1029

1029. Two City Scheduling (Medium)

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

Companies:
Bloomberg

Related Topics:
Array, Greedy, Sorting

Solution 1. Bottom-up DP

Let dp[i + 1][j] be the min cost arranging the first i + 1 people and when j of them go to city A, 0 <= i < N, 0 <= j <= i + 1.

For dp[i + 1][j], we have two options:

• The i-th person goes to city A. We get dp[i][j - 1] + A[i][0] (j - 1 >= 0).
• The i-th person goes to city B. We get dp[i][j] + A[i][1] (j <= i).

dp[i + 1][j] = min(
                    j - 1 >= 0 ? dp[i][j - 1] + A[i][0] : INF,       // the i-th person goes to city A
                    j <= i ? dp[i][j] + A[i][1] : INF                // the i-th person goes to city B
                  )
dp[0][0] = 0

The answer is dp[2N][N].

// OJ: https://leetcode.com/problems/two-city-scheduling/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& A) {
        int N = A.size() / 2;
        vector<vector<int>> dp(2 * N + 1, vector<int>(N + 1, INT_MAX));
        dp[0][0] = 0;
        for (int i = 0; i < 2 * N; ++i) {
            for (int j = 0; j <= min(i + 1, N); ++j) {
                dp[i + 1][j] = min(j - 1 >= 0 ? dp[i][j - 1] + A[i][0] : INT_MAX, j <= i ? dp[i][j] + A[i][1] : INT_MAX);
            }
        }
        return dp[2 * N][N];
    }
};

Solution 2. DP with Space Optimization

// OJ: https://leetcode.com/problems/two-city-scheduling/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& A) {
        int N = A.size() / 2;
        vector<int> dp(N + 1, INT_MAX);
        dp[0] = 0;
        for (int i = 0; i < 2 * N; ++i) {
            for (int j = min(i + 1, N); j >= 0; --j) {
                dp[j] = min(j - 1 >= 0 ? dp[j - 1] + A[i][0] : INT_MAX, j <= i ? dp[j] + A[i][1] : INT_MAX);
            }
        }
        return dp[N];
    }
};

Solution 3. Greedy

The smaller cost[i][0] - cost[i][1] is, the more likely i-th person should go to city A.

So we can sort the array in ascending order of cost[i][0] - cost[i][1]. The first half goes to city A, the second half goes to city B.

// OJ: https://leetcode.com/problems/two-city-scheduling/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& A) {
        int N = A.size() / 2, ans = 0;
        sort(begin(A), end(A), [](auto &a, auto &b) { return (a[0] - a[1]) < (b[0] - b[1]); });
        for (int i = 0; i < 2 * N; ++i) ans += A[i][i >= N];
        return ans;
    }
};