0942MediumMathCombinatorics

Placing k Markers With None Adjacent

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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 has n slots in a row, and k markers are to be placed in them, at most one per slot.

No two markers may sit in neighbouring slots.

Return how many arrangements there are, modulo 10^9 + 7.

Examples

Example 1

Input: n = 9, k = 3
Output: 35

Three markers over nine slots with none touching leaves seven slots to choose three positions from, which is 35 arrangements.

Example 2

Input: n = 3, k = 2
Output: 1

The only arrangement puts markers in the first and last slots.

Example 3

Input: n = 2, k = 2
Output: 0

Two markers cannot both fit in two neighbouring slots, so there is no arrangement.

Constraints

1 <= n <= 5 * 10^5
1 <= k <= n

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_valid_sequences(n: int, k: int) -> int:

Java

public int countValidSequences(int n, int k)

The intended approach

Time O(n) to build the factorial tables, then constant work.Space O(n) for the tables.

Think of laying the markers out with the empty slots filled in around them. Every arrangement is settled by how many empty slots go before the first marker, between each neighbouring pair, and after the last. There are `k + 1` such runs, the `k - 1` runs between markers must each hold at least one empty slot, and the two on the ends may be empty. Setting aside one empty slot for each in-between run leaves `n - k - (k - 1)` empty slots to spread freely over the `k + 1` runs. Spreading that many indistinguishable slots over that many runs is a standard count, and it works out to choosing `k` places from `n - k + 1`. When `k` is larger than `n - k + 1` there is no room and the answer is nothing. Working that out under the modulus means factorials and their inverses up to `n`, built once by walking upwards and then inverting the last one and walking back down.

Pythonentry point: count_valid_sequences

def count_valid_sequences(n, k):
    MOD = 1000000007
    top = n - k + 1
    if k > top:
        return 0
    factorial = [1] * (top + 1)
    for i in range(1, top + 1):
        factorial[i] = factorial[i - 1] * i % MOD
    inverse = [1] * (top + 1)
    inverse[top] = pow(factorial[top], MOD - 2, MOD)
    for i in range(top, 0, -1):
        inverse[i - 1] = inverse[i] * i % MOD
    return factorial[top] * inverse[k] % MOD * inverse[top - k] % MOD

The obvious version

Fill the slots one at a time, carrying how many markers are still to place and whether the slot just behind holds one, and add up the ways. Costs work proportional to the slots times the markers and makes no combinatorial argument at all.

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_valid_sequences(n, k):
    MOD = 1000000007
    # ways[used][last] : last is 1 when the previous slot holds a marker
    ways = [[0, 0] for _ in range(k + 1)]
    ways[0][0] = 1
    for _ in range(n):
        nxt = [[0, 0] for _ in range(k + 1)]
        for used in range(k + 1):
            for last in (0, 1):
                count = ways[used][last]
                if count == 0:
                    continue
                nxt[used][0] = (nxt[used][0] + count) % MOD
                if last == 0 and used < k:
                    nxt[used + 1][1] = (nxt[used + 1][1] + count) % MOD
        ways = nxt
    return (ways[k][0] + ways[k][1]) % MOD

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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