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1029. Two City Scheduling.py
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1029. Two City Scheduling.py
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1029. Two City Scheduling
Medium/1587/182
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
"""
Send everyone to city a.
Compute the cost for each person to city B. cost[i][1] - cost[i][0]
sort by smallest first, then take the first n, and "subtract" their cost from the total.
time: n log n
space: n
"""
class Solution:
def twoCitySchedCost(self, costs: List[List[int]]) -> int:
tc = 0
n = len(costs)
move = [0] * n
for i in range(n):
tc += costs[i][0]
move[i] = costs[i][1] - costs[i][0]
move.sort()
for i in range(int(n/2)):
tc += move[i]
return tc