0171MediumArrayHash TableMathSliding WindowPrefix Sum

Runs With K Odd Readings

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Trains the technique from

LeetCode 1248Count Number of Nice 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 flow meter logs one positive whole number per minute, and the array readings holds the shift in order. A run is any block of consecutive minutes, so it is fixed by the minute it opens on and the minute it closes on, and it always covers at least one minute.

The lab audits a run by counting how many of its readings are odd numbers. A run passes the audit when that count comes to exactly k; a run holding fewer odd readings fails, and so does one holding more.

Tally how many runs of readings pass the audit. Two runs count separately whenever they open or close on different minutes, even if the numbers they hold happen to match.

Examples

Example 1

Input: readings = [7, 12, 5, 4, 4, 9], k = 2
Output: 5

Odd readings sit at minutes 0, 2 and 5. The runs holding exactly two of them are [7, 12, 5], [7, 12, 5, 4], [7, 12, 5, 4, 4], [12, 5, 4, 4, 9] and [5, 4, 4, 9].

Example 2

Input: readings = [4, 8, 12], k = 1
Output: 0

Every reading on this shift is even, so no run can reach an odd count of one.

Example 3

Input: readings = [3, 5, 7], k = 2
Output: 2

All three readings are odd, so a passing run has to cover exactly two neighbouring minutes: [3, 5] or [5, 7].

Constraints

1 <= readings.length <= 50000
1 <= readings[i] <= 10^5
1 <= k <= readings.length

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 runs_with_k_odd_readings(readings: list[int], k: int) -> int:

Java

public int runsWithKOddReadings(int[] readings, int k)

The intended approach

Time O(n)Space O(n)

Only the parity of each reading matters, so carry a running count of odd readings seen from the start of the shift. A run that closes at the current minute passes the audit exactly when some earlier prefix ended with a running count of k less than the current one, because the readings between those two points then hold k odd values. Keep a map from running count to how many prefixes ended on it, seeded with one empty prefix at zero so runs opening on minute 0 are counted, and add the matching tally at every minute.

Pythonentry point: runs_with_k_odd_readings

def runs_with_k_odd_readings(readings, k):
    # Time: O(n) -- one pass, with constant-time map work per reading.
    # Space: O(n) -- at most one map entry per distinct running odd count.
    seen = {0: 1}
    odd_so_far = 0
    total = 0
    for value in readings:
        odd_so_far += value % 2
        total += seen.get(odd_so_far - k, 0)
        seen[odd_so_far] = seen.get(odd_so_far, 0) + 1
    return total

The obvious version

Fix each opening minute, walk forward while tallying odd readings, count every closing minute whose tally lands on k, and abandon the walk once the tally overshoots.

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_runs_with_k_odd_readings(readings, k):
    n = len(readings)
    total = 0
    for start in range(n):
        odds = 0
        for end in range(start, n):
            if readings[end] % 2 == 1:
                odds += 1
            if odds == k:
                total += 1
            elif odds > k:
                break
    return total

How this problem was checked

Reference solution executed against 14 test cases — 14 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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