1060EasyString

The Longest Run of One Mark

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Trains the technique from

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 tape of lowercase letters reads tape.

Return the length of the longest stretch of neighbouring positions all holding the same letter.

Examples

Example 1

Input: tape = "aabbbcc"
Output: 3

The three b's in the middle form the longest stretch; the a's and the c's manage only two apiece.

Example 2

Input: tape = "zzzz"
Output: 4

The whole tape is one letter repeated, so the stretch runs the full length.

Example 3

Input: tape = "abc"
Output: 1

No two neighbouring letters match, so every stretch holds a single letter.

Constraints

1 <= tape.length <= 500
The tape is made of lowercase English letters.

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 max_power(tape: str) -> int:

Java

public int maxPower(String tape)

The intended approach

Time O(n) over the tape's length.Space O(1).

Walk the tape once carrying how long the stretch ending here is. Each letter either matches the one before it, which lengthens the stretch by one, or differs, which starts a fresh stretch of one. Keep the longest stretch seen against that running length. The first letter starts a stretch of one, so the answer is never below one for a non-empty tape. One pass and two numbers.

Pythonentry point: max_power

def max_power(tape):
    best = 1
    run = 1
    for at in range(1, len(tape)):
        if tape[at] == tape[at - 1]:
            run += 1
        else:
            run = 1
        if run > best:
            best = run
    return best

The obvious version

Try every stretch of the tape: for each starting place walk forward while the letter stays the same, measuring how far it reaches, and keep the longest. Costs the tape's length squared rather than one pass.

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_max_power(tape):
    n = len(tape)
    best = 0
    for start in range(n):
        end = start
        while end < n and tape[end] == tape[start]:
            end += 1
        if end - start > best:
            best = end - start
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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