# 2848. Points That Intersect With Cars

## Description

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.

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 <= starti <= endi <= 100

## Solutions

Solution 1: Difference Array

We create a difference array $d$ of length $110$, then traverse the given array. For each interval $[a, b]$, we increase $d[a]$ by $1$ and decrease $d[b + 1]$ by $1$. Finally, we traverse the difference array $d$, calculate the prefix sum $s$ at each position. If $s > 0$, it means that the position is covered, and we increase the answer by $1$.

The time complexity is $O(n)$, and the space complexity is $O(M)$. Here, $n$ is the length of the given array, and $M$ is the maximum value in the array.

• class Solution {
public int numberOfPoints(List<List<Integer>> nums) {
int[] d = new int[110];
for (var e : nums) {
d[e.get(0)]++;
d[e.get(1) + 1]--;
}
int ans = 0, s = 0;
for (int x : d) {
s += x;
if (s > 0) {
ans++;
}
}
return ans;
}
}

• class Solution {
public:
int numberOfPoints(vector<vector<int>>& nums) {
int d[110]{};
for (auto& e : nums) {
d[e[0]]++;
d[e[1] + 1]--;
}
int ans = 0, s = 0;
for (int x : d) {
s += x;
ans += s > 0;
}
return ans;
}
};

• class Solution:
def numberOfPoints(self, nums: List[List[int]]) -> int:
d = [0] * 110
for a, b in nums:
d[a] += 1
d[b + 1] -= 1
return sum(s > 0 for s in accumulate(d))


• func numberOfPoints(nums [][]int) (ans int) {
d := [110]int{}
for _, e := range nums {
d[e[0]]++
d[e[1]+1]--
}
s := 0
for _, x := range d {
s += x
if s > 0 {
ans++
}
}
return
}

• function numberOfPoints(nums: number[][]): number {
const d: number[] = Array(110).fill(0);
for (const [a, b] of nums) {
d[a]++;
d[b + 1]--;
}
let ans = 0;
let s = 0;
for (const x of d) {
s += x;
if (s > 0) {
ans++;
}
}
return ans;
}