0930HardArrayMathDynamic ProgrammingMemoizationNumber Theory

Selections of Throws Scoring Exactly k

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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 run of dice throws is given as nums, each a number from 1 to 6.

A selection takes some of the throws, keeping their order, and its score is the product of the throws taken. Two selections are different when they take different positions, even if the throws taken read the same.

Return how many non-empty selections score exactly k.

Examples

Example 1

Input: nums = [2, 3, 4, 6, 1, 5], k = 24
Output: 4

Selections scoring 24 include the throws 4 and 6, the throws 2, 3 and 4, and each of those again with the throw of 1 taken as well, since taking it changes nothing about the product but does make a different selection.

Example 2

Input: nums = [1, 1, 1], k = 1
Output: 7

Any non-empty selection of throws of one scores one, and there are seven of them.

Example 3

Input: nums = [6], k = 5
Output: 0

The only selection scores 6, so nothing scores 5.

Constraints

1 <= nums.length <= 19
1 <= nums[i] <= 6
1 <= k <= 10^15

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_sequences(nums: list[int], k: int) -> int:

Java

public int countSequences(int[] nums, long k)

The intended approach

Time O(n * d) for n throws and d divisors of k that the tally ever holds.Space O(d) for the tally.

Trying every selection means half a million of them at the largest size, which is more than necessary. Instead carry a tally of scores: how many selections so far reach each score. Reading a new throw, every score already reachable can be multiplied by it, so the tally grows by folding itself forward. What keeps the tally small is that a score can only ever grow, since every throw is at least one, so a score that does not divide `k` can never become `k` and may be dropped on the spot. That leaves only divisors of `k` in the tally, of which there are few, and since the throws are all at most six their prime factors are only twos, threes and fives. Reading the tally's entry for `k` at the end gives the answer, less the empty selection when `k` is one, since the empty selection has an empty product.

Pythonentry point: count_sequences

def count_sequences(nums, k):
    ways = {1: 1}
    for v in nums:
        added = {}
        for score in ways:
            nxt = score * v
            if k % nxt == 0:
                added[nxt] = added.get(nxt, 0) + ways[score]
        for score in added:
            ways[score] = ways.get(score, 0) + added[score]
    total = ways.get(k, 0)
    if k == 1:
        total -= 1
    return total

The obvious version

Try every selection of the throws, multiply out each one, and count those scoring exactly `k`. Exponential in the throw count and it never prunes a score that cannot get there.

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_sequences(nums, k):
    n = len(nums)
    total = 0
    for mask in range(1, 1 << n):
        score = 1
        for i in range(n):
            if mask >> i & 1:
                score *= nums[i]
        if score == k:
            total += 1
    return total

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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