Two City Scheduling - LeetCode
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
- code
class Solution:
def twoCitySchedCost(self, costs: List[List[int]]) -> int:
diff = [[v[0] - v[1], i] for i, v in enumerate(costs)]
diff.sort()
res = 0
for i in range(len(diff)//2):
res += costs[diff[i][1]][0]
for i in range(len(diff)//2, len(diff)):
res += costs[diff[i][1]][1]
return res
- code
class Solution:
def twoCitySchedCost(self, costs: List[List[int]]) -> int:
costs.sort(key = lambda x : x[0] - x[1])
total = 0
n = len(costs) // 2
for i in range(n):
total += costs[i][0] + costs[i + n][1]
return total