0515EasyMathSorting

Two Wheels on the Dial

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Trains the technique from

LeetCode 3536Maximum Product of Two Digits

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 mechanical counter shows the value reading on a row of digit wheels, one wheel per decimal digit, with no leading wheel showing a zero.

An inspector picks two different wheels and multiplies the two digits they show. Two wheels showing the same digit are still two different wheels, so a repeated digit may be used twice as long as it comes from two separate wheels.

Return the largest product the inspector can obtain.

Examples

Example 1

Input: reading = 7744
Output: 49

The wheels show 7, 7, 4 and 4. The two wheels showing 7 are different wheels, so their digits multiply to 49.

Example 2

Input: reading = 45067
Output: 42

The wheels show 4, 5, 0, 6 and 7, and the wheels showing 7 and 6 multiply to 42.

Example 3

Input: reading = 8069
Output: 72

The wheels show 8, 0, 6 and 9, and the wheels showing 9 and 8 multiply to 72.

Constraints

10 <= reading <= 10^9

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_dial_product(reading: int) -> int:

Java

public int largestDialProduct(int reading)

The intended approach

Time O(d) where d is the number of wheels, so O(log reading).Space O(1): two running digits.

Raising either factor of a product of two non-negative digits never lowers the product, so the two largest digits on the row are the pair to take, counted with multiplicity so a repeated digit may supply both factors. One pass that peels digits off with remainder and division by ten, keeping the best and second best, is enough.

Pythonentry point: largest_dial_product

def largest_dial_product(reading):
    # Only the two biggest digits can win, since replacing either factor of a pair
    # by something bigger cannot lower the product. Peel the digits off with
    # division by ten and keep the best two seen.
    # O(d) time in the number of wheels, O(1) space.
    best = -1
    runner = -1
    rest = reading
    while rest > 0:
        digit = rest % 10
        if digit > best:
            runner = best
            best = digit
        elif digit > runner:
            runner = digit
        rest //= 10
    return best * runner

The obvious version

Spell the reading out, then try every ordered pair of distinct wheel positions and keep the largest product. O(d^2).

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_largest_dial_product(reading):
    digits = [int(ch) for ch in str(reading)]
    best = 0
    for i in range(len(digits)):
        for j in range(len(digits)):
            if i == j:
                continue
            if digits[i] * digits[j] > best:
                best = digits[i] * digits[j]
    return best

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

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