# 1029. Two City Scheduling

## Description

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

## Solutions

Greedy.

• class Solution {
public int twoCitySchedCost(int[][] costs) {
Arrays.sort(costs, (a, b) -> { return a[0] - a[1] - (b[0] - b[1]); });
int ans = 0;
int n = costs.length >> 1;
for (int i = 0; i < n; ++i) {
ans += costs[i][0] + costs[i + n][1];
}
return ans;
}
}

• class Solution {
public:
int twoCitySchedCost(vector<vector<int>>& costs) {
sort(costs.begin(), costs.end(), [](const vector<int>& a, const vector<int>& b) {
return a[0] - a[1] < b[0] - b[1];
});
int n = costs.size() / 2;
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += costs[i][0] + costs[i + n][1];
}
return ans;
}
};


• class Solution:
def twoCitySchedCost(self, costs: List[List[int]]) -> int:
costs.sort(key=lambda x: x[0] - x[1])
n = len(costs) >> 1
return sum(costs[i][0] + costs[i + n][1] for i in range(n))


• func twoCitySchedCost(costs [][]int) (ans int) {
sort.Slice(costs, func(i, j int) bool {
return costs[i][0]-costs[i][1] < costs[j][0]-costs[j][1]
})
n := len(costs) >> 1
for i, a := range costs[:n] {
ans += a[0] + costs[i+n][1]
}
return
}

• function twoCitySchedCost(costs: number[][]): number {
costs.sort((a, b) => a[0] - a[1] - (b[0] - b[1]));
const n = costs.length >> 1;
let ans = 0;
for (let i = 0; i < n; ++i) {
ans += costs[i][0] + costs[i + n][1];
}
return ans;
}