Formatted question description: https://leetcode.ca/all/1109.html

1109. Corporate Flight Bookings (Medium)

There are n flights, and they are labeled from 1 to n.

We have a list of flight bookings.  The i-th booking bookings[i] = [i, j, k] means that we booked k seats from flights labeled i to j inclusive.

Return an array answer of length n, representing the number of seats booked on each flight in order of their label.

 

Example 1:

Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
Output: [10,55,45,25,25]

 

Constraints:

  • 1 <= bookings.length <= 20000
  • 1 <= bookings[i][0] <= bookings[i][1] <= n <= 20000
  • 1 <= bookings[i][2] <= 10000

Related Topics:
Array, Math

Solution 1. Heap

// OJ: https://leetcode.com/problems/corporate-flight-bookings/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> corpFlightBookings(vector<vector<int>>& A, int n) {
        vector<int> ans(n);
        sort(begin(A), end(A));
        auto cmp = [&](int a, int b) { return A[a][1] > A[b][1]; };
        priority_queue<int, vector<int>, decltype(cmp)> q(cmp);
        for (int i = 0, j = 0, sum = 0; i < n; ++i) {
            while (q.size() && A[q.top()][1] < i + 1) {
                sum -= A[q.top()][2];
                q.pop();
            }
            while (j < A.size() && A[j][0] <= i + 1 && A[j][1] >= i + 1) {
                q.push(j);
                sum += A[j++][2];
            }
            ans[i] = sum;
        }
        return ans;
    }
};

Solution 2. Diff Array

// OJ: https://leetcode.com/problems/corporate-flight-bookings/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> corpFlightBookings(vector<vector<int>>& A, int n) {
        vector<int> ans(n);
        for (auto &v : A) {
            ans[v[0] - 1] += v[2];
            if (v[1] < n) ans[v[1]] -= v[2];
        }
        for (int i = 1; i < n; ++i) ans[i] += ans[i - 1];
        return ans;
    }
};

Java

  • class Solution {
        public int[] corpFlightBookings(int[][] bookings, int n) {
            int[] seats = new int[n];
            for (int[] booking : bookings) {
                int start = booking[0], end = booking[1], count = booking[2];
                seats[start - 1] += count;
                if (end < n)
                    seats[end] -= count;
            }
            for (int i = 1; i < n; i++)
                seats[i] += seats[i - 1];
            return seats;
        }
    }
    
  • // OJ: https://leetcode.com/problems/corporate-flight-bookings/
    // Time: O(NlogN)
    // Space: O(N)
    class Solution {
    public:
        vector<int> corpFlightBookings(vector<vector<int>>& A, int n) {
            vector<int> ans(n);
            sort(begin(A), end(A));
            auto cmp = [&](int a, int b) { return A[a][1] > A[b][1]; };
            priority_queue<int, vector<int>, decltype(cmp)> q(cmp);
            for (int i = 0, j = 0, sum = 0; i < n; ++i) {
                while (q.size() && A[q.top()][1] < i + 1) {
                    sum -= A[q.top()][2];
                    q.pop();
                }
                while (j < A.size() && A[j][0] <= i + 1 && A[j][1] >= i + 1) {
                    q.push(j);
                    sum += A[j++][2];
                }
                ans[i] = sum;
            }
            return ans;
        }
    };
    
  • # 1109. Corporate Flight Bookings
    # https://leetcode.com/problems/corporate-flight-bookings/
    
    class Solution:
        def corpFlightBookings(self, bookings: List[List[int]], n: int) -> List[int]:
            res = [0] * (n + 1)
            
            for start,end,seats in bookings:
                res[start - 1] += seats
                res[end] -= seats
            
            for i in range(1, n):
                res[i] += res[i - 1]
            
            return res[:-1]
    
    

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