Stealth Interview
  • Features
  • Pricing
  • Blog
  • Login
  • Sign up

Leetcode #2848: Points That Intersect With Cars

In this guide, we solve Leetcode #2848 Points That Intersect With Cars 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

You are given a 0-indexed 2D integer array nums representing the coordinates of the cars parking on a number line. For any index i, nums[i] = [starti, endi] where starti is the starting point of the ith car and endi is the ending point of the ith car.

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: Array, Hash Table, Prefix Sum

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

Input: nums = [[3,6],[1,5],[4,7]] Output: 7 Explanation: All the points from 1 to 7 intersect at least one car, therefore the answer would be 7.

Python Solution

class Solution: def numberOfPoints(self, nums: List[List[int]]) -> int: m = 102 d = [0] * m for start, end in nums: d[start] += 1 d[end + 1] -= 1 return sum(s > 0 for s in accumulate(d))

Complexity

The time complexity is O(n+m)O(n + m)O(n+m), and the space complexity is O(m)O(m)O(m), where nnn is the length of the given array, and mmm is the maximum value in the array. The space complexity is O(m)O(m)O(m), where nnn is the length of the given array, and mmm is the maximum value in the array.

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.


Ace your next coding interview

We're here to help you ace your next coding interview.

Subscribe
Stealth Interview
© 2026 Stealth Interview®Stealth Interview is a registered trademark. All rights reserved.
Product
  • Blog
  • Pricing
Company
  • Terms of Service
  • Privacy Policy