0530HardMathDynamic Programming

Total Turning Points in a Number Range

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Trains the technique from

LeetCode 3753Total Waviness of Numbers in Range II

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.

Write a whole number in base ten with no leading zeros. One of its digits is a turning point when it has a digit on each side and it is either strictly larger than both of those neighbours or strictly smaller than both of them. The waviness of a number is how many of its digits are turning points, so any number below 100 has waviness 0.

Take 36269 as an illustration. Its middle digit 2 is smaller than the 6 on either side of it, and the first 6 is larger than the 3 before it and the 2 after it, so 36269 has waviness 2. The last 6 is not a turning point, since it is above the 2 before it but below the 9 after it.

Given low and high, add up the waviness of every whole number from low to high inclusive and return that sum reduced modulo 1000000007.

Examples

Example 1

Input: low = 121, high = 121
Output: 1

The digits of 121 are 1, 2 and 1. The middle digit is larger than both neighbours, so it is a turning point and the range totals 1.

Example 2

Input: low = 95, high = 105
Output: 5

The numbers from 95 to 99 have only two digits, so they add nothing. 100 has no turning point, while each of 101, 102, 103, 104 and 105 has its middle 0 below both neighbours, which totals 5.

Example 3

Input: low = 10101, high = 10101
Output: 3

In 10101 the second, third and fourth digits each sit below or above both of their neighbours, so the waviness of this single number is 3.

Constraints

1 <= low <= high <= 10^15
The answer is the sum reduced modulo 1000000007, so it stays below 10^9 + 7.

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 total_waviness(low: int, high: int) -> int:

Java

public long totalWaviness(long low, long high)

The intended approach

Time O(d * 2200) where d <= 16 is the digit count of high: each position sweeps a fixed table of at most a few hundred states over ten digit choices.Space O(1): the table holds at most a few hundred states regardless of the bounds.

Write a helper that totals the waviness of every number from 1 up to a bound, then subtract the helper at `low - 1` from the helper at `high`. The helper fills the digits of the bound left to right, keeping a table whose key records the two digits placed most recently, whether any non-zero digit has been placed yet, and whether the prefix still matches the bound exactly. Each table entry carries how many numbers reached that state and the running total of turning points they hold between them, so placing a new digit only needs to test the previous two digits against it and add one turning point per number in the entry when the middle digit is an extreme.

Pythonentry point: total_waviness

def total_waviness(low, high):
    # Digit dynamic programme: walk the digit positions of the bound once,
    # carrying how many numbers sit in each state and how many turns they have
    # racked up between them.  O(d) passes over a fixed-size state set.
    MOD = 1000000007
    return (_upto(high) - _upto(low - 1)) % MOD


def _upto(x):
    if x <= 0:
        return 0
    text = str(x)
    states = {(0, -1, -1, 1): (1, 0)}
    for ch in text:
        limit = int(ch)
        nxt = {}
        for key, val in states.items():
            started, p2, p1, tight = key
            cnt, tot = val
            top = limit if tight else 9
            for d in range(top + 1):
                ntight = 1 if (tight and d == top) else 0
                add = 0
                if not started and d == 0:
                    nkey = (0, -1, -1, ntight)
                else:
                    if p2 != -1 and p1 != -1:
                        if (p1 > p2 and p1 > d) or (p1 < p2 and p1 < d):
                            add = 1
                    nkey = (1, p1, d, ntight)
                slot = nxt.get(nkey)
                if slot is None:
                    nxt[nkey] = [cnt, tot + add * cnt]
                else:
                    slot[0] += cnt
                    slot[1] += tot + add * cnt
        states = dict((k, (v[0], v[1])) for k, v in nxt.items())
    total = 0
    for key, val in states.items():
        if key[0] == 1:
            total += val[1]
    return total

The obvious version

Visit every number in the range, turn it into text and test each interior digit against the two beside it, adding up the turning points found.

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_total_waviness(low, high):
    MOD = 1000000007
    running = 0
    for value in range(low, high + 1):
        digits = str(value)
        for i in range(1, len(digits) - 1):
            left = digits[i - 1]
            mid = digits[i]
            right = digits[i + 1]
            if mid > left and mid > right:
                running += 1
            elif mid < left and mid < right:
                running += 1
    return running % MOD

How this problem was checked

Reference solution executed against 16 test cases — 16 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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