# 1094. Car Pooling



T: O(nlogn)

S: O(n)

```java
class Solution {

    public boolean carPooling(int[][] trips, int capacity) {
        TreeMap<Integer, Integer> map = new TreeMap<>();
        for (int[] trip : trips) {
            map.put(trip[1], map.getOrDefault(trip[1], 0) + trip[0]);
            map.put(trip[2], map.getOrDefault(trip[2], 0) - trip[0]);
        }
        int max = 0;
        int count = 0;
        for (int value : map.values()) {
            count += value;
            max = Math.max(max, count);
            if (max > capacity) {
                return false;
            }
        }
        return true;
    }
}

/*
car goes with one direction

capacity
an array trips where trip[i] = [numPassengersi, fromi, toi]


from to pick and drop

  in here the sweep line shows, it takes 2 + 3 = 5 people, 5 > 4, so it's fail
    |
  2 |
1.  |  5
  3.|     7
    |3
when 3, pick up 3 more people, it's fail

  2+3
*/
```



## `bucket sort idea`

because this Q's time range is small

* `1 <= trips.length <= 1000`

`so we can use bucket sort idea`

`T: O(n + 1001),  n is trips length`

`S: O(n)`

```java
class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        int[] bucket = new int[1001];
        for (int i = 0; i < trips.length; i++) {
            int[] trip = trips[i];
            bucket[trip[1]] += trip[0];
            bucket[trip[2]] -= trip[0];
        }
        int max = 0;
        int count = 0;
        for (int i = 0; i < bucket.length; i++) {
            count += bucket[i];
            max = Math.max(max, count);
            if (max > capacity) {
                return false;
            }
        }
        return true;
    }
}
```
