1095. Two City Scheduling¶

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1095. 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

Solution¶

import java.util.Arrays;

class Solution {
    private int dp[][][];

    public int twoCitySchedCost(int[][] costs) {
        int n = costs.length;
        dp = new int[n + 1][n / 2 + 1][n / 2 + 1];
        for (int current[][] : dp)
            for (int current1[] : current)
                Arrays.fill(current1, -1);
        int res = solve(0, n / 2, n / 2, costs);
        return res;
    }

    private int solve(int ind, int aCount, int bCount, int cost[][]) {
        if (ind >= cost.length) {
            if (aCount > 0 || bCount > 0)
                return Integer.MAX_VALUE / 10;
            return 0;
        }
        if (dp[ind][aCount][bCount] != -1)
            return dp[ind][aCount][bCount];
        /* Need to decide whether to move this person to city A or B */
        int op1 = Integer.MAX_VALUE / 10, op2 = Integer.MAX_VALUE / 10;
        if (aCount > 0)
            op1 = cost[ind][0] + solve(ind + 1, aCount - 1, bCount, cost);
        if (bCount > 0)
            op2 = cost[ind][1] + solve(ind + 1, aCount, bCount - 1, cost);
        return dp[ind][aCount][bCount] = Math.min(op1, op2);
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here