0336HardDynamic ProgrammingPrefix Sum

Counting Sawtooth Brightness Programs

Tracked in this browser only

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 light bar carries n cells in a row. A program assigns every cell an integer brightness in the closed range [l, r], so a program is an array a of length n with l <= a[i] <= r.

A program is a sawtooth when the brightness turns around at every cell it can:

no two neighbouring cells share a brightness, so a[i] != a[i - 1] for every i >= 1;
the direction alternates, so for every i with 1 <= i <= n - 2 exactly one of a[i - 1] < a[i] > a[i + 1] and a[i - 1] > a[i] < a[i + 1] holds.

Both opening directions are allowed: a program may start by going up or by going down.

Return how many sawtooth programs exist, taken modulo 10^9 + 7.

Examples

Example 1

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

Only [1, 2, 1] and [2, 1, 2] turn around at the middle cell; the six other programs over two brightness values repeat a neighbour somewhere.

Example 2

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

The count includes peaks such as [1, 3, 2] and troughs such as [3, 1, 2], and the reported figure is already below the modulus.

Example 3

Input: n = 4, l = 2, r = 4
Output: 16

One valid program is [2, 4, 3, 4], which rises, falls, rises; the figure returned counts every such program of four cells over the brightness values 2, 3 and 4.

Example 4

Input: n = 6, l = 1, r = 2
Output: 2

The two programs counted are [1, 2, 1, 2, 1, 2] and [2, 1, 2, 1, 2, 1]; each of the six cells sits in range and every step reverses the previous one.

Constraints

3 <= n <= 2000
1 <= l < r <= 2000

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 sawtooth_programs(length: int, low: int, high: int) -> int:

Java

public int sawtoothPrograms(int length, int low, int high)

The intended approach

Time O(n * (r - l + 1))Space O(r - l + 1)

Keep two counts per brightness value: how many programs of the current length end at that value on a rising step, and how many end there on a falling step. Growing the program by one cell turns rising into falling and vice versa, and the new count at a value is the total of the opposite counts strictly below it (or strictly above it). Those totals come from a single running prefix sum per row, and only the previous pair of rows is needed.

Pythonentry point: sawtooth_programs

def sawtooth_programs(length, low, high):
    # Two rolling rows over the brightness values: rising[v] counts programs
    # built so far whose last step went up into v, falling[v] the ones whose last
    # step came down into v. Extending swaps the roles and needs a running total
    # of everything strictly below (or above) v, so prefix sums do each row in
    # one pass instead of re-summing per value.
    # Time O(length * (high - low + 1)). Space O(high - low + 1).
    MOD = 1000000007
    span = high - low + 1
    rising = list(range(span))
    falling = [span - 1 - i for i in range(span)]
    for _ in range(3, length + 1):
        nxt_rise = [0] * span
        run = 0
        for i in range(span):
            nxt_rise[i] = run
            run += falling[i]
            if run >= MOD:
                run -= MOD
        nxt_fall = [0] * span
        run = 0
        for i in range(span - 1, -1, -1):
            nxt_fall[i] = run
            run += rising[i]
            if run >= MOD:
                run -= MOD
        rising = nxt_rise
        falling = nxt_fall
    return (sum(rising) + sum(falling)) % MOD

The obvious version

Enumerate every array of the given length over the brightness range and test the two sawtooth conditions directly.

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_sawtooth_programs(length, low, high):
    from itertools import product
    MOD = 1000000007
    total = 0
    for arr in product(range(low, high + 1), repeat=length):
        ok = True
        for i in range(1, length):
            if arr[i] == arr[i - 1]:
                ok = False
                break
        if ok:
            for i in range(1, length - 1):
                if (arr[i] > arr[i - 1]) == (arr[i + 1] > arr[i]):
                    ok = False
                    break
        if ok:
            total += 1
    return total % MOD

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
4 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in