0924HardMathDynamic Programming

Counting Alternating Runs of a Given Length

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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 reading log is a list of n whole numbers, each between l and r inclusive.

The log alternates when every step from one reading to the next changes direction: it goes up, then down, then up, and so on, or down, then up, then down, and so on. No two neighbouring readings may be equal.

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

Examples

Example 1

Input: n = 3, l = 1, r = 3
Output: 10

Of the twenty-seven logs of three readings over 1, 2 and 3, ten alternate: five that go up then down and five that go down then up.

Example 2

Input: n = 3, l = 1, r = 2
Output: 2

Only 1, 2, 1 and 2, 1, 2 alternate.

Example 3

Input: n = 7, l = 4, r = 9
Output: 16544

Six readings are allowed and the log holds seven of them, and the count comes out of carrying the two families forward six steps.

Constraints

3 <= n <= 10^9
1 <= l < r <= 75

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 zig_zag_arrays(n: int, l: int, r: int) -> int:

Java

public int zigZagArrays(int n, int l, int r)

The intended approach

Time O(m^3 log n) for m allowed readings, the matrix power dominating.Space O(m^2) for the matrix.

Only the number of allowed readings matters, not what they are, so let there be `m` of them, numbered from zero. Count the logs by what happens at the end: hold, for each reading, how many alternating logs of the current length finish on it having just stepped up, and likewise for having just stepped down. Extending a log by one reading turns an up-ending log into a down-ending one and the other way round, which makes the counts a linear map from one length to the next. Reversing the order of the readings swaps the two families, so the down-ending counts are the up-ending counts read backwards. That halves the state: one vector of `m` numbers suffices, and one `m` by `m` matrix carries it forward a step, where the entry for a reading gathers the counts from the readings at or above its mirror position. Since `n` can be enormous, raise that matrix to the power of one less than `n` by repeated squaring, apply it to the all-ones vector that stands for logs of length one, and add up the result. Doubling it accounts for both families, which partition the logs of length two or more.

Pythonentry point: zig_zag_arrays

def zig_zag_arrays(n, l, r):
    MOD = 1000000007
    m = r - l + 1
    step = [[0] * m for _ in range(m)]
    for v in range(m):
        for w in range(m):
            if w >= m - v:
                step[v][w] = 1

    def multiply(a, b):
        out = [[0] * m for _ in range(m)]
        for i in range(m):
            row = a[i]
            target = out[i]
            for t in range(m):
                if row[t] == 0:
                    continue
                factor = row[t]
                other = b[t]
                for j in range(m):
                    target[j] = (target[j] + factor * other[j]) % MOD
        return out

    power = n - 1
    result = [[1 if i == j else 0 for j in range(m)] for i in range(m)]
    base = step
    while power > 0:
        if power & 1:
            result = multiply(result, base)
        base = multiply(base, base)
        power >>= 1
    total = 0
    for i in range(m):
        row = 0
        for j in range(m):
            row += result[i][j]
        total = (total + row) % MOD
    return total * 2 % MOD

The obvious version

Carry the counts forward one length at a time, keeping both families separately and adding up the counts below or above each reading at every step. Linear in the log's length, so hopeless at the declared size, but it makes no argument about symmetry and no use of matrix powers.

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_zig_zag_arrays(n, l, r):
    MOD = 1000000007
    m = r - l + 1
    up = [1] * m
    down = [1] * m
    for _ in range(n - 1):
        next_up = [0] * m
        next_down = [0] * m
        running = 0
        for v in range(m):
            next_up[v] = running
            running = (running + down[v]) % MOD
        running = 0
        for v in range(m - 1, -1, -1):
            next_down[v] = running
            running = (running + up[v]) % MOD
        up = next_up
        down = next_down
    total = 0
    for v in range(m):
        total = (total + up[v] + down[v]) % MOD
    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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