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Formatted question description: https://leetcode.ca/all/1029.html

1029. Two City Scheduling (Easy)

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.

 

Note:

  1. 1 <= costs.length <= 100
  2. It is guaranteed that costs.length is even.
  3. 1 <= costs[i][0], costs[i][1] <= 1000

Related Topics:
Greedy

Solution 1. DP

Let dp[i + 1][j] be the min cost arranging the first i + 1 people and when j people 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].
  • The i-th person goes to city B. We get dp[i][j] + A[i][1]. Note that i + 1 > j because otherwise we don’t have the spot for the i-th person to go to city B.
dp[i + 1][j] = min(
                    dp[i][j - 1] + A[i][0],                     // the i-th person goes to city A
                    i + 1> j  ? dp[i][j] + A[i][1] : INF        // the i-th person goes to city B
                  )
dp[i + 1][0] = sum( A[k][1] | 0 <= k <= i )

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

Solution 2. DP with Space Optimization

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

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 pairs <cost[i][0] - cost[i][1], i> in ascending order. The first half goes to city A, the second half goes to city B.

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

  • class Solution {
    public int twoCitySchedCost(int[][] costs) {
        int length = costs.length;
        int sum = 0;
        int[] differences = new int[length];
        for (int i = 0; i < length; i++) {
            sum += costs[i][0];
            differences[i] = costs[i][1] - costs[i][0];
        }
        Arrays.sort(differences);
        int halfLength = length / 2;
        for (int i = 0; i < halfLength; i++)
            sum += differences[i];
        return sum;
    }
}

############

class Solution {
    public int twoCitySchedCost(int[][] costs) {
        Arrays.sort(costs, (a, b) -> { return a[0] - a[1] - (b[0] - b[1]); });
        int ans = 0;
        int n = costs.length >> 1;
        for (int i = 0; i < n; ++i) {
            ans += costs[i][0] + costs[i + n][1];
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/two-city-scheduling/
// 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];
    }
};

  • class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        costs.sort(key=lambda x: x[0] - x[1])
        n = len(costs) >> 1
        return sum(costs[i][0] + costs[i + n][1] for i in range(n))

############

# 1029. Two City Scheduling
# https://leetcode.com/problems/two-city-scheduling/

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        n = len(costs) // 2
        firstCity = [a for a, b in costs]
        diff = sorted([b - a for a, b in costs])
        
        return sum(firstCity) + sum(diff[:n])


  • func twoCitySchedCost(costs [][]int) int {
	sort.Slice(costs, func(i, j int) bool {
		return costs[i][0]-costs[i][1] < costs[j][0]-costs[j][1]
	})
	ans, n := 0, len(costs)>>1
	for i := 0; i < n; i++ {
		ans += costs[i][0] + costs[i+n][1]
	}
	return ans
}

  • function twoCitySchedCost(costs: number[][]): number {
    costs.sort((a, b) => a[0] - a[1] - (b[0] - b[1]));
    const n = costs.length >> 1;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans += costs[i][0] + costs[i + n][1];
    }
    return ans;
}

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