0551MediumArrayHash TableMathPrefix SumPigeonhole Principle

Billing Window on the Water Meter

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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 smart water meter reports the litres drawn on each of a run of consecutive days, given as nums, so nums[i] is the reading for day i. A quiet day reports 0.

The utility settles accounts in windows. A window is a stretch of consecutive days holding two days or more.

Return true when at least one window has a total that is a whole multiple of k, and false when none does. A total of 0 counts as a whole multiple of k.

Examples

Example 1

Input: nums = [8, 3, 9], k = 12
Output: true

Days 1 and 2 form a window of two days totalling 3 + 9 = 12, which is one whole multiple of 12.

Example 2

Input: nums = [9, 2, 5], k = 100
Output: false

The windows available total 11, 7 and 16, and none of those is a multiple of 100.

Example 3

Input: nums = [7, 4], k = 7
Output: false

The only window of two days or more is the whole log, totalling 11, which is not a multiple of 7. Day 0 on its own reads 7 but a single day is too short to be a window.

Constraints

1 <= nums.length <= 10^5
0 <= nums[i] <= 10^9
0 <= sum(nums) <= 2^31 - 1
1 <= k <= 2^31 - 1

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 check_subarray_sum(nums: list[int], k: int) -> bool:

Java

public boolean checkSubarraySum(int[] nums, int k)

The intended approach

Time O(n): one pass with a constant amount of work per day.Space O(min(n, k)) for the map of remainders.

Keep a running total from day zero. The total of the window covering days `a + 1` through `b` is the running total after day `b` minus the running total after day `a`, so that window is a multiple of `k` exactly when the two running totals leave the same remainder against `k`. Record, for each remainder, the earliest day index that produced it, with remainder `0` recorded at index `-1` to stand for the empty prefix. When a day's remainder has been seen before and the earliest sighting is at least two days back, the window between them is long enough and the answer is yes. Recording only the earliest sighting is what makes the gap as wide as it can be.

Pythonentry point: check_subarray_sum

def check_subarray_sum(nums, k):
    # O(n) time. A window's total is a multiple of k exactly when the running
    # totals just before and just after it leave the same remainder against k.
    # So record the earliest day index at which each remainder appeared, seeding
    # remainder 0 at index -1 for the empty prefix, and a later day with the same
    # remainder at least two days along answers yes.
    first = {0: -1}
    running = 0
    for i, litres in enumerate(nums):
        running += litres
        r = running % k
        if r in first:
            if i - first[r] >= 2:
                return True
        else:
            first[r] = i
    return False

The obvious version

Take every possible first day, extend the window one day at a time, and test each total of two or more days for divisibility by k. Report yes at the first success.

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_check_subarray_sum(nums, k):
    n = len(nums)
    for start in range(n):
        total = nums[start]
        for end in range(start + 1, n):
            total += nums[end]
            if total % k == 0:
                return True
    return False

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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