README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2800-2899/2848.Points%20That%20Intersect%20With%20Cars/README_EN.md	1229	Weekly Contest 362 Q1	Array Hash Table Prefix Sum

2848. Points That Intersect With Cars

中文文档

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

According to the problem description, we need to add one vehicle to each interval $[\textit{start}_i, \textit{end}_i]$. We can use a difference array to achieve this.

We define an array $d$ of length 102. For each interval $[\textit{start}_i, \textit{end}_i]$, we increment $d[\textit{start}_i]$ by 1 and decrement $d[\textit{end}_i + 1]$ by 1.

Finally, we perform a prefix sum operation on $d$ and count the number of elements in the prefix sum that are greater than 0.

The time complexity is $O(n + m)$, and the space complexity is $O(m)$, where $n$ is the length of the given array, and $m$ is the maximum value in the array. In this problem, $m \leq 102$.

Python3

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))

Java

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;
    }
}

C++

class Solution {
public:
    int numberOfPoints(vector<vector<int>>& nums) {
        int d[102]{};
        for (const auto& e : nums) {
            int start = e[0], end = e[1];
            ++d[start];
            --d[end + 1];
        }
        int ans = 0, s = 0;
        for (int x : d) {
            s += x;
            ans += s > 0;
        }
        return ans;
    }
};

Go

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
}

TypeScript

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;
}

Solution 2: Hash Table + Difference Array + Sorting

If the range of intervals in the problem is large, we can use a hash table to store the start and end points of the intervals. Then, we sort the keys of the hash table and perform prefix sum statistics.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the given array.

Python3

class Solution:
    def numberOfPoints(self, nums: List[List[int]]) -> int:
        d = defaultdict(int)
        for start, end in nums:
            d[start] += 1
            d[end + 1] -= 1
        ans = s = last = 0
        for cur, v in sorted(d.items()):
            if s > 0:
                ans += cur - last
            s += v
            last = cur
        return ans

Java

class Solution {
    public int numberOfPoints(List<List<Integer>> nums) {
        TreeMap<Integer, Integer> d = new TreeMap<>();
        for (var e : nums) {
            int start = e.get(0), end = e.get(1);
            d.merge(start, 1, Integer::sum);
            d.merge(end + 1, -1, Integer::sum);
        }
        int ans = 0, s = 0, last = 0;
        for (var e : d.entrySet()) {
            int cur = e.getKey(), v = e.getValue();
            if (s > 0) {
                ans += cur - last;
            }
            s += v;
            last = cur;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int numberOfPoints(vector<vector<int>>& nums) {
        map<int, int> d;
        for (const auto& e : nums) {
            int start = e[0], end = e[1];
            ++d[start];
            --d[end + 1];
        }
        int ans = 0, s = 0, last = 0;
        for (const auto& [cur, v] : d) {
            if (s > 0) {
                ans += cur - last;
            }
            s += v;
            last = cur;
        }
        return ans;
    }
};

Go

func numberOfPoints(nums [][]int) (ans int) {
	d := map[int]int{}
	for _, e := range nums {
		start, end := e[0], e[1]
		d[start]++
		d[end+1]--
	}
	keys := []int{}
	for k := range d {
		keys = append(keys, k)
	}
	s, last := 0, 0
	sort.Ints(keys)
	for _, cur := range keys {
		if s > 0 {
			ans += cur - last
		}
		s += d[cur]
		last = cur
	}
	return
}

TypeScript

function numberOfPoints(nums: number[][]): number {
    const d = new Map<number, number>();
    for (const [start, end] of nums) {
        d.set(start, (d.get(start) || 0) + 1);
        d.set(end + 1, (d.get(end + 1) || 0) - 1);
    }
    const keys = [...d.keys()].sort((a, b) => a - b);
    let [ans, s, last] = [0, 0, 0];
    for (const cur of keys) {
        if (s > 0) {
            ans += cur - last;
        }
        s += d.get(cur)!;
        last = cur;
    }
    return ans;
}