Leetcode #2237: Count Positions on Street With Required Brightness
In this guide, we solve Leetcode #2237 Count Positions on Street With Required Brightness in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer n. A perfectly straight street is represented by a number line ranging from 0 to n - 1.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Prefix Sum
Intuition
Range queries become simple once we precompute cumulative sums.
We can transform subarray conditions into prefix comparisons.
Approach
Compute prefix sums and use a map to find matching prefixes.
This avoids nested loops while keeping the logic clear.
Steps:
- Compute prefix sums.
- Use a map to find valid ranges.
- Update the answer.
Example
Input: n = 5, lights = [[0,1],[2,1],[3,2]], requirement = [0,2,1,4,1]
Output: 4
Explanation:
- The first street lamp lights up the area from [max(0, 0 - 1), min(n - 1, 0 + 1)] = [0, 1] (inclusive).
- The second street lamp lights up the area from [max(0, 2 - 1), min(n - 1, 2 + 1)] = [1, 3] (inclusive).
- The third street lamp lights up the area from [max(0, 3 - 2), min(n - 1, 3 + 2)] = [1, 4] (inclusive).
- Position 0 is covered by the first street lamp. It is covered by 1 street lamp which is greater than requirement[0].
- Position 1 is covered by the first, second, and third street lamps. It is covered by 3 street lamps which is greater than requirement[1].
- Position 2 is covered by the second and third street lamps. It is covered by 2 street lamps which is greater than requirement[2].
- Position 3 is covered by the second and third street lamps. It is covered by 2 street lamps which is less than requirement[3].
- Position 4 is covered by the third street lamp. It is covered by 1 street lamp which is equal to requirement[4].
Positions 0, 1, 2, and 4 meet the requirement so we return 4.
Python Solution
class Solution:
def meetRequirement(
self, n: int, lights: List[List[int]], requirement: List[int]
) -> int:
d = [0] * (n + 1)
for p, r in lights:
i, j = max(0, p - r), min(n - 1, p + r)
d[i] += 1
d[j + 1] -= 1
return sum(s >= r for s, r in zip(accumulate(d), requirement))
Complexity
The time complexity is , and the space complexity is , where is the number of streetlights. The space complexity is , where is the number of streetlights.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.