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.

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 , and the space complexity is , where is the length of the given array, and is the maximum value in the array. The space complexity is , where is the length of the given array, and 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.