0259MediumArrayGreedySortingHeap (Priority Queue)

Most Inspections Booked

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Trains the technique from

LeetCode 1353Maximum Number of Events That Can Be Attended

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 safety office has one inspector and a list of sites waiting to be inspected. Entry windows[i] = [opens, closes] means site i may be inspected on any single whole day from day opens to day closes inclusive.

The inspector carries out at most one inspection per day and needs the whole day for it. A site is either inspected on one day inside its window or not inspected at all, and days may be left empty.

Return the largest number of sites the inspector can get through.

Examples

Example 1

Input: windows = [[1, 2], [1, 1]]
Output: 2

The second site is inspected on day 1, its only day, and the first site on day 2, which is inside its window of days 1 to 2.

Example 2

Input: windows = [[1, 3], [1, 3], [1, 3]]
Output: 3

All three windows span days 1 to 3, so the inspector takes one site on each of those days.

Example 3

Input: windows = [[1, 1], [1, 1], [2, 2]]
Output: 2

Day 1 takes one of the two sites whose window is that single day, and day 2 takes the third site. The remaining site has no day left inside its window.

Constraints

1 <= windows.length <= 10^5
windows[i].length == 2
1 <= windows[i][0] <= windows[i][1] <= 10^5

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 most_inspections(windows: list[list[int]]) -> int:

Java

public int mostInspections(int[][] windows)

The intended approach

Time O(n log n) for n sites: each site is pushed and popped at most once and the sort dominates.Space O(n) for the sorted order and the heap.

Sort the sites by the day their window opens and walk the calendar forward, skipping straight to the next opening day whenever nothing is waiting. Every site whose window has opened by the current day goes into a heap keyed on its closing day. Sites whose windows have already shut are dropped from the heap, and the inspector spends the day on whichever of the rest shuts soonest, since any other choice keeps a site that has more days left waiting.

Pythonentry point: most_inspections

def most_inspections(windows):
    import heapq

    ordered = sorted(windows, key=lambda w: w[0])
    total = len(ordered)
    deadlines = []
    booked = 0
    i = 0
    day = 0
    while i < total or deadlines:
        if not deadlines and day < ordered[i][0]:
            day = ordered[i][0]
        while i < total and ordered[i][0] <= day:
            heapq.heappush(deadlines, ordered[i][1])
            i += 1
        while deadlines and deadlines[0] < day:
            heapq.heappop(deadlines)
        if deadlines:
            heapq.heappop(deadlines)
            booked += 1
        day += 1
    return booked

The obvious version

Step through every day from 1 to the last closing day and rescan all the sites, booking the not-yet-inspected site with the earliest closing day whose window covers that day. O(d * n) for d days.

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_most_inspections(windows):
    total = len(windows)
    used = [False] * total
    last = 0
    for window in windows:
        if window[1] > last:
            last = window[1]
    booked = 0
    for day in range(1, last + 1):
        pick = -1
        for i in range(total):
            if used[i] or windows[i][0] > day or windows[i][1] < day:
                continue
            if pick == -1 or windows[i][1] < windows[pick][1]:
                pick = i
        if pick != -1:
            used[pick] = True
            booked += 1
    return booked

How this problem was checked

Reference solution executed against 12 test cases — 12 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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