## 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.length`
• `2 <= costs.length <= 100`
• `costs.length` is 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 - o1 - (o2 - o2));
int total = 0;
int n = costs.length / 2;
for (int i = 0; i < n; ++i) {
total += costs[i] + costs[i + n];
}
}
}
``````
Copy The Code &

Input

cmd
costs = [[10,20],[30,200],[400,50],[30,20]]

Output

cmd
110

### #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] = dp[i - 1] + costs[i - 1]
}
for (let j = 1; j <= N; j++) {
dp[j] = dp[j - 1] + costs[j - 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],
dp[i][j - 1] + costs[i + j - 1]
)
}
}
return dp[N][N]
}
``````
Copy The Code &

Input

cmd
costs = [[10,20],[30,200],[400,50],[30,20]]

Output

cmd
110

### #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 - x))
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
``````
Copy The Code &

Input

cmd
costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]

Output

cmd
1859

### #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 - a) - (b - b));

var n = costs.Length / 2;
var result = 0;
for (int i = 0; i < n; i++)
result += costs[i] + costs[i + n];

return result;
}
}
}
``````
Copy The Code &

Input

cmd
costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]

Output

cmd
1859