LeetCode #2848 — EASY

Points That Intersect With Cars

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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] = [start_i, end_i] where start_i is the starting point of the i^th car and end_i is the ending point of the i^th car.

Return the number of integer points on the line that are covered with any part of a car.

Example 1:

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.

Example 2:

Input: nums = [[1,3],[5,8]]
Output: 7
Explanation: Points intersecting at least one car are 1, 2, 3, 5, 6, 7, 8. There are a total of 7 points, therefore the answer would be 7.

Constraints:

1 <= nums.length <= 100
nums[i].length == 2
1 <= start_i <= end_i <= 100

Step 01

Brute Force Baseline

Problem summary: 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. Return the number of integer points on the line that are covered with any part of a car.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[[3,6],[1,5],[4,7]]

Example 2

[[1,3],[5,8]]

Core Insight

What unlocks the optimal approach

Sort the array according to first element and then starting from the <code>0<sup>th</sup></code> index remove the overlapping parts and return the count of non-overlapping points.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2848: Points That Intersect With Cars
class Solution {
    public int numberOfPoints(List<List<Integer>> nums) {
        int[] d = new int[102];
        for (var e : nums) {
            int start = e.get(0), end = e.get(1);
            ++d[start];
            --d[end + 1];
        }
        int ans = 0, s = 0;
        for (int x : d) {
            s += x;
            if (s > 0) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2848: Points That Intersect With Cars
func numberOfPoints(nums [][]int) (ans int) {
	d := [102]int{}
	for _, e := range nums {
		start, end := e[0], e[1]
		d[start]++
		d[end+1]--
	}
	s := 0
	for _, x := range d {
		s += x
		if s > 0 {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2848: Points That Intersect With Cars
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))

// Accepted solution for LeetCode #2848: Points That Intersect With Cars
fn number_of_points(nums: Vec<Vec<i32>>) -> i32 {
    let mut nums = nums;

    nums.sort_by(|a, b| {
        if a[0] == b[0] {
            b[1].cmp(&a[1])
        } else {
            a[0].cmp(&b[0])
        }
    });

    let mut ret = 0;
    let mut s = nums[0][0];
    let mut e = nums[0][1];

    for v in nums.into_iter().skip(1) {
        if v[0] > e {
            ret += e - s + 1;
            s = v[0];
            e = v[1];
        } else {
            s = std::cmp::min(s, v[0]);
            e = std::cmp::max(e, v[1]);
        }
    }

    ret += e - s + 1;

    ret
}

fn main() {
    let nums = vec![vec![1, 3], vec![5, 8]];
    let ret = number_of_points(nums);
    println!("ret={ret}");
}

#[test]
fn test_number_of_points() {
    {
        let nums = vec![vec![3, 6], vec![1, 5], vec![4, 7]];
        let ret = number_of_points(nums);
        assert_eq!(ret, 7);
    }
    {
        let nums = vec![vec![1, 3], vec![5, 8]];
        let ret = number_of_points(nums);
        assert_eq!(ret, 7);
    }
    {
        let nums = vec![vec![1, 2], vec![3, 4], vec![5, 6]];
        let ret = number_of_points(nums);
        assert_eq!(ret, 6);
    }
    {
        let nums = vec![vec![1, 10], vec![3, 4], vec![5, 6]];
        let ret = number_of_points(nums);
        assert_eq!(ret, 10);
    }
}

// Accepted solution for LeetCode #2848: Points That Intersect With Cars
function numberOfPoints(nums: number[][]): number {
    const d: number[] = Array(102).fill(0);
    for (const [start, end] of nums) {
        ++d[start];
        --d[end + 1];
    }
    let ans = 0;
    let s = 0;
    for (const x of d) {
        s += x;
        ans += s > 0 ? 1 : 0;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + m)

Space

O(m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.