0078EasyHash TableMathString

Harborside Tally Value

Tracked in this browser only

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.

Harborside dockhands write whole quantities using a shorthand built from seven marks:

S stands for 1
H stands for 5
K stands for 10
P stands for 50
T stands for 100
W stands for 500
Z stands for 1000

A tally is normally written heaviest mark first, and its quantity is then just the total of its marks, so PKKSS reads 50 + 10 + 10 + 1 + 1 = 72. To keep a tally short a dockhand may put one lighter mark directly in front of a heavier one, and that lighter mark is then taken off the heavier one instead of added to it. Exactly six such pairings are ever written: SH for 4, SK for 9, KP for 40, KT for 90, TW for 400 and TZ for 900.

Given a tally code that a dockhand wrote correctly, return the quantity it stands for.

Examples

Example 1

Input: code = "KKSH"
Output: 24

Two K marks contribute 10 each, then S sits in front of the heavier H, so that pairing contributes 4.

Example 2

Input: code = "ZWTKSH"
Output: 1614

The heavy marks add up to 1000 + 500 + 100 + 10, and the trailing pairing adds 4.

Example 3

Input: code = "PKKSS"
Output: 72

No mark is lighter than the mark to its right, so every mark is simply added.

Constraints

1 <= code.length <= 15
code contains only the characters ('S', 'H', 'K', 'P', 'T', 'W', 'Z')
code is guaranteed to be a correctly written tally for a quantity in [1, 3999]

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 tally_value(code: str) -> int:

Java

public int tallyValue(String code)

The intended approach

Time O(n)Space O(1)

A mark is part of a lighter-in-front pairing exactly when its weight is less than the weight of the mark immediately to its right, and that is decidable from those two marks alone. So walk the tally once, look up each mark's weight, and subtract it when the next mark outweighs it, otherwise add it. The final mark has nothing to its right, so it is always added.

Pythonentry point: tally_value

def tally_value(code):
    weight = {"S": 1, "H": 5, "K": 10, "P": 50, "T": 100, "W": 500, "Z": 1000}
    total = 0
    n = len(code)
    for i in range(n):
        here = weight[code[i]]
        if i + 1 < n and here < weight[code[i + 1]]:
            total -= here
        else:
            total += here
    return total

The obvious version

Work the other way round: write out the tally for every quantity from 1 to 3999 by greedily peeling off the heaviest rung that fits, cache that as a lookup table, then read the answer straight out of it.

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

_TALLY_LOOKUP = {}


def bf_tally_value(code):
    if not _TALLY_LOOKUP:
        rungs = [(1000, "Z"), (900, "TZ"), (500, "W"), (400, "TW"),
                 (100, "T"), (90, "KT"), (50, "P"), (40, "KP"),
                 (10, "K"), (9, "SK"), (5, "H"), (4, "SH"), (1, "S")]
        for quantity in range(1, 4000):
            left = quantity
            marks = []
            for value, mark in rungs:
                while left >= value:
                    marks.append(mark)
                    left -= value
            _TALLY_LOOKUP["".join(marks)] = quantity
    return _TALLY_LOOKUP[code]

How this problem was checked

Reference solution executed against 14 test cases — 14 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