human
I found an interesting problem on https://leetcode.com/problems/two-city-scheduling/:
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

I tried it in Python, but could not do it. Can you solve it?

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What is a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        lc = len(costs)
        c = lc//2
        cost1 = 0
        cost2 = 0
        al = c
        bl = c
        for i in range(lc):
            for j in range(i+1, lc):
                if abs(costs[i][0]-costs[i][1]) < abs(costs[j][0]-costs[j][1]):
                    costs[i], costs[j] = costs[j], costs[i]
        
        for i in range(lc):
            if (costs[i][0] <= costs[i][1] and al>0) or bl==0:
                cost1 += costs[i][0]
                al -= 1
            elif bl>0:
                cost1 += costs[i][1]
                bl -= 1
        al = c
        bl = c
        for i in range(lc):
            if (costs[i][0] < costs[i][1] and al>0) or bl == 0:
                cost2 += costs[i][0]
                al -= 1
            elif bl>0:
                cost2 += costs[i][1]
                bl -= 1
        return min(cost1,cost2)