# 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], and the cost of flying the i-th person to city B is costs[i].
Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.
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.
Approach 1: find the difference for each person by going to A instead of B, and sort the differences.
# Code (Python)
def twoCitySchedCost(self, costs: List[List[int]]) -> int: money_saved_at_a = list(map(lambda cost: (cost - cost, cost, cost), 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] else: total += money_saved_at_a[i] return total
# Code (C++)