Algorithm
Problem Name: 1029. 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.length2 <= costs.length <= 100costs.lengthis even.1 <= aCosti, bCosti <= 1000
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int twoCitySchedCost(int[][] costs) {
Arrays.sort(costs, (o1, o2) -> o1[0] - o1[1] - (o2[0] - o2[1]));
int total = 0;
int n = costs.length / 2;
for (int i = 0; i < n; ++i) {
total += costs[i][0] + costs[i + n][1];
}
return total;
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const twoCitySchedCost = function(costs) {
const N = costs.length / 2
const dp = Array.from({ length: N + 1 }, () => new Array(N + 1).fill(0))
for (let i = 1; i < = N; i++) {
dp[i][0] = dp[i - 1][0] + costs[i - 1][0]
}
for (let j = 1; j < = N; j++) {
dp[0][j] = dp[0][j - 1] + costs[j - 1][1]
}
for (let i = 1; i < = N; i++) {
for (let j = 1; j < = N; j++) {
dp[i][j] = Math.min(
dp[i - 1][j] + costs[i + j - 1][0],
dp[i][j - 1] + costs[i + j - 1][1]
)
}
}
return dp[N][N]
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def twoCitySchedCost(self, costs: List[List[int]]) -> int:
costs.sort(key = lambda x: abs(x[0] - x[1]))
a = b = 0
N = len(costs) // 2
c = 0
for c1, c2 in costs[::-1]:
if c1 <= c2 and a < N or b >= N:
c += c1
a += 1
else:
c += c2
b += 1
return c
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _1029_TwoCityScheduling
{
public int TwoCitySchedCost(int[][] costs)
{
Array.Sort(costs, (a, b) => (a[0] - a[1]) - (b[0] - b[1]));
var n = costs.Length / 2;
var result = 0;
for (int i = 0; i < n; i++)
result += costs[i][0] + costs[i + n][1];
return result;
}
}
}
Input
Output