1153EasyString

The Roughness of a Strip of Letters

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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 strip strip holds lowercase letters. Its roughness is the total, taken over every pair of neighbouring letters, of how far apart the two sit in the alphabet.

Return the roughness.

Examples

Example 1

Input: strip = "abc"
Output: 2

The two neighbouring pairs each sit one letter apart, so the roughness is 2.

Example 2

Input: strip = "az"
Output: 25

The two letters sit at opposite ends of the alphabet, twenty-five places apart.

Example 3

Input: strip = "aaaa"
Output: 0

Every letter is the same, so no pair is any distance apart at all.

Constraints

2 <= strip.length <= 100
strip holds only 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 score_of_string(strip: str) -> int:

Java

public int scoreOfString(String strip)

The intended approach

Time O(n) over the strip.Space O(1).

Each neighbouring pair contributes on its own, so one pass over the strip settles the whole thing. Take the difference between a letter's alphabet position and the next letter's, throw the sign away, and add it in. Only neighbouring pairs count, so a strip of `n` letters has `n - 1` of them and both ends take part in exactly one pair each. Throwing the sign away matters: without it the total would collapse to the gap between the first letter and the last, since the middle terms would cancel.

Pythonentry point: score_of_string

def score_of_string(strip):
    total = 0
    for i in range(len(strip) - 1):
        gap = ord(strip[i]) - ord(strip[i + 1])
        if gap < 0:
            gap = -gap
        total += gap
    return total

The obvious version

Turn the whole strip into a list of alphabet positions first, then walk that list adding the size of each step. Costs an extra pass and a list rather than working straight off the letters.

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_score_of_string(strip):
    spots = []
    for ch in strip:
        spots.append(ord(ch) - ord("a"))
    total = 0
    for i in range(1, len(spots)):
        total += abs(spots[i] - spots[i - 1])
    return total

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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