0446HardArrayBinary SearchQueueSliding WindowHeap (Priority Queue)Prefix SumMonotonic Queue

Shortest Run Meeting the Grid Quota

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Trains the technique from

LeetCode 862Shortest Subarray with Sum at Least K

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 tidal power station logs its net contribution to the grid once a minute. net[i] is the number of kilowatt-hours the station delivered during minute i, and the figure is negative for a minute in which the station drew more from the grid than it delivered, which happens while the sluices are being refilled around slack water.

A run is a block of one or more consecutive minutes, and its yield is the total of the logged figures across those minutes.

The operator has to show one run whose yield is at least quota, and wants the shortest such run. Return the number of minutes in the shortest run whose yield is at least quota. If no run reaches quota at all, return -1.

Examples

Example 1

Input: net = [4, -1, 4], quota = 7
Output: 3

The three minutes together yield 4 - 1 + 4 = 7, which meets the quota; neither single minute nor either pair of neighbouring minutes reaches 7.

Example 2

Input: net = [1, 1, 100], quota = 2
Output: 1

The last minute on its own yields 100, comfortably past a quota of 2, and a run cannot be shorter than one minute.

Example 3

Input: net = [17], quota = 18
Output: -1

The log holds one run, and its yield of 17 falls short of the quota.

Example 4

Input: net = [3, -2, 6, -1, 4], quota = 9
Output: 3

The last three minutes yield 6 - 1 + 4 = 9, exactly the quota, while no single minute and no neighbouring pair reaches 9.

Constraints

1 <= net.length <= 10^5
-10^5 <= net[i] <= 10^5
1 <= quota <= 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 shortest_run(net: list[int], quota: int) -> int:

Java

public int shortestRun(int[] net, int quota)

The intended approach

Time O(n): each index is appended to the queue once and removed once.Space O(n) for the running totals and the queue of candidate left ends.

Build the running totals so the yield of a run is one total minus an earlier total, then sweep the right end forward keeping a queue of the left ends still worth trying. A left end that already served the current right end can only give longer runs afterwards, so it is retired; and a left end whose running total is not below that of a later one is useless, since the later one sits closer and starts no higher. The queue therefore holds left ends with strictly rising totals, every index joins and leaves it at most once, and the shortest run is found while retiring left ends.

Pythonentry point: shortest_run

def shortest_run(net, quota):
    # Work with running totals: the yield of the run covering minutes a..b-1 is
    # total[b] - total[a], so for each right end b we want the largest a whose
    # total is at most total[b] - quota. Two facts prune the candidates for a:
    # once a candidate has served some right end it can never give a shorter run
    # later, so it is discarded; and a candidate whose total is not below that of
    # a later candidate is useless, because the later one is both closer and no
    # higher. What survives is a queue of candidate left ends with strictly rising
    # totals, and each index enters and leaves it once.
    # O(n) time, O(n) space for the running totals and the queue.
    from collections import deque
    size = len(net)
    total = [0] * (size + 1)
    for i in range(size):
        total[i + 1] = total[i] + net[i]
    best = size + 1
    lefts = deque()
    for right in range(size + 1):
        while lefts and total[right] - total[lefts[0]] >= quota:
            span = right - lefts.popleft()
            if span < best:
                best = span
        while lefts and total[lefts[-1]] >= total[right]:
            lefts.pop()
        lefts.append(right)
    return best if best <= size else -1

The obvious version

Start a run at each minute in turn, extend it a minute at a time adding the figures up, and stop that start as soon as the quota is met, keeping the shortest span found over all starts.

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_shortest_run(net, quota):
    size = len(net)
    best = -1
    for start in range(size):
        running = 0
        for end in range(start, size):
            running += net[end]
            if running >= quota:
                span = end - start + 1
                if best < 0 or span < best:
                    best = span
                break
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
4 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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