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1029 - Two City Scheduling (Medium)

https://leetcode.com/problems/two-city-scheduling/

Problem Statement​

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti.

Return the minimum cost to fly every person to a city such that exactly n people arrive in each city.

Example 1:

Input: costs = [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

Example 2:

Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
Output: 1859

Example 3:

Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
Output: 3086

Constraints:

  • 2 * n == costs.length
  • 2 <= costs.length <= 100
  • costs.length is even.
  • 1 <= aCosti, bCosti <= 1000

Approach 1: Sorting + Greedy​

If we need to spend more money to fly to city B, then it's better to send it to city A, and vice versa. Therefore, we sort the input by their difference (i.e. how much a company can save) and take the first nn to city A and the rest of them to city B.

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs) {
        int n = costs.size(), ans = 0;
        // sort by the diff
        sort(costs.begin(), costs.end(), [&](const vector<int>& x, const vector<int>& y) {
           return x[1] - x[0] > y[1] - y[0]; 
        });
        for (int i = 0; i < n / 2; i++) {
            // cost to fly i-th person to city A
            ans += costs[i][0];
            // cost to fly (n - 1 - i)-th person to city B
            ans += costs[n - 1 - i][1];
        }
        return ans;
    }
};