1109

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/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// 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;
    }
};