0791EasyHash TableStringSliding WindowCounting

Three Clean Pigment Windows

Tracked in this browser only

Trains the technique from

LeetCode 1876Substrings of Size Three with Distinct Characters

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A paint line mixes one batch at a time and records which pigment each batch used as a lowercase letter. The letters are written down in mixing order, giving the string batch.

The washing crew looks at every window of three consecutive batches. A window is clean when its three batches used three different pigments. Windows starting at different positions are counted separately, even when they use the same three pigments.

Return the number of clean windows in batch.

Examples

Example 1

Input: batch = "kltklm"
Output: 4

The windows are `klt`, `ltk`, `tkl` and `klm`. Each one uses three different pigments, so all four are clean.

Example 2

Input: batch = "nnpq"
Output: 1

The windows are `nnp`, which repeats pigment `n`, and `npq`, whose three pigments differ, so one window is clean.

Example 3

Input: batch = "rsr"
Output: 0

The log holds a single window, `rsr`, and its first and third batches both used pigment `r`.

Constraints

1 <= batch.length <= 100
Every character of batch is a lowercase English letter.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_clean_windows(batch: str) -> int:

Java

public int countCleanWindows(String batch)

The intended approach

Time O(n) for a log of n batches: one constant-time check per window.Space O(1): three letters and a counter.

Slide a window of width three across the log once. For each position compare the three letters against each other, all three pairs, and add one to the tally when no pair matches. The loop stops at the last position that still has three batches ahead of it.

Pythonentry point: count_clean_windows

def count_clean_windows(batch):
    # slide a width-three window once across the log
    total = 0
    for i in range(len(batch) - 2):
        a, b, c = batch[i], batch[i + 1], batch[i + 2]
        if a != b and b != c and a != c:
            total += 1
    return total

The obvious version

Cut out every substring of the log, keep the ones that happen to hold three batches, and tally each pigment in the piece with a dictionary to see whether three different pigments were used. O(n^2) pieces.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_clean_windows(batch):
    n = len(batch)
    total = 0
    for i in range(n):
        for j in range(i, n):
            piece = batch[i:j + 1]
            if len(piece) != 3:
                continue
            tally = {}
            for ch in piece:
                tally[ch] = tally.get(ch, 0) + 1
            if len(tally) == 3:
                total += 1
    return total

How this problem was checked

Reference solution executed against 12 test cases — 12 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in