0140MediumArrayBinary SearchSliding WindowPrefix Sum

Fewest Shifts to Clear the Quota

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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 field engineer logs the hours worked on each day of a tour, day by day, in hours.

Payroll wants to credit one unbroken run of days whose logged hours add up to quota or more, and it wants that run to cover as few days as it can. The run has to be consecutive days: skipping a day inside it is not allowed.

Return the number of days in the shortest qualifying run. When no run of consecutive days reaches quota, return 0.

Examples

Example 1

Input: quota = 11, hours = [4, 2, 5, 1, 6, 3]
Output: 3

The days logging 5, 1 and 6 add up to 12, which clears the quota with three days. No pair of neighbouring days gets past 9, so two days are never enough.

Example 2

Input: quota = 10, hours = [1, 1, 10]
Output: 1

The final day alone logs 10 and clears the quota, so a single day is the shortest run even though the whole tour also qualifies.

Example 3

Input: quota = 100, hours = [1, 2, 3]
Output: 0

The entire tour logs only 6 hours, so no run can reach 100 and the answer falls back to 0.

Constraints

1 <= quota <= 10^9
1 <= hours.length <= 10^5
1 <= hours[i] <= 10^4
Return 0 when no run of consecutive days reaches quota

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 fewest_shifts_for_quota(quota: int, hours: list[int]) -> int:

Java

public int fewestShiftsForQuota(int quota, int[] hours)

The intended approach

Time O(n)Space O(1)

Hold a window over the tour together with the hours inside it. Take in one more day on the right, then, while the window already clears the quota, record its length and drop days off the left. Since every day logs at least one hour the total only grows as the window widens, so each edge moves forward at most once per day and every day is the right end of exactly one shortest qualifying window.

Pythonentry point: fewest_shifts_for_quota

def fewest_shifts_for_quota(quota, hours):
    best = 0
    total = 0
    left = 0
    for right in range(len(hours)):
        total += hours[right]
        while total >= quota:
            span = right - left + 1
            if best == 0 or span < best:
                best = span
            total -= hours[left]
            left += 1
    return best

The obvious version

Build running totals up front, then test every pair of endpoints and keep the shortest span that clears the quota, O(n^2).

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_fewest_shifts(quota, hours):
    n = len(hours)
    running = [0] * (n + 1)
    for i in range(n):
        running[i + 1] = running[i] + hours[i]
    best = 0
    for i in range(n):
        for j in range(i + 1, n + 1):
            if running[j] - running[i] >= quota:
                span = j - i
                if best == 0 or span < best:
                    best = span
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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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