LeetCode 1029. Two City Scheduling Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1

PythonJavaC++GoTypeScript

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))(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)