Frequency sliding window
Repaint a Signal Run
01. The question
A beacon log uses uppercase ASCII letters. Repaint at most k positions so one continuous run has a single letter. Return the largest run length, not the repainted text. Empty logs return 0.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input"ABABBA", parameter = "1"
Output4
- "ABABB" needs two repaintings, so a budget of one cannot keep all five.
- Shrink its left edge to "BABB": three B letters need one repainting.
- The longest affordable window has length four.
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: "ABABBA". Parameter: "1". 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 downloads only when you run code, using compressed runtime assets. Code is not saved or uploaded. Enable JavaScript to run it.
Printed output
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
- Count letters in the current window.
- Its repaint cost is window length minus its most frequent count.
- Shrink until the cost fits; remember the largest valid length.
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
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 solve(data, parameter=''):
budget = int(parameter)
counts = Counter()
left = best = 0
for right, letter in enumerate(data):
counts[letter] += 1
while right - left + 1 - max(counts.values()) > budget:
counts[data[left]] -= 1
left += 1
best = max(best, right - left + 1)
return bestHow 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 solve(data, parameter=''):
budget = int(parameter)
counts = Counter()
left = best = 0
for right, letter in enumerate(data):
counts[letter] += 1
while right - left + 1 - max(counts.values()) > budget:
counts[data[left]] -= 1
left += 1
best = max(best, right - left + 1)
return best06. Cost & edge cases
- Time
- O(26n) = O(n) for the fixed uppercase alphabet
- Space
- O(26) = O(1)
- Budget 0 permits only existing equal-letter runs.
- A budget at least the log length permits the whole log.
- Counts are recomputed exactly; this version does not use a historical maximum.