Points That Intersect With Cars - Practice Coding Problems

Points That Intersect With Cars - Problem

Imagine a long street where cars are parked at various segments. Each car occupies a continuous section of the street from its starting point to its ending point (inclusive).

You are given a 0-indexed 2D integer array nums where nums[i] = [start_i, end_i] represents the ith car parked from position start_i to position end_i on the number line.

Your task: Count how many integer points on the street are covered by at least one parked car.

Example: If cars are parked at [[3,6], [1,5], [4,7]], then points 1,2,3,4,5,6,7 are all covered, so the answer is 7.

Input & Output

example_1.py — Basic Case

$ Input: nums = [[3,6],[1,5],[4,7]]

› Output: 7

💡 Note: Car 1 covers points [3,4,5,6], Car 2 covers [1,2,3,4,5], Car 3 covers [4,5,6,7]. Combined unique points: {1,2,3,4,5,6,7} = 7 points.

example_2.py — No Overlap

$ Input: nums = [[1,3],[5,8]]

› Output: 7

💡 Note: Car 1 covers [1,2,3] and Car 2 covers [5,6,7,8]. No overlap, so total is 3 + 4 = 7 points.

example_3.py — Single Car

$ Input: nums = [[2,5]]

› Output: 4

💡 Note: Only one car covering points [2,3,4,5], which gives us 4 points total.

Constraints

1 ≤ nums.length ≤ 100
nums[i].length == 2
1 ≤ start_i ≤ end_i ≤ 100
All coordinates are positive integers

Visualization

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Understanding the Visualization

Park the Cars

Each car occupies a continuous range of spots

Mark Occupied Spots

Add each occupied spot to our set

Count Unique Spots

The set size gives us the answer

Key Takeaway

🎯 Key Insight: Use a set data structure to automatically eliminate duplicate points when cars overlap, making the solution both simple and efficient.

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal approach uses a hash set to collect all covered points from each car. Since sets automatically handle duplicates, overlapping ranges are managed naturally. Simply iterate through each car, add all points in its range to the set, and return the set size. Time complexity is O(n × k) where k is the average range size.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Range Checking)	O(n × m)	O(1)	For each possible point, check if any car covers it
Hash Set (Optimal)	O(n × k)	O(m)	Use a set to collect all covered points, automatically handling duplicates

Brute Force (Range Checking) — Algorithm Steps

Find the overall minimum and maximum points from all car ranges
Iterate through each point from min to max
For each point, check if any car's range includes this point
Count points that are covered by at least one car

Visualization

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Step-by-Step Walkthrough

Find Range

Determine min and max points from all cars

Check Points

For each point, check if any car covers it

Count Coverage

Increment counter when point is covered

Code -

solution.c — C

int numberOfPointsIntersect(int** nums, int numsSize, int* numsColSize) {
    if (numsSize == 0) return 0;
    
    // Find the overall range
    int minPoint = nums[0][0], maxPoint = nums[0][1];
    for (int i = 0; i < numsSize; i++) {
        if (nums[i][0] < minPoint) minPoint = nums[i][0];
        if (nums[i][1] > maxPoint) maxPoint = nums[i][1];
    }
    
    int count = 0;
    // Check each point in the range
    for (int point = minPoint; point <= maxPoint; point++) {
        // Check if this point is covered by any car
        for (int i = 0; i < numsSize; i++) {
            if (nums[i][0] <= point && point <= nums[i][1]) {
                count++;
                break; // Found one car covering this point
            }
        }
    }
    
    return count;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

Where n is number of cars and m is the range size (max - min)

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track min, max, and count

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Range Checking) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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