Minimum covering window

Find the Smallest Supply Label

Original & variantMinimum Window Substring ↗

Original examples and wording. The question below states the exact input format and output contract.

01. The question

A lowercase label must contain every requested letter, including repeated requests. Return its shortest continuous piece that covers the request. Ties keep the earliest piece; no covering piece returns "".

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

02. Worked example

Inputlabel = "cabbac", request = "aac"

Output"cabba"

  1. The complete prefix "cabba" covers two a letters and one c.
  2. Removing its c breaks coverage.
  3. The final c lets the left edge shrink to "abbac", tied in length; the earlier "cabba" should win.

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: "cabbac". Parameter: "aac". 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. Count required letters, including multiplicities.
  2. Extend right until every required occurrence is covered.
  3. While covered, save improvements and shrink from the left.

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

Label: 0-30 lowercase letters; request: 0-12 lowercase letters. Repeated requests count.

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

from collections import Counter

def supply_label(label, request):
    if not request:
        return ""
    need = Counter(request)
    missing = len(request)
    left = best_start = 0
    best_length = len(label) + 1
    for right, letter in enumerate(label):
        if need[letter] > 0:
            missing -= 1
        need[letter] -= 1
        while missing == 0:
            length = right - left + 1
            if length < best_length:
                best_start, best_length = left, length
            outgoing = label[left]
            need[outgoing] += 1
            if need[outgoing] > 0:
                missing += 1
            left += 1
    return "" if best_length > len(label) else label[best_start:best_start + best_length]
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

from collections import Counter

def supply_label(label, request):
    if not request:
        return ""
    need = Counter(request)
    missing = len(request)
    left = best_start = 0
    best_length = len(label) + 1
    for right, letter in enumerate(label):
        if need[letter] > 0:
            missing -= 1
        need[letter] -= 1
        while missing == 0:
            length = right - left + 1
            if length < best_length:
                best_start, best_length = left, length
            outgoing = label[left]
            need[outgoing] += 1
            if need[outgoing] > 0:
                missing += 1
            left += 1
    return "" if best_length > len(label) else label[best_start:best_start + best_length]

06. Cost & edge cases

Time
O(n + m) expected
Space
O(alphabet size)
  • An empty request returns "".
  • Repeated requested letters need separate occurrences.
  • An earlier window wins a length tie.

Keep going