0028MediumArrayTwo PointersGreedySortingHeap (Priority Queue)Prefix Sum

Peak Charging Bay Demand

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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 depot is sizing a bank of identical charging bays before opening day. You are given sessions, where sessions[i] = [arrive_i, leave_i] is the minute a vehicle plugs in and the minute it unplugs.

A bay serves one vehicle at a time. A vehicle holds its bay over the half-open span from arrive_i up to leave_i, so a vehicle unplugging at minute t hands the bay straight to a vehicle arriving at minute t. Two sessions that merely touch at an endpoint therefore never contend for a bay.

Return the smallest bay count that lets the depot honour every session in sessions.

Examples

Example 1

Input: sessions = [[4, 9], [9, 14], [14, 20]]
Output: 1

Each vehicle unplugs exactly as the next one arrives, so one bay is handed down the whole chain.

Example 2

Input: sessions = [[0, 12], [3, 6], [4, 8]]
Output: 3

At minute 4 all three vehicles are plugged in at once, so the depot cannot get by with fewer than 3 bays.

Example 3

Input: sessions = [[10, 20], [30, 40], [15, 35]]
Output: 2

The long session spanning minutes 15 to 35 doubles up first with the 10-to-20 session and later with the 30-to-40 one, but the first and second sessions never coincide, so 2 bays suffice.

Constraints

1 <= sessions.length <= 10^4
sessions[i].length == 2
0 <= arrive_i < leave_i <= 10^6

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 peak_bay_demand(sessions: list[list[int]]) -> int:

Java

public int peakBayDemand(int[][] sessions)

The intended approach

Time O(n log n)Space O(n)

Sort the arrival minutes and the departure minutes into two separate ascending lists. Sweep the arrivals in order while a second cursor retires every departure at or before the current arrival, which keeps a running count of vehicles plugged in at that instant. Because a departure at minute t is retired before an arrival at minute t is counted, touching sessions never inflate the tally, and the largest running count is the answer.

Pythonentry point: peak_bay_demand

def peak_bay_demand(sessions):
    starts = sorted(a for a, _ in sessions)
    ends = sorted(b for _, b in sessions)
    best = 0
    active = 0
    freed = 0
    for arrive in starts:
        while freed < len(ends) and ends[freed] <= arrive:
            active -= 1
            freed += 1
        active += 1
        if active > best:
            best = active
    return best

The obvious version

Concurrency can only peak at the instant some vehicle arrives, so for every arrival minute count how many sessions cover it directly with a nested scan, 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_peak_bay_demand(sessions):
    best = 0
    for arrive, _ in sessions:
        busy = 0
        for a, b in sessions:
            if a <= arrive < b:
                busy += 1
        if busy > best:
            best = busy
    return best

How this problem was checked

Reference solution executed against 9 test cases — 9 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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