1123MediumArrayMath

Stretches Holding Nothing but Zeros

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Trains the technique from

LeetCode 2348Number of Zero-Filled Subarrays

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 log holds the readings readings.

Return how many stretches of one or more neighbouring readings hold nothing but zeros.

Examples

Example 1

Input: readings = [0, 0, 0]
Output: 6

Three single zeros, two neighbouring pairs and the whole run come to six stretches.

Example 2

Input: readings = [0, 1, 0]
Output: 2

Each zero stands on its own, so only the two one-reading stretches count.

Example 3

Input: readings = [0, 0, 1, 0, 0, 0]
Output: 9

The run of two zeros gives three stretches and the run of three gives six.

Constraints

1 <= readings.length <= 10^5
-10^9 <= readings[i] <= 10^9

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 zero_filled_subarray(readings: list[int]) -> int:

Java

public long zeroFilledSubarray(int[] readings)

The intended approach

Time O(n) over the log.Space O(1).

An all-zero stretch lies wholly inside one unbroken run of zeros, so the runs can be handled one at a time. A run of length `r` holds `r` stretches of one reading, `r - 1` of two, and so on, which adds up to `r * (r + 1) / 2`. Rather than measure the runs and then work that out, notice how the total grows as a run lengthens. Stretching a run from `r - 1` to `r` adds exactly `r` new stretches, namely every stretch whose right end is the newest zero. So walk the log carrying the length of the run ending at the current reading: a zero makes it one longer and contributes that many to the total, while anything else drops it back to nothing. That handles every run and their boundaries in one pass with a single counter. The total grows quadratically with the log, so it runs past a thirty-two-bit integer well inside the allowed size.

Pythonentry point: zero_filled_subarray

def zero_filled_subarray(readings):
    total = 0
    run = 0
    for value in readings:
        if value == 0:
            run += 1
            total += run
        else:
            run = 0
    return total

The obvious version

Take every stretch outright: fix a start and walk forward while the readings stay zero, counting each step and stopping that start at the first non-zero. Costs the log 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_zero_filled_subarray(readings):
    n = len(readings)
    total = 0
    for start in range(n):
        for end in range(start, n):
            if readings[end] != 0:
                break
            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

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