Skip to main content

Sweep Line - Question 2

1109. 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.

Constraints:

1 <= n <= 2 * 10^4

1 <= bookings.length <= 2 * 10^4

bookings[i].length == 3

1 <= firsti <= lasti <= n

1 <= seatsi <= 10^4


Analysis:

The naïve method is to fill in every fight with the number of seats booked. The overall time complexity is O(N^2), which cannot pass the online OJ.

We do not have to do this. Instead, we can just label the starting and ending flights (since all the flights between them are the same).

Or in the terminology of Sweep line, we can treat the flights range as an interval: the start flight is the start of the interval, and end is the end of the interval.

After labeling each interval, we just need to a one-time sweeping of all the flights. The accumulated number of seats at each flight are the number of seat booked for that flight.

Time complexity: O(N)

Space complexity: O(N)


See the code below:

class Solution {
public:
    vector<int> corpFlightBookings(vector<vector<int>>& bookings, int n) {
        vector<int> res(n, 0), data(n+1, 0);
        for(auto &a : bookings) {
            data[a[0]-1] += a[2];
            data[a[1]] -= a[2];
        }
        for(int i=0; i<n; ++i) {
            res[i] = (i==0?0:res[i-1]) + data[i];
        }
        return res;
    }
};



Upper Layer


Comments

Post a Comment

Popular posts from this blog

Brute Force - Question 2

2105. Watering Plants II Alice and Bob want to water n plants in their garden. The plants are arranged in a row and are labeled from 0 to n - 1 from left to right where the ith plant is located at x = i. Each plant needs a specific amount of water. Alice and Bob have a watering can each, initially full. They water the plants in the following way: Alice waters the plants in order from left to right, starting from the 0th plant. Bob waters the plants in order from right to left, starting from the (n - 1)th plant. They begin watering the plants simultaneously. It takes the same amount of time to water each plant regardless of how much water it needs. Alice/Bob must water the plant if they have enough in their can to fully water it. Otherwise, they first refill their can (instantaneously) then water the plant. In case both Alice and Bob reach the same plant, the one with more water currently in his/her watering can should water this plant. If they have the same amount of water, then Alice ...
Post a Comment
Read more

Sliding Window - Question 1

Leetcode 76. Minimum Window Substring Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "". The testcases will be generated such that the answer is unique. A substring is a contiguous sequence of characters within the string. Constraints: m == s.length n == t.length 1 <= m, n <= 10^5 s and t consist of uppercase and lowercase English letters. Analysis: The first step to do is to count the letters in string t, since the relative order of chars does not matter. After the counting, we can use two pointers method on string s: starting with beginning of the string s, the second pointer can continue to move until the sub-string [i, j) has all the chars in t, where i the index of the first pointer, and j is the index of the second pointer. We are looking for the sub-string with the minimum leng...
Post a Comment
Read more