## DEV Community

Abhishek Chaudhary

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# 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:

• `2 * n == costs.length`
• `2 <= costs.length <= 100`
• `costs.length` is even.
• `1 <= aCosti, bCosti <= 1000`

SOLUTION:

``````from functools import cmp_to_key

class Solution:
def compare(self, cost1, cost2):
a, b = cost1
c, d = cost2
if a + d <= b + c:
return -1
return 1

def twoCitySchedCost(self, costs: List[List[int]]) -> int:
n = len(costs)
costs.sort(key = cmp_to_key(self.compare))
cost = 0
for i in range(n // 2):
cost += costs[i][0]
for i in range(n // 2, n):
cost += costs[i][1]
return cost
``````