Algorithm Notesworking through it, one step at a time

Monotonic stack

MediumPractice ↗

Wait for a Lower Quote

01. The question

A workshop receives one material-price quote per day. For each day, return how many days it takes to see a strictly lower quote. Return 0 if no later lower quote exists. Equal prices do not qualify.

Before revealing the idea, predict the result and describe what information you would need to track.

02. Worked example

Inputprices = [7, 7, 5, 6, 3]

Output[2, 1, 2, 1, 0]

  1. The second 7 does not resolve the first day: equal is not lower.
  2. Price 5 resolves both 7s, after 2 days and 1 day.
  3. Price 3 resolves the pending 6 and 5. The final day has no future quote, so stays 0.

Try your code

Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.

Edit your test cases

Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.

Input contract & reference code

Example data: [7,7,5,6,3]. Parameter: "". Numeric parameters arrive as strings; convert them with int(parameter) when needed.

The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.

Python runtime notices

Python downloads only when you run code, using compressed runtime assets. Code is not saved or uploaded. Enable JavaScript to run it.

Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.

03. The idea

Try sketching your own approach first. What would you track, and why would it work?

Reveal the idea
  1. Store the positions whose first lower quote is still unknown.
  2. When a new price is lower than the stack top, pop that position and record the day difference.
  3. Repeat until the top is no greater than the new price, then push the new position.

04. Watch it happen

Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.

Input & settings

Up to 30 integer prices from 0 to 1,000,000. Only a strictly lower future price qualifies.

Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.

Working state / what changes

Ready to begin

Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.

Python / why it changes

The line behind the step

def lower_quotes(prices):
    waits = [0] * len(prices)
    pending = []
    for day, price in enumerate(prices):
        while pending and prices[pending[-1]] > price:
            earlier = pending.pop()
            waits[earlier] = day - earlier
        pending.append(day)
    return waits
How to use this walkthrough

Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.

05. Python solution

def lower_quotes(prices):
    waits = [0] * len(prices)
    pending = []
    for day, price in enumerate(prices):
        while pending and prices[pending[-1]] > price:
            earlier = pending.pop()
            waits[earlier] = day - earlier
        pending.append(day)
    return waits

06. Cost & edge cases

Time
O(n): each position is pushed and popped at most once
Space
O(n)
  • Empty input returns [].
  • Equal prices remain pending.
  • Increasing prices produce all zeros; decreasing prices produce 1s except for the final 0.

Keep going

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