0818EasyArrayHash TablePrefix Sum

Patrol Coverage Audit

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Trains the technique from

LeetCode 1893Check if All the Integers in a Range Are Covered

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 numbers its patrol posts with consecutive integers. The night rota is given as shifts, where shifts[i] = [from_i, to_i] means guard i walks every post numbered from_i through to_i. Both ends of a walk are included, so a guard with [4, 4] walks post 4 and nothing else.

An auditor picks the span of posts numbered first through last, again including both ends, and wants to know whether the rota leaves no hole: every post in that span must be walked by at least one guard. Different guards may cover different parts of the span, and walks may overlap or sit side by side.

Return true if every post from first to last is walked by at least one guard, and false otherwise.

Examples

Example 1

Input: shifts = [[2, 5], [6, 9]], first = 3, last = 8
Output: true

The audited posts are 3, 4, 5, 6, 7 and 8. The first guard walks 3, 4 and 5, and the second walks 6, 7 and 8, so none of the six posts is left without a guard.

Example 2

Input: shifts = [[2, 5], [7, 9]], first = 4, last = 8
Output: false

Post 6 lies inside the audited span, but the first guard stops at post 5 and the second starts at post 7, so post 6 is walked by nobody.

Example 3

Input: shifts = [[7, 7]], first = 8, last = 8
Output: false

The audited span is the single post 8, and the only guard on the rota walks post 7 alone.

Constraints

1 <= shifts.length <= 50
shifts[i].length == 2
1 <= shifts[i][j] <= 50
shifts[i][0] <= shifts[i][1]
1 <= first <= last <= 50

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 posts_all_covered(shifts: list[list[int]], first: int, last: int) -> bool:

Java

public boolean postsAllCovered(int[][] shifts, int first, int last)

The intended approach

Time O(m + v) for m shifts and a largest post number v.Space O(v) for the array of marks.

Build one array of marks over the post numbers: a walk adds one at its first post and subtracts one at the post just past its last. Sweeping the post numbers upward while accumulating those marks gives, at each post, the number of guards walking it, so the answer is false as soon as a post inside the audited span accumulates to zero.

Pythonentry point: posts_all_covered

def posts_all_covered(shifts, first, last):
    # Record +1 where a walk starts and -1 just past where it ends, then sweep the
    # post numbers in order accumulating those marks: the running total is how many
    # guards walk the current post. Any post inside the audited span whose total is
    # zero settles the answer.
    # Time: O(m + v) for m shifts and a largest post number v.
    # Space: O(v) for the marks.
    top = last
    for pair in shifts:
        if pair[1] > top:
            top = pair[1]
    marks = [0] * (top + 2)
    for pair in shifts:
        marks[pair[0]] += 1
        marks[pair[1] + 1] -= 1
    walkers = 0
    for post in range(1, top + 1):
        walkers += marks[post]
        if first <= post <= last and walkers == 0:
            return False
    return True

The obvious version

Handle the audited posts one by one, and for each of them scan the entire rota looking for a walk that contains it. Stops at the first post no walk contains.

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_posts_all_covered(shifts, first, last):
    """Take the audited posts one at a time and scan the whole rota for each."""
    for post in range(first, last + 1):
        found = False
        for pair in shifts:
            if pair[0] <= post <= pair[1]:
                found = True
                break
        if not found:
            return False
    return True

How this problem was checked

Reference solution executed against 20 test cases — 20 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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