0006MediumHash TableStringSliding Window

Largest Clean Tap Block

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Trains the technique from

LeetCode 3Longest Substring Without Repeating 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 lobby turnstile appends one character to a log string every time somebody taps a badge. That character is the person's short code, and codes are drawn from the full printable range: upper and lower case letters, digits, punctuation marks, and the space reserved for visitor passes. Codes are compared exactly, so K and k are two different people.

Security calls a block of back-to-back taps clean when no code shows up twice inside it. Given the log tape, return how many taps sit in the largest clean block. If nobody has tapped yet, report 0.

A block has to be an unbroken run of taps. You may not drop taps out of the middle of a block to keep it clean.

Examples

Example 1

Input: tape = "K7K7Q9"
Output: 4

The last four taps K, 7, Q and 9 are all different codes. Any block of five taps here picks up a second 7.

Example 2

Input: tape = "PQQP"
Output: 2

The trailing pair QP is clean. Stretching it one tap to the left gives QQP, which repeats Q, so 2 is the ceiling.

Example 3

Input: tape = "Q7! Q7!x"
Output: 5

The visitor-pass space starts the winning block, which then runs Q, 7, ! and x for five distinct codes.

Constraints

0 <= tape.length <= 10^5
Every character of `tape` is a printable code: an English letter, a digit, a punctuation symbol or a space
A space counts as a code like any other, so two spaces in one block make it unclean
Codes are case sensitive: an upper case letter and its lower case form are distinct codes

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 largest_clean_tap_block(tape: str) -> int:

Java

public int largestCleanTapBlock(String tape)

The intended approach

Time O(n)Space O(k), where k is the number of distinct codes on the tape and is capped by the size of the printable character set

Sweep a window across the tape and keep a table mapping each code to the most recent position where it was tapped. When the current code was already tapped at or after the window's left edge, jump that edge just past the earlier tap; a code seen only before the edge is already outside the window and must not move it. The best width observed during the sweep is the answer.

Pythonentry point: largest_clean_tap_block

def largest_clean_tap_block(tape):
    last_seen = {}
    start = 0
    best = 0
    for i, code in enumerate(tape):
        prev = last_seen.get(code, -1)
        if prev >= start:
            start = prev + 1
        last_seen[code] = i
        width = i - start + 1
        if width > best:
            best = width
    return best

The obvious version

Test every substring for distinctness, O(n^3)-ish.

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_window(tape):
    best = 0
    for i in range(len(tape)):
        for j in range(i, len(tape)):
            seg = tape[i:j + 1]
            if len(set(seg)) == len(seg) and len(seg) > best:
                best = len(seg)
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 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

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