0560MediumMathRecursion

Shelf Label Codes

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Trains the technique from

LeetCode 1922Count Good Numbers

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 warehouse prints shelf labels. A label is a row of exactly n digit slots, numbered from 0 on the left. The printer accepts a label only when

every slot whose number is even carries a digit that is a multiple of two, that is one of 0, 2, 4, 6, 8, and
every slot whose number is odd carries a prime digit, that is one of 2, 3, 5, 7.

Leading zeros are ordinary digits here; a label is just a row of slots, not a number.

Given n, count the labels the printer accepts. The count grows far beyond what a 64-bit integer holds, so return it modulo 1000000007.

Examples

Example 1

Input: n = 7
Output: 40000

Slots 0, 2, 4 and 6 each accept five digits and slots 1, 3 and 5 each accept four, so the printer accepts 625 * 64 = 40000 labels, which is already below the modulus.

Example 2

Input: n = 26
Output: 426560007

There are 13 even-numbered slots and 13 odd-numbered ones, and the resulting count reduced modulo 1000000007 is 426560007.

Example 3

Input: n = 900000000000000
Output: 676101114

The label has 450000000000000 slots of each kind, and the count reduced modulo 1000000007 is 676101114.

Constraints

1 <= n <= 10^15
The answer is returned modulo 1000000007, so it is always below that modulus.

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 count_good_numbers(n: int) -> int:

Java

public int countGoodNumbers(long n)

The intended approach

Time O(log n) modular multiplications.Space O(1).

The slots are chosen independently, so the total is the number of choices per slot multiplied together: five options on each even-numbered slot and four on each odd-numbered one. With `n` up to 10^15 the two powers have to be raised by repeated squaring, halving the exponent each step and reducing under the modulus at every multiplication. There are `(n + 1) // 2` even slots and `n // 2` odd ones.

Pythonentry point: count_good_numbers

def count_good_numbers(n):
    # O(log n) time, O(1) space: the slots split into two independent groups, so
    # the count is a product of two powers, each raised by repeated squaring
    # under the modulus.
    MOD = 1000000007

    def power(base, exp):
        result = 1
        base %= MOD
        while exp > 0:
            if exp & 1:
                result = result * base % MOD
            base = base * base % MOD
            exp >>= 1
        return result

    even_slots = (n + 1) // 2
    odd_slots = n // 2
    return power(5, even_slots) * power(4, odd_slots) % MOD

The obvious version

Walk the slots one at a time from left to right, multiplying a running count by five on an even-numbered slot and by four on an odd-numbered one, reducing under the modulus after each step.

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_count_good_numbers(n):
    MOD = 1000000007
    total = 1
    for i in range(n):
        if i % 2 == 0:
            total = total * 5 % MOD
        else:
            total = total * 4 % MOD
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 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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