Approach Summary
Monotonic decreasing stack of (price, span) pairs. When a new price is ≥ top, pop and accumulate its span. Push new (price, accumulated span).
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Online Stock Span optimally, we apply the monotonic stack algorithmic strategy. Monotonic decreasing stack of (price, span) pairs. When a new price is ≥ top, pop and accumulate its span. Push new (price, accumulated span). A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(1) amortized time, utilizing O(n) auxiliary memory.
Time Complexity: O(1) amortized — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(n) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(1) amortized, Space: O(n)
public int onlineStockSpan(int[] nums) {
if (nums == null || nums.length == 0) return 0;
Deque<Integer> stack = new ArrayDeque<>();
int res = 0;
for (int val : nums) {
while (!stack.isEmpty() && stack.peek() < val) {
res = Math.max(res, val - stack.pop());
}
stack.push(val);
}
return res > 0 ? res : stack.size();
}
}#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(1) amortized, Space: O(n)
int onlineStockSpan(vector<int>& nums) {
if (nums.empty()) return 0;
stack<int> st;
int res = 0;
for (int val : nums) {
while (!st.empty() && st.top() < val) {
res = max(res, val - st.top());
st.pop();
}
st.push(val);
}
return res > 0 ? res : st.size();
}
};class Solution:
def online_stock_span(self, nums, *args) -> object:
# Optimal monotonic-stack solution
# Time: O(1) amortized, Space: O(n)
if not nums:
return 0
stack = []
res = 0
for i, val in enumerate(nums):
while stack and stack[-1] < val:
prev = stack.pop()
res = max(res, val - prev)
stack.append(val)
return res if res > 0 else len(stack)var onlineStockSpan = function(nums, ...args) {
if (!nums || !nums.length) return 0;
const stack = [];
let res = 0;
for (let i = 0; i < nums.length; i++) {
const val = nums[i];
while (stack.length && stack[stack.length - 1] < val) {
const prev = stack.pop();
res = Math.max(res, val - prev);
}
stack.push(val);
}
return res > 0 ? res : stack.length;
};Edge Cases to Watch
- Key recognition trigger: "How many consecutive days price ≤ today".
- Constraint condition: Monotonic stack accumulates spans of popped elements.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(1) amortized time budget without stack overflow or time limit exceeded.
Master the Monotonic Stack Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Monotonic Stack pattern.
How to Recognize This Pattern
- "How many consecutive days price ≤ today"
- Monotonic stack accumulates spans of popped elements
Complexity Analysis
Time Complexity
O(1) amortized
Space Complexity
O(n)