Leetcode #2021: Brightest Position on Street

In this guide, we solve Leetcode #2021 Brightest Position on Street 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.

Leetcode

Problem Statement

A perfectly straight street is represented by a number line. The street has street lamp(s) on it and is represented by a 2D integer array lights.

Quick Facts

Difficulty: Medium
Premium: Yes
Tags: Array, Ordered Set, Prefix Sum, Sorting

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: lights = [[-3,2],[1,2],[3,3]]
Output: -1
Explanation:
The first street lamp lights up the area from [(-3) - 2, (-3) + 2] = [-5, -1].
The second street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
The third street lamp lights up the area from [3 - 3, 3 + 3] = [0, 6].

Position -1 has a brightness of 2, illuminated by the first and second street light.
Positions 0, 1, 2, and 3 have a brightness of 2, illuminated by the second and third street light.
Out of all these positions, -1 is the smallest, so return it.

Python Solution

class Solution:
    def brightestPosition(self, lights: List[List[int]]) -> int:
        d = defaultdict(int)
        for i, j in lights:
            l, r = i - j, i + j
            d[l] += 1
            d[r + 1] -= 1
        ans = s = mx = 0
        for k in sorted(d):
            s += d[k]
            if mx < s:
                mx = s
                ans = k
        return ans

Complexity

The time complexity is $O(n \times \log n)$ , and the space complexity is $O(n)$ . The space complexity is $O(n)$ .

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.

Example

Input: lights = [[-3,2],[1,2],[3,3]]
Output: -1
Explanation:
The first street lamp lights up the area from [(-3) - 2, (-3) + 2] = [-5, -1].
The second street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3].
The third street lamp lights up the area from [3 - 3, 3 + 3] = [0, 6].

Position -1 has a brightness of 2, illuminated by the first and second street light.
Positions 0, 1, 2, and 3 have a brightness of 2, illuminated by the second and third street light.
Out of all these positions, -1 is the smallest, so return it.

Python Solution

class Solution:
    def brightestPosition(self, lights: List[List[int]]) -> int:
        d = defaultdict(int)
        for i, j in lights:
            l, r = i - j, i + j
            d[l] += 1
            d[r + 1] -= 1
        ans = s = mx = 0
        for k in sorted(d):
            s += d[k]
            if mx < s:
                mx = s
                ans = k
        return ans

Leetcode #2021: Brightest Position on Street

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2021: Brightest Position on Street

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview