1151MediumHash TableStringCountingPrefix Sum

Stretches Whose Total Divides Their Length

Tracked in this browser only

Trains the technique from

LeetCode 2950Number of Divisible Substrings

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.

Every lowercase letter carries a number. The first three letters carry 1, the next three carry 2, and so on in groups of three, so the last group, holding only y and z, carries 9.

A stretch of neighbouring letters is even-handed when the total of its letters' numbers divides exactly by how many letters the stretch holds.

Return how many even-handed stretches the word word holds.

Examples

Example 1

Input: word = "az"
Output: 3

Each letter on its own is even-handed, and together they total 10 across two letters, which divides evenly.

Example 2

Input: word = "zzz"
Output: 6

Every letter carries 9, so every stretch totals nine times its own length.

Example 3

Input: word = "abz"
Output: 5

The three single letters count, and so do the first two together totalling 2 and the last two totalling 10. All three together total 11, which three does not divide.

Constraints

1 <= word.length <= 2000
word holds only lowercase English letters

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_divisible_substrings(word: str) -> int:

Java

public int countDivisibleSubstrings(String word)

The intended approach

Time O(9n) over the word.Space O(n) for the running totals seen.

Being even-handed means the total divided by the length comes out whole, and that quotient is the stretch's average. Every letter carries between `1` and `9`, so the average can only be one of those nine values, which is what makes this cheap. Fix an average `a` and subtract `a` from every letter's number. A stretch has average `a` exactly when the shifted numbers across it add to nothing, and stretches adding to nothing are counted the usual way: sweep the word carrying the running total of the shifted numbers, and each time a total is met again, every earlier place holding that same total marks a stretch. Counting the empty running total as already seen once is what lets a stretch starting at the very first letter be found. Doing that for each of the nine averages counts every even-handed stretch exactly once, since a stretch has only one average.

Pythonentry point: count_divisible_substrings

def count_divisible_substrings(word):
    values = []
    for ch in word:
        values.append((ord(ch) - 97) // 3 + 1)
    total = 0
    for avg in range(1, 10):
        seen = {0: 1}
        run = 0
        for value in values:
            run += value - avg
            total += seen.get(run, 0)
            seen[run] = seen.get(run, 0) + 1
    return total

The obvious version

Take every stretch outright: fix a starting letter, walk forward keeping the running total, and after each step check whether the total divides by the letters counted so far. Costs the word length squared.

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_divisible_substrings(word):
    n = len(word)
    values = []
    for ch in word:
        values.append((ord(ch) - 97) // 3 + 1)
    total = 0
    for start in range(n):
        run = 0
        for end in range(start, n):
            run += values[end]
            if run % (end - start + 1) == 0:
                total += 1
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 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

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