# 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: