source: https://leetcode.com/problems/two-city-scheduling/description/
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
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class Solution { public int twoCitySchedCost(int[][] costs) { Integer[][] memo = new Integer[costs.length/2+1][costs.length/2+1]; return dfs(costs, 0, costs.length/2, costs.length/2, memo); } private int dfs(int[][] costs, int index, int a, int b, Integer[][] memo){ if(index>=costs.length){ return 0; } if(a>0 && b > 0 && memo[a][b]!=null) return memo[a][b]; int minA = Integer.MAX_VALUE; int minB = Integer.MAX_VALUE; //if a person is sent to city a if(a>0){ minA = costs[index][0]+dfs(costs, index+1, a-1, b, memo); } //if a person is sent to cisty B if(b>0){ minB = costs[index][1]+dfs(costs, index+1, a, b-1, memo); } memo[a][b] = Math.min(minA, minB); return memo[a][b]; } } |