Algorithm


Problem Name: 1029. 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

 

Constraints:

  • 2 * n == costs.length
  • 2 <= costs.length <= 100
  • costs.length is even.
  • 1 <= aCosti, bCosti <= 1000

Code Examples

#1 Code Example with Java Programming

Code - Java Programming


class Solution {
  public int twoCitySchedCost(int[][] costs) {
    Arrays.sort(costs, (o1, o2) -> o1[0] - o1[1] - (o2[0] - o2[1]));
    int total = 0;
    int n = costs.length / 2;
    for (int i = 0; i  <  n; ++i) {
      total += costs[i][0] + costs[i + n][1];
    }
    return total;
  }
}
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Input

x
+
cmd
costs = [[10,20],[30,200],[400,50],[30,20]]

Output

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110

#2 Code Example with Javascript Programming

Code - Javascript Programming


const twoCitySchedCost = function(costs) {
  const N = costs.length / 2
  const dp = Array.from({ length: N + 1 }, () => new Array(N + 1).fill(0))
  for (let i = 1; i  < = N; i++) {
    dp[i][0] = dp[i - 1][0] + costs[i - 1][0]
  }
  for (let j = 1; j  < = N; j++) {
    dp[0][j] = dp[0][j - 1] + costs[j - 1][1]
  }
  for (let i = 1; i  < = N; i++) {
    for (let j = 1; j  < = N; j++) {
      dp[i][j] = Math.min(
        dp[i - 1][j] + costs[i + j - 1][0],
        dp[i][j - 1] + costs[i + j - 1][1]
      )
    }
  }
  return dp[N][N]
}
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Input

x
+
cmd
costs = [[10,20],[30,200],[400,50],[30,20]]

Output

x
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cmd
110

#3 Code Example with Python Programming

Code - Python Programming


class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        costs.sort(key = lambda x: abs(x[0] - x[1]))
        a = b = 0
        N = len(costs) // 2
        c = 0
        for c1, c2 in costs[::-1]:
            if c1 <= c2 and a < N or b >= N:
                c += c1
                a += 1
            else:
                c += c2
                b += 1
        return c
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Input

x
+
cmd
costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]

Output

x
+
cmd
1859

#4 Code Example with C# Programming

Code - C# Programming


using System;

namespace LeetCode
{
    public class _1029_TwoCityScheduling
    {
        public int TwoCitySchedCost(int[][] costs)
        {
            Array.Sort(costs, (a, b) => (a[0] - a[1]) - (b[0] - b[1]));

            var n = costs.Length / 2;
            var result = 0;
            for (int i = 0; i  <  n; i++)
                result += costs[i][0] + costs[i + n][1];

            return result;
        }
    }
}
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Input

x
+
cmd
costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]

Output

x
+
cmd
1859
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