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1029. Two City Scheduling

Description

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

Solutions

Greedy.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• class Solution {
  public:
      int twoCitySchedCost(vector<vector<int>>& costs) {
          sort(costs.begin(), costs.end(), [](const vector<int>& a, const vector<int>& b) {
              return a[0] - a[1] < b[0] - b[1];
          });
          int n = costs.size() / 2;
          int ans = 0;
          for (int i = 0; i < n; ++i) {
              ans += costs[i][0] + costs[i + n][1];
          }
          return ans;
      }
  };
  
  

• 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))
  
  

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

• 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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