pop()[1];
}
this.stk.push([price, cnt]);
return cnt;
}
}
/**
* Your StockSpanner object will be instantiated and called as such:
* var obj = new StockSpanner()
* var param_1 = obj.next(price)
*/
Use this to step through a reusable interview workflow for this problem.
For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.
Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.