Question

Formatted question description: https://leetcode.ca/all/1094.html

 1094. Car Pooling

 You are driving a vehicle that has capacity empty seats initially available for passengers.
 The vehicle only drives east (ie. it cannot turn around and drive west.)

 Given a list of trips, trip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip:
 the number of passengers that must be picked up, and the locations to pick them up and drop them off.
 The locations are given as the number of kilometers due east from your vehicle's initial location.

 Return true if and only if it is possible to pick up and drop off all passengers for all the given trips.

 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

 Example 3:

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

 Example 4:

 Input: trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11
 Output: true

Algorithm

For each interval, at start, there are passangers getting on, at end, there are passagers getting off.

Then arrange original interval as start with positive integer of passager count, end with negative integer of passage count.

Sort them based on time. Let negative number come first as they get off, could minimize current passager count.

If current passage count > capacity, return false.

Time Complexity: O(nlogn). n = trips.length.

Space: O(n).

Code

Java

public class Car_Pooling {

    class Solution {
        public boolean carPooling(int[][] trips, int capacity) {
            if(trips == null || trips.length == 0){
                return true;
            }

            List<int []> list = new ArrayList<>();
            for(int [] trip : trips){
                list.add(new int[]{trip[1], trip[0]});
                list.add(new int[]{trip[2], -trip[0]});
            }

            // same location, off bus first, then onboard new.
            // so, negative value first, then positive value
            Collections.sort(list, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
            int count = 0;
            for(int [] arr : list){
                count += arr[1];
                if(count > capacity){
                    return false;
                }
            }

            return true;
        }
    }
}

// OJ: https://leetcode.com/problems/car-pooling/
// Time: O(T + P)
// Space: O(P)
class Solution {
public:
    bool carPooling(vector<vector<int>>& trips, int capacity) {
        int diff[1001] = {};
        for (auto &v : trips) {
            diff[v[1]] += v[0];
            diff[v[2]] -= v[0];
        }
        int cnt = 0;
        for (int d : diff) {
            if ((cnt += d) > capacity) return false;
        }
        return true;
    }
};

class Solution:
    def carPooling(self, trips: List[List[int]], capacity: int) -> bool:
        delta = [0] * 1001
        for num, start, end in trips:
            delta[start] += num
            delta[end] -= num
        return all(s <= capacity for s in accumulate(delta))

############

# 1094. Car Pooling
# https://leetcode.com/problems/car-pooling/

class Solution:
    def carPooling(self, trips: List[List[int]], capacity: int) -> bool:
        trips.sort(key = lambda x:(x[1], x[2], x[0]))
​
        dest = []
        
        for num, start, end in trips:
            capacity -= num
            
            while dest and dest[0][0] <= start:
                item = heapq.heappop(dest)
                capacity += item[1]
                
            if capacity < 0: return False
            heapq.heappush(dest, (end, num))
        
        return capacity >= 0

Java

class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        PriorityQueue<int[]> priorityQueue = new PriorityQueue<int[]>(new Comparator<int[]>() {
            public int compare(int[] array1, int[] array2) {
                if (array1[0] != array2[0])
                    return array1[0] - array2[0];
                else
                    return array1[1] - array2[1];
            }
        });
        for (int[] trip : trips) {
            int passengersCount = trip[0], start = trip[1], end = trip[2];
            int[] passengersStart = {start, passengersCount};
            int[] passengersEnd = {end, -passengersCount};
            priorityQueue.offer(passengersStart);
            priorityQueue.offer(passengersEnd);
        }
        int passengers = 0;
        while (!priorityQueue.isEmpty()) {
            int[] passengersChange = priorityQueue.poll();
            passengers += passengersChange[1];
            if (passengers > capacity)
                return false;
        }
        return true;
    }
}

// OJ: https://leetcode.com/problems/car-pooling/
// Time: O(T + P)
// Space: O(P)
class Solution {
public:
    bool carPooling(vector<vector<int>>& trips, int capacity) {
        int diff[1001] = {};
        for (auto &v : trips) {
            diff[v[1]] += v[0];
            diff[v[2]] -= v[0];
        }
        int cnt = 0;
        for (int d : diff) {
            if ((cnt += d) > capacity) return false;
        }
        return true;
    }
};

class Solution:
    def carPooling(self, trips: List[List[int]], capacity: int) -> bool:
        delta = [0] * 1001
        for num, start, end in trips:
            delta[start] += num
            delta[end] -= num
        return all(s <= capacity for s in accumulate(delta))

############

# 1094. Car Pooling
# https://leetcode.com/problems/car-pooling/

class Solution:
    def carPooling(self, trips: List[List[int]], capacity: int) -> bool:
        trips.sort(key = lambda x:(x[1], x[2], x[0]))
​
        dest = []
        
        for num, start, end in trips:
            capacity -= num
            
            while dest and dest[0][0] <= start:
                item = heapq.heappop(dest)
                capacity += item[1]
                
            if capacity < 0: return False
            heapq.heappush(dest, (end, num))
        
        return capacity >= 0

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