# 1094. Car Pooling

## Description

There is a car with capacity empty seats. The vehicle only drives east (i.e., it cannot turn around and drive west).

You are given the integer capacity and an array trips where trips[i] = [numPassengersi, fromi, toi] indicates that the ith trip has numPassengersi passengers and the locations to pick them up and drop them off are fromi and toi respectively. The locations are given as the number of kilometers due east from the car's initial location.

Return true if it is possible to pick up and drop off all passengers for all the given trips, or false otherwise.

Example 1:

Input: trips = [[2,1,5],[3,3,7]], capacity = 4
Output: false


Example 2:

Input: trips = [[2,1,5],[3,3,7]], capacity = 5
Output: true


Constraints:

• 1 <= trips.length <= 1000
• trips[i].length == 3
• 1 <= numPassengersi <= 100
• 0 <= fromi < toi <= 1000
• 1 <= capacity <= 105

## Solutions

Solution 1: Difference Array

We can use the idea of a difference array, adding the number of passengers to the starting point of each trip and subtracting from the end point. Finally, we just need to check whether the prefix sum of the difference array does not exceed the maximum passenger capacity of the car.

The time complexity is $O(n)$, and the space complexity is $O(M)$. Here, $n$ is the number of trips, and $M$ is the maximum end point in the trips. In this problem, $M \le 1000$.

• class Solution {
public boolean carPooling(int[][] trips, int capacity) {
int[] d = new int[1001];
for (var trip : trips) {
int x = trip[0], f = trip[1], t = trip[2];
d[f] += x;
d[t] -= x;
}
int s = 0;
for (int x : d) {
s += x;
if (s > capacity) {
return false;
}
}
return true;
}
}

• class Solution {
public:
bool carPooling(vector<vector<int>>& trips, int capacity) {
int d[1001]{};
for (auto& trip : trips) {
int x = trip[0], f = trip[1], t = trip[2];
d[f] += x;
d[t] -= x;
}
int s = 0;
for (int x : d) {
s += x;
if (s > capacity) {
return false;
}
}
return true;
}
};

• class Solution:
def carPooling(self, trips: List[List[int]], capacity: int) -> bool:
mx = max(e[2] for e in trips)
d = [0] * (mx + 1)
for x, f, t in trips:
d[f] += x
d[t] -= x
return all(s <= capacity for s in accumulate(d))


• func carPooling(trips [][]int, capacity int) bool {
d := [1001]int{}
for _, trip := range trips {
x, f, t := trip[0], trip[1], trip[2]
d[f] += x
d[t] -= x
}
s := 0
for _, x := range d {
s += x
if s > capacity {
return false
}
}
return true
}

• function carPooling(trips: number[][], capacity: number): boolean {
const mx = Math.max(...trips.map(([, , t]) => t));
const d = Array(mx + 1).fill(0);
for (const [x, f, t] of trips) {
d[f] += x;
d[t] -= x;
}
let s = 0;
for (const x of d) {
s += x;
if (s > capacity) {
return false;
}
}
return true;
}


• /**
* @param {number[][]} trips
* @param {number} capacity
* @return {boolean}
*/
var carPooling = function (trips, capacity) {
const mx = Math.max(...trips.map(([, , t]) => t));
const d = Array(mx + 1).fill(0);
for (const [x, f, t] of trips) {
d[f] += x;
d[t] -= x;
}
let s = 0;
for (const x of d) {
s += x;
if (s > capacity) {
return false;
}
}
return true;
};


• public class Solution {
public bool CarPooling(int[][] trips, int capacity) {
int mx = trips.Max(x => x[2]);
int[] d = new int[mx + 1];
foreach (var trip in trips) {
int x = trip[0], f = trip[1], t = trip[2];
d[f] += x;
d[t] -= x;
}
int s = 0;
foreach (var x in d) {
s += x;
if (s > capacity) {
return false;
}
}
return true;
}
}


• impl Solution {
pub fn car_pooling(trips: Vec<Vec<i32>>, capacity: i32) -> bool {
let mx = trips
.iter()
.map(|e| e[2])
.max()
.unwrap_or(0) as usize;
let mut d = vec![0; mx + 1];
for trip in &trips {
let (x, f, t) = (trip[0], trip[1] as usize, trip[2] as usize);
d[f] += x;
d[t] -= x;
}
d.iter()
.scan(0, |acc, &x| {
*acc += x;
Some(*acc)
})
.all(|s| s <= capacity)
}
}