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:
2n == costs.length
2 <= costs.length <= 100
costs.length is even.
1 <= aCosti, bCosti <= 1000
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        N = len(costs)
        dp = [0]+[float('inf')]*(N//2)
        
        for i in range(1, N+1):
            
            for j in range(N//2, 0, -1):
                dp[j] = min(dp[j]+costs[i-1][1], dp[j-1]+costs[i-1][0])
            dp[0] += costs[i-1][1]
        # print(dp)
        return dp[-1]
class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        N = len(costs)
        dp = [0]+[float('inf')]*(N//2)
        
        for i in range(1, N+1):
            
            for j in range(N//2, 0, -1):
                dp[j] = min(dp[j]+costs[i-1][1], dp[j-1]+costs[i-1][0])
            dp[0] += costs[i-1][1]
        # print(dp)
        return dp[-1]