# 1029. Two City Scheduling

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

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

Example:

Input: [[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.

# Solution

Approach 1: find the difference for each person by going to A instead of B, and sort the differences.

# Code (Python)

Approach 1:

    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        money_saved_at_a = list(map(lambda cost: (cost[1] - cost[0], cost[0], cost[1]), costs))
        money_saved_at_a.sort()
        total = 0
        for i in range(len(money_saved_at_a)):
            if i < len(money_saved_at_a) // 2:
                total += money_saved_at_a[i][2]
            else:
                total += money_saved_at_a[i][1]
        return total

# Code (C++)

Approach 1:

Approach 2: