# 32. Corporate Flight Bookings

There are `n` flights that are labeled from `1` to `n`.

You are given an array of flight bookings `bookings`, where `bookings[i] = [firsti, lasti, seatsi]` represents a booking for flights `firsti` through `lasti` (**inclusive**) with `seatsi` seats reserved for **each flight** in the range.

Return _an array_ `answer` _of length_ `n`_, where_ `answer[i]` _is the total number of seats reserved for flight_ `i`.

**Example 1:**

```
Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
Output: [10,55,45,25,25]
Explanation:
Flight labels:        1   2   3   4   5
Booking 1 reserved:  10  10
Booking 2 reserved:      20  20
Booking 3 reserved:      25  25  25  25
Total seats:         10  55  45  25  25
Hence, answer = [10,55,45,25,25]
```

**Example 2:**

```
Input: bookings = [[1,2,10],[2,2,15]], n = 2
Output: [10,25]
Explanation:
Flight labels:        1   2
Booking 1 reserved:  10  10
Booking 2 reserved:      15
Total seats:         10  25
Hence, answer = [10,25]
```

## Solution: (Update Range Query in O(1) )

**Approach:**

![image](https://assets.leetcode.com/users/votrubac/image_1562473681.png)

```cpp
class Solution
{
public:
    vector<int> corpFlightBookings(vector<vector<int>> &bookings, int n)
    {
        vector<int> res(n + 2);
        vector<int> arr(n, 0);
        for (int i = 0; i < bookings.size(); i++)
        {
            int start = bookings[i][0];
            int end = bookings[i][1];
            int price = bookings[i][2];

            res[start] += price;
            res[end + 1] -= price;
        }

        int count = 0;

        for (int i = 0; i < n; i++)
        {
            count += res[i + 1];

            arr[i] = count;
        }

        return arr;
    }
};
```

**Time Complexity: O(n)**
