0853MediumArrayPrefix Sum

Posts Meeting Their Light Quota

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Trains the technique from

LeetCode 2237Count Positions on Street With Required Brightness

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 lane has n posts numbered 0 through n - 1. Each entry lights[j] = [at, reach] is a lamp standing at post at that lights every post from at - reach to at + reach inclusive, clipped to the lane. Several lamps may stand at the same post.

The brightness of a post is how many lamps light it. requirement[i] is the brightness post i must have.

Return how many posts have brightness at least their requirement.

Examples

Example 1

Input: n = 5, lights = [[0, 1], [2, 1], [3, 2]], requirement = [0, 2, 1, 4, 1]
Output: 4

The posts end up with brightness 1, 2, 2, 2 and 2. Posts 0, 1, 2 and 4 each reach their requirement, while post 3 needs 4 and has 2.

Example 2

Input: n = 4, lights = [[0, 100000]], requirement = [1, 1, 1, 1]
Output: 4

The single lamp reaches far beyond both ends of the lane, so every post is lit once and all four requirements of 1 are met.

Example 3

Input: n = 3, lights = [[1, 0]], requirement = [0, 0, 0]
Output: 3

Every post needs no light at all, so all three qualify whatever the lamp reaches.

Constraints

1 <= n <= 10^5
1 <= lights.length <= 10^5
lights[j].length == 2
0 <= lights[j][0] <= 99999
0 <= lights[j][1] <= 10^5
requirement.length == n
0 <= requirement[i] <= 10^5
Every lamp stands at a post inside the lane

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_meeting_quota(n: int, lights: list[list[int]], requirement: list[int]) -> int:

Java

public int postsMeetingQuota(int n, int[][] lights, int[] requirement)

The intended approach

Time O(n + L) for n posts and L lamps.Space O(n) for the marks.

Walking every post a lamp lights would cost far too much when the reaches are wide, so record each lamp as two marks instead: one where its light begins and one just past where it ends, both clipped to the lane. Sweeping the lane while adding up the marks seen so far gives each post's brightness, because a lamp starts contributing at its first post and its closing mark cancels the contribution as soon as the post is out of reach. Compare each brightness against the post's requirement as the sweep passes it.

Pythonentry point: posts_meeting_quota

def posts_meeting_quota(n, lights, requirement):
    # one spare slot so a lamp reaching the last post needs no special case
    marks = [0] * (n + 1)
    for at, reach in lights:
        lo = at - reach
        if lo < 0:
            lo = 0
        hi = at + reach
        if hi > n - 1:
            hi = n - 1
        marks[lo] += 1
        marks[hi + 1] -= 1
    met = 0
    running = 0
    for i in range(n):
        running += marks[i]
        if running >= requirement[i]:
            met += 1
    return met

The obvious version

Give every post a tally and, for each lamp, walk every post it reaches adding one. Then compare each tally against its requirement. Quadratic in the worst case, but it applies the definition of brightness literally.

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_meeting_quota(n, lights, requirement):
    bright = [0] * n
    for at, reach in lights:
        for i in range(at - reach, at + reach + 1):
            if 0 <= i < n:
                bright[i] += 1
    met = 0
    for i in range(n):
        if bright[i] >= requirement[i]:
            met += 1
    return met

How this problem was checked

Reference solution executed against 11 test cases — 11 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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