Leetcode - Two City Scheduling

https://leetcode.com/problems/two-city-scheduling/

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.

Example 1:

Input: [[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.

func twoCitySchedCost(costs [][]int) int {
    n := len(costs)/2
    dp := make([][]int, n+1)
    for i := 0; i < len(dp); i++ {
        dp[i] = make([]int, n+1)
    }
    for i := 1; i <= n; i++ {
        dp[i][0] = dp[i-1][0] + costs[i-1][0]
    }
    for j := 1; j <= n; j++ {
        dp[0][j] = dp[0][j-1] + costs[j-1][1]
    }
    for i := 1; i <= n; i++ {
        for j := 1; j <= n; j++ {
            dp[i][j] = min(dp[i-1][j] + costs[i+j-1][0], dp[i][j-1] + costs[i+j-1][1])
        }
    }
    return dp[n][n]
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

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