Hex.*

LC1109 - Corporate Flight Bookings

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Leetcode: 124/2658 (52+59+13)

Description

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]

Constraints:

1 <= n <= 2 * 104
1 <= bookings.length <= 2 * 104
bookings[i].length == 3
1 <= firsti <= lasti <= n
1 <= seatsi <= 104

Solution

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#前缀和 O(n^2) time | O(1) space
class Solution:
    def corpFlightBookings(self, bookings: List[List[int]], n: int) -> List[int]:
        answer = [0] * (n+1)
        for start, end, value in bookings:
            for i in range(start, end+1):
                answer[i] += value
        return answer[1:]

The idea of the difference method:

  • Start index first, accumulate the number of seats
  • At the end of index last+1, subtract the current number of seats
  • Because the flight after the last, there is no current seat
  • When subtracting first, and then adding prefixes, they will actually cancel each other out
  • It is easy to understand how the idea of difference solves this problem by drawing an array. 
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    #O(n) time | O(n) space
    class Solution:
        def corpFlightBookings(self, bookings: List[List[int]], n: int) -> List[int]:
            diff = [0] * (n+1)

            for start, end, value in bookings:
                diff[start] += value
                if end < n:
                    diff[end+1] -= value

            for i in range(1, len(diff)):
                diff[i] += diff[i-1] 
            return diff[1:]